Javascript Regex: How to put a variable inside a regular expression?
up vote
146
down vote
favorite
23
So for example:
function(input){
var testVar = input;
string = ...
string.replace(/ReGeX + testVar + ReGeX/, "replacement")
}
But this is of course not working :)
Is there any way to do this?
javascript regex variables
edited Dec 8 '13 at 22:55
Josh Crozier
152k35268221
asked Oct 27 '10 at 0:30
Adam
20.4k52121187
add a comment |
up vote
146
down vote
favorite
23
So for example:
function(input){
var testVar = input;
string = ...
string.replace(/ReGeX + testVar + ReGeX/, "replacement")
}
But this is of course not working :)
Is there any way to do this?
javascript regex variables
edited Dec 8 '13 at 22:55
Josh Crozier
152k35268221
asked Oct 27 '10 at 0:30
Adam
20.4k52121187
9
Be aware that if you let the user supply this variable, it's easy for a malicious user to crash your application via catastrophic backtracking.
– Tim Pietzcker
Oct 27 '10 at 6:23
1
what is "ReGeX"? is it a variable name? it makes answers confusing by implying that it is a part of RegExp construction (in case a reader did not see those weird characters in the question).
– Eduard
Oct 17 '17 at 13:12
add a comment |
up vote
146
down vote
favorite
23
up vote
146
down vote
favorite
23
23
So for example:
function(input){
var testVar = input;
string = ...
string.replace(/ReGeX + testVar + ReGeX/, "replacement")
}
But this is of course not working :)
Is there any way to do this?
javascript regex variables
edited Dec 8 '13 at 22:55
Josh Crozier
152k35268221
asked Oct 27 '10 at 0:30
Adam
20.4k52121187
So for example:
function(input){
var testVar = input;
string = ...
string.replace(/ReGeX + testVar + ReGeX/, "replacement")
}
But this is of course not working :)
Is there any way to do this?
javascript regex variables
javascript regex variables
edited Dec 8 '13 at 22:55
Josh Crozier
152k35268221
asked Oct 27 '10 at 0:30
Adam
20.4k52121187
edited Dec 8 '13 at 22:55
Josh Crozier
152k35268221
asked Oct 27 '10 at 0:30
Adam
20.4k52121187
edited Dec 8 '13 at 22:55
Josh Crozier
152k35268221
edited Dec 8 '13 at 22:55
Josh Crozier
152k35268221
edited Dec 8 '13 at 22:55
Josh Crozier
152k35268221
152k35268221
asked Oct 27 '10 at 0:30
Adam
20.4k52121187
asked Oct 27 '10 at 0:30
Adam
20.4k52121187
asked Oct 27 '10 at 0:30
Adam
20.4k52121187
20.4k52121187
9
Be aware that if you let the user supply this variable, it's easy for a malicious user to crash your application via catastrophic backtracking.
– Tim Pietzcker
Oct 27 '10 at 6:23
1
what is "ReGeX"? is it a variable name? it makes answers confusing by implying that it is a part of RegExp construction (in case a reader did not see those weird characters in the question).
– Eduard
Oct 17 '17 at 13:12
add a comment |
9
Be aware that if you let the user supply this variable, it's easy for a malicious user to crash your application via catastrophic backtracking.
– Tim Pietzcker
Oct 27 '10 at 6:23
1
what is "ReGeX"? is it a variable name? it makes answers confusing by implying that it is a part of RegExp construction (in case a reader did not see those weird characters in the question).
– Eduard
Oct 17 '17 at 13:12
9
9
Be aware that if you let the user supply this variable, it's easy for a malicious user to crash your application via catastrophic backtracking.
– Tim Pietzcker
Oct 27 '10 at 6:23
Be aware that if you let the user supply this variable, it's easy for a malicious user to crash your application via catastrophic backtracking.
– Tim Pietzcker
Oct 27 '10 at 6:23
1
1
what is "ReGeX"? is it a variable name? it makes answers confusing by implying that it is a part of RegExp construction (in case a reader did not see those weird characters in the question).
– Eduard
Oct 17 '17 at 13:12
what is "ReGeX"? is it a variable name? it makes answers confusing by implying that it is a part of RegExp construction (in case a reader did not see those weird characters in the question).
– Eduard
Oct 17 '17 at 13:12
add a comment |
7 Answers
7
active
oldest
votes
up vote
208
down vote
accepted
var regex = new RegExp("ReGeX" + testVar + "ReGeX");
...
string.replace(regex, "replacement");
Update
Per some of the comments, it's important to note that you may want to escape the variable if there is potential for malicious content (e.g. the variable comes from user input)
edited May 23 '17 at 11:55
Community♦
11
answered Oct 27 '10 at 0:33
Jason McCreary
56.4k19104152
5
Make sure to escape your string! See @zzzzBov answer
– jtpereyda
Oct 6 '15 at 21:18
This is not working in my case, can you please have a look this jsfiddle and let me know if what i have to do if I want to set a country code dynamic intoregexexpression
– Kirankumar Dafda
Jun 10 '16 at 8:30
6
One important thing to remember is that in regular strings the character needs to be escaped while in the regex literal (usually) the / character needs to be escaped. So/w+//ibecomesnew RegExp("\w+/", "i")
– Ali
Jan 16 '17 at 14:46
1
How about the /g?
– Elgs Qian Chen
Mar 8 '17 at 3:27
add a comment |
up vote
67
down vote
You can use the RegExp object:
var regexstring = "whatever";
var regexp = new RegExp(regexstring, "gi");
var str = "whateverTest";
var str2 = str.replace(regexp, "other");
document.write(str2);
Then you can construct regexstring in any way you want.
You can read more about it here.
answered Oct 27 '10 at 0:40
steinar
7,77511730
5
+1 for the full sample, RegExp params, and link.
– Jason McCreary
Oct 27 '10 at 0:45
This does not work for me on chrome. :(
– Ankit Tanna
May 8 '15 at 9:15
var characterCount = parseInt(attrs.limitCharacterCount); console.log(characterCount); var specialCharactersValidation = new RegExp("/^[a-zA-Z0-9]{"+characterCount+"}$/"); //^[a-zA-Z0-9]{7}$// requestShipmentDirectives.js:76 false main.js:46 Uncaught TypeError: Cannot read property 'offsetWidth' of null This is what is happening when I make the RegEx Dynamic.
– Ankit Tanna
May 8 '15 at 9:19
add a comment |
up vote
28
down vote
To build a regular expression from a variable in JavaScript, you'll need to use the RegExp constructor with a string parameter.
function reg(input) {
var flags;
//could be any combination of 'g', 'i', and 'm'
flags = 'g';
return new RegExp('ReGeX' + input + 'ReGeX', flags);
}
of course, this is a very naive example. It assumes that input is has been properly escaped for a regular expression. If you're dealing with user-input, or simply want to make it more convenient to match special characters, you'll need to escape special characters:
function regexEscape(str) {
return str.replace(/[-/\^$*+?.()|[]{}]/g, '\$&')
}
function reg(input) {
var flags;
//could be any combination of 'g', 'i', and 'm'
flags = 'g';
input = regexEscape(input);
return new RegExp('ReGeX' + input + 'ReGeX', flags);
}
edited May 23 '17 at 11:55
Community♦
11
answered Jan 16 '13 at 13:29
zzzzBov
129k33261305
The only way this worked for me was if I deletedReGeXtext. So I used:return new RegExp(input, flags);
– Michael Rader
Dec 30 '16 at 1:47
1
@MichaelRader, yes, the "ReGeX" text was part of the question and used as an example, not as something that's required to make the code work.
– zzzzBov
Dec 30 '16 at 3:49
lol just realized that was just your placeholder, but as a noob this took me awhile to figure out. Thanks.
– Michael Rader
Dec 30 '16 at 4:43
add a comment |
up vote
6
down vote
You can always give regular expression as string, i.e. "ReGeX" + testVar + "ReGeX". You'll possibly have to escape some characters inside your string (e.g., double quote), but for most cases it's equivalent.
You can also use RegExp constructor to pass flags in (see the docs).
answered Oct 27 '10 at 0:33
Nikita Rybak
56.2k18137163
add a comment |
up vote
3
down vote
if you're using es6 template literals are an option...
string.replace(new RegExp(`ReGeX${testVar}ReGeX`), "replacement")
answered Apr 5 at 10:19
shunryu111
2,72531513
add a comment |
up vote
2
down vote
accepted answer doesn't work for me and doesn't follow MDN examples
see the 'Description' section in above link
I'd go with the following it's working for me:
let stringThatIsGoingToChange = 'findMe';
let flagsYouWant = 'gi' //simple string with flags
let dynamicRegExp = new RegExp(`${stringThatIsGoingToChange}`, flagsYouWant)
// that makes dynamicRegExp = /findMe/gi
answered Jun 7 at 22:52
Timothy Mitchell
213
It still works. The issue is that 'ReGeX' + testVar +' ReGeX' makes the solution confusing. Adding an actually example would clear things up in my opinion. var testVar = '321'; var regex = new RegExp( "[" + testVar + "]{2}", "g" ); // above equivalent to /[321]{2}/g
– HelloWorldPeace
Jun 23 at 19:38
flagsYouWant is a string, not a variable, so that won't work
– macasas
Sep 14 at 11:24
@macasas right you are i think i fixed thanks!
– Timothy Mitchell
Oct 4 at 15:40
add a comment |
up vote
1
down vote
Here's an pretty useless function that return values wrapped by specific characters. :)
jsfiddle: https://jsfiddle.net/squadjot/43agwo6x/
function getValsWrappedIn(str,c1,c2){
var rg = new RegExp("(?<=\"+c1+")(.*?)(?=\"+c2+")","g");
return str.match(rg);
}
var exampleStr = "Something (5) or some time (19) or maybe a (thingy)";
var results = getValsWrappedIn(exampleStr,"(",")")
// Will return array ["5","19","thingy"]
console.log(results)
answered Feb 17 at 1:20
Jakob Sternberg
1,2171010
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f4029109%2fjavascript-regex-how-to-put-a-variable-inside-a-regular-expression%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
208
down vote
accepted
var regex = new RegExp("ReGeX" + testVar + "ReGeX");
...
string.replace(regex, "replacement");
Update
Per some of the comments, it's important to note that you may want to escape the variable if there is potential for malicious content (e.g. the variable comes from user input)
edited May 23 '17 at 11:55
Community♦
11
answered Oct 27 '10 at 0:33
Jason McCreary
56.4k19104152
5
Make sure to escape your string! See @zzzzBov answer
– jtpereyda
Oct 6 '15 at 21:18
This is not working in my case, can you please have a look this jsfiddle and let me know if what i have to do if I want to set a country code dynamic intoregexexpression
– Kirankumar Dafda
Jun 10 '16 at 8:30
6
One important thing to remember is that in regular strings the character needs to be escaped while in the regex literal (usually) the / character needs to be escaped. So/w+//ibecomesnew RegExp("\w+/", "i")
– Ali
Jan 16 '17 at 14:46
1
How about the /g?
– Elgs Qian Chen
Mar 8 '17 at 3:27
add a comment |
up vote
208
down vote
accepted
var regex = new RegExp("ReGeX" + testVar + "ReGeX");
...
string.replace(regex, "replacement");
Update
Per some of the comments, it's important to note that you may want to escape the variable if there is potential for malicious content (e.g. the variable comes from user input)
edited May 23 '17 at 11:55
Community♦
11
answered Oct 27 '10 at 0:33
Jason McCreary
56.4k19104152
5
Make sure to escape your string! See @zzzzBov answer
– jtpereyda
Oct 6 '15 at 21:18
This is not working in my case, can you please have a look this jsfiddle and let me know if what i have to do if I want to set a country code dynamic intoregexexpression
– Kirankumar Dafda
Jun 10 '16 at 8:30
6
One important thing to remember is that in regular strings the character needs to be escaped while in the regex literal (usually) the / character needs to be escaped. So/w+//ibecomesnew RegExp("\w+/", "i")
– Ali
Jan 16 '17 at 14:46
1
How about the /g?
– Elgs Qian Chen
Mar 8 '17 at 3:27
add a comment |
up vote
208
down vote
accepted
up vote
208
down vote
accepted
var regex = new RegExp("ReGeX" + testVar + "ReGeX");
...
string.replace(regex, "replacement");
Update
Per some of the comments, it's important to note that you may want to escape the variable if there is potential for malicious content (e.g. the variable comes from user input)
edited May 23 '17 at 11:55
Community♦
11
answered Oct 27 '10 at 0:33
Jason McCreary
56.4k19104152
var regex = new RegExp("ReGeX" + testVar + "ReGeX");
...
string.replace(regex, "replacement");
Update
Per some of the comments, it's important to note that you may want to escape the variable if there is potential for malicious content (e.g. the variable comes from user input)
edited May 23 '17 at 11:55
Community♦
11
answered Oct 27 '10 at 0:33
Jason McCreary
56.4k19104152
edited May 23 '17 at 11:55
Community♦
11
edited May 23 '17 at 11:55
Community♦
11
edited May 23 '17 at 11:55
Community♦
11
11
answered Oct 27 '10 at 0:33
Jason McCreary
56.4k19104152
answered Oct 27 '10 at 0:33
Jason McCreary
56.4k19104152
answered Oct 27 '10 at 0:33
Jason McCreary
56.4k19104152
56.4k19104152
5
Make sure to escape your string! See @zzzzBov answer
– jtpereyda
Oct 6 '15 at 21:18
This is not working in my case, can you please have a look this jsfiddle and let me know if what i have to do if I want to set a country code dynamic intoregexexpression
– Kirankumar Dafda
Jun 10 '16 at 8:30
6
One important thing to remember is that in regular strings the character needs to be escaped while in the regex literal (usually) the / character needs to be escaped. So/w+//ibecomesnew RegExp("\w+/", "i")
– Ali
Jan 16 '17 at 14:46
1
How about the /g?
– Elgs Qian Chen
Mar 8 '17 at 3:27
add a comment |
5
Make sure to escape your string! See @zzzzBov answer
– jtpereyda
Oct 6 '15 at 21:18
This is not working in my case, can you please have a look this jsfiddle and let me know if what i have to do if I want to set a country code dynamic intoregexexpression
– Kirankumar Dafda
Jun 10 '16 at 8:30
6
One important thing to remember is that in regular strings the character needs to be escaped while in the regex literal (usually) the / character needs to be escaped. So/w+//ibecomesnew RegExp("\w+/", "i")
– Ali
Jan 16 '17 at 14:46
1
How about the /g?
– Elgs Qian Chen
Mar 8 '17 at 3:27
5
5
Make sure to escape your string! See @zzzzBov answer
– jtpereyda
Oct 6 '15 at 21:18
Make sure to escape your string! See @zzzzBov answer
– jtpereyda
Oct 6 '15 at 21:18
This is not working in my case, can you please have a look this jsfiddle and let me know if what i have to do if I want to set a country code dynamic into
regex expression– Kirankumar Dafda
Jun 10 '16 at 8:30
This is not working in my case, can you please have a look this jsfiddle and let me know if what i have to do if I want to set a country code dynamic into
regex expression– Kirankumar Dafda
Jun 10 '16 at 8:30
6
6
One important thing to remember is that in regular strings the character needs to be escaped while in the regex literal (usually) the / character needs to be escaped. So
/w+//i becomes new RegExp("\w+/", "i")– Ali
Jan 16 '17 at 14:46
One important thing to remember is that in regular strings the character needs to be escaped while in the regex literal (usually) the / character needs to be escaped. So
/w+//i becomes new RegExp("\w+/", "i")– Ali
Jan 16 '17 at 14:46
1
1
How about the /g?
– Elgs Qian Chen
Mar 8 '17 at 3:27
How about the /g?
– Elgs Qian Chen
Mar 8 '17 at 3:27
add a comment |
up vote
67
down vote
You can use the RegExp object:
var regexstring = "whatever";
var regexp = new RegExp(regexstring, "gi");
var str = "whateverTest";
var str2 = str.replace(regexp, "other");
document.write(str2);
Then you can construct regexstring in any way you want.
You can read more about it here.
answered Oct 27 '10 at 0:40
steinar
7,77511730
5
+1 for the full sample, RegExp params, and link.
– Jason McCreary
Oct 27 '10 at 0:45
This does not work for me on chrome. :(
– Ankit Tanna
May 8 '15 at 9:15
var characterCount = parseInt(attrs.limitCharacterCount); console.log(characterCount); var specialCharactersValidation = new RegExp("/^[a-zA-Z0-9]{"+characterCount+"}$/"); //^[a-zA-Z0-9]{7}$// requestShipmentDirectives.js:76 false main.js:46 Uncaught TypeError: Cannot read property 'offsetWidth' of null This is what is happening when I make the RegEx Dynamic.
– Ankit Tanna
May 8 '15 at 9:19
add a comment |
up vote
67
down vote
You can use the RegExp object:
var regexstring = "whatever";
var regexp = new RegExp(regexstring, "gi");
var str = "whateverTest";
var str2 = str.replace(regexp, "other");
document.write(str2);
Then you can construct regexstring in any way you want.
You can read more about it here.
answered Oct 27 '10 at 0:40
steinar
7,77511730
5
+1 for the full sample, RegExp params, and link.
– Jason McCreary
Oct 27 '10 at 0:45
This does not work for me on chrome. :(
– Ankit Tanna
May 8 '15 at 9:15
var characterCount = parseInt(attrs.limitCharacterCount); console.log(characterCount); var specialCharactersValidation = new RegExp("/^[a-zA-Z0-9]{"+characterCount+"}$/"); //^[a-zA-Z0-9]{7}$// requestShipmentDirectives.js:76 false main.js:46 Uncaught TypeError: Cannot read property 'offsetWidth' of null This is what is happening when I make the RegEx Dynamic.
– Ankit Tanna
May 8 '15 at 9:19
add a comment |
up vote
67
down vote
up vote
67
down vote
You can use the RegExp object:
var regexstring = "whatever";
var regexp = new RegExp(regexstring, "gi");
var str = "whateverTest";
var str2 = str.replace(regexp, "other");
document.write(str2);
Then you can construct regexstring in any way you want.
You can read more about it here.
answered Oct 27 '10 at 0:40
steinar
7,77511730
You can use the RegExp object:
var regexstring = "whatever";
var regexp = new RegExp(regexstring, "gi");
var str = "whateverTest";
var str2 = str.replace(regexp, "other");
document.write(str2);
Then you can construct regexstring in any way you want.
You can read more about it here.
answered Oct 27 '10 at 0:40
steinar
7,77511730
answered Oct 27 '10 at 0:40
steinar
7,77511730
answered Oct 27 '10 at 0:40
steinar
7,77511730
answered Oct 27 '10 at 0:40
steinar
7,77511730
7,77511730
5
+1 for the full sample, RegExp params, and link.
– Jason McCreary
Oct 27 '10 at 0:45
This does not work for me on chrome. :(
– Ankit Tanna
May 8 '15 at 9:15
var characterCount = parseInt(attrs.limitCharacterCount); console.log(characterCount); var specialCharactersValidation = new RegExp("/^[a-zA-Z0-9]{"+characterCount+"}$/"); //^[a-zA-Z0-9]{7}$// requestShipmentDirectives.js:76 false main.js:46 Uncaught TypeError: Cannot read property 'offsetWidth' of null This is what is happening when I make the RegEx Dynamic.
– Ankit Tanna
May 8 '15 at 9:19
add a comment |
5
+1 for the full sample, RegExp params, and link.
– Jason McCreary
Oct 27 '10 at 0:45
This does not work for me on chrome. :(
– Ankit Tanna
May 8 '15 at 9:15
var characterCount = parseInt(attrs.limitCharacterCount); console.log(characterCount); var specialCharactersValidation = new RegExp("/^[a-zA-Z0-9]{"+characterCount+"}$/"); //^[a-zA-Z0-9]{7}$// requestShipmentDirectives.js:76 false main.js:46 Uncaught TypeError: Cannot read property 'offsetWidth' of null This is what is happening when I make the RegEx Dynamic.
– Ankit Tanna
May 8 '15 at 9:19
5
5
+1 for the full sample, RegExp params, and link.
– Jason McCreary
Oct 27 '10 at 0:45
+1 for the full sample, RegExp params, and link.
– Jason McCreary
Oct 27 '10 at 0:45
This does not work for me on chrome. :(
– Ankit Tanna
May 8 '15 at 9:15
This does not work for me on chrome. :(
– Ankit Tanna
May 8 '15 at 9:15
var characterCount = parseInt(attrs.limitCharacterCount); console.log(characterCount); var specialCharactersValidation = new RegExp("/^[a-zA-Z0-9]{"+characterCount+"}$/"); //^[a-zA-Z0-9]{7}$// requestShipmentDirectives.js:76 false main.js:46 Uncaught TypeError: Cannot read property 'offsetWidth' of null This is what is happening when I make the RegEx Dynamic.
– Ankit Tanna
May 8 '15 at 9:19
var characterCount = parseInt(attrs.limitCharacterCount); console.log(characterCount); var specialCharactersValidation = new RegExp("/^[a-zA-Z0-9]{"+characterCount+"}$/"); //^[a-zA-Z0-9]{7}$// requestShipmentDirectives.js:76 false main.js:46 Uncaught TypeError: Cannot read property 'offsetWidth' of null This is what is happening when I make the RegEx Dynamic.
– Ankit Tanna
May 8 '15 at 9:19
add a comment |
up vote
28
down vote
To build a regular expression from a variable in JavaScript, you'll need to use the RegExp constructor with a string parameter.
function reg(input) {
var flags;
//could be any combination of 'g', 'i', and 'm'
flags = 'g';
return new RegExp('ReGeX' + input + 'ReGeX', flags);
}
of course, this is a very naive example. It assumes that input is has been properly escaped for a regular expression. If you're dealing with user-input, or simply want to make it more convenient to match special characters, you'll need to escape special characters:
function regexEscape(str) {
return str.replace(/[-/\^$*+?.()|[]{}]/g, '\$&')
}
function reg(input) {
var flags;
//could be any combination of 'g', 'i', and 'm'
flags = 'g';
input = regexEscape(input);
return new RegExp('ReGeX' + input + 'ReGeX', flags);
}
edited May 23 '17 at 11:55
Community♦
11
answered Jan 16 '13 at 13:29
zzzzBov
129k33261305
The only way this worked for me was if I deletedReGeXtext. So I used:return new RegExp(input, flags);
– Michael Rader
Dec 30 '16 at 1:47
1
@MichaelRader, yes, the "ReGeX" text was part of the question and used as an example, not as something that's required to make the code work.
– zzzzBov
Dec 30 '16 at 3:49
lol just realized that was just your placeholder, but as a noob this took me awhile to figure out. Thanks.
– Michael Rader
Dec 30 '16 at 4:43
add a comment |
up vote
28
down vote
To build a regular expression from a variable in JavaScript, you'll need to use the RegExp constructor with a string parameter.
function reg(input) {
var flags;
//could be any combination of 'g', 'i', and 'm'
flags = 'g';
return new RegExp('ReGeX' + input + 'ReGeX', flags);
}
of course, this is a very naive example. It assumes that input is has been properly escaped for a regular expression. If you're dealing with user-input, or simply want to make it more convenient to match special characters, you'll need to escape special characters:
function regexEscape(str) {
return str.replace(/[-/\^$*+?.()|[]{}]/g, '\$&')
}
function reg(input) {
var flags;
//could be any combination of 'g', 'i', and 'm'
flags = 'g';
input = regexEscape(input);
return new RegExp('ReGeX' + input + 'ReGeX', flags);
}
edited May 23 '17 at 11:55
Community♦
11
answered Jan 16 '13 at 13:29
zzzzBov
129k33261305
The only way this worked for me was if I deletedReGeXtext. So I used:return new RegExp(input, flags);
– Michael Rader
Dec 30 '16 at 1:47
1
@MichaelRader, yes, the "ReGeX" text was part of the question and used as an example, not as something that's required to make the code work.
– zzzzBov
Dec 30 '16 at 3:49
lol just realized that was just your placeholder, but as a noob this took me awhile to figure out. Thanks.
– Michael Rader
Dec 30 '16 at 4:43
add a comment |
up vote
28
down vote
up vote
28
down vote
To build a regular expression from a variable in JavaScript, you'll need to use the RegExp constructor with a string parameter.
function reg(input) {
var flags;
//could be any combination of 'g', 'i', and 'm'
flags = 'g';
return new RegExp('ReGeX' + input + 'ReGeX', flags);
}
of course, this is a very naive example. It assumes that input is has been properly escaped for a regular expression. If you're dealing with user-input, or simply want to make it more convenient to match special characters, you'll need to escape special characters:
function regexEscape(str) {
return str.replace(/[-/\^$*+?.()|[]{}]/g, '\$&')
}
function reg(input) {
var flags;
//could be any combination of 'g', 'i', and 'm'
flags = 'g';
input = regexEscape(input);
return new RegExp('ReGeX' + input + 'ReGeX', flags);
}
edited May 23 '17 at 11:55
Community♦
11
answered Jan 16 '13 at 13:29
zzzzBov
129k33261305
To build a regular expression from a variable in JavaScript, you'll need to use the RegExp constructor with a string parameter.
function reg(input) {
var flags;
//could be any combination of 'g', 'i', and 'm'
flags = 'g';
return new RegExp('ReGeX' + input + 'ReGeX', flags);
}
of course, this is a very naive example. It assumes that input is has been properly escaped for a regular expression. If you're dealing with user-input, or simply want to make it more convenient to match special characters, you'll need to escape special characters:
function regexEscape(str) {
return str.replace(/[-/\^$*+?.()|[]{}]/g, '\$&')
}
function reg(input) {
var flags;
//could be any combination of 'g', 'i', and 'm'
flags = 'g';
input = regexEscape(input);
return new RegExp('ReGeX' + input + 'ReGeX', flags);
}
edited May 23 '17 at 11:55
Community♦
11
answered Jan 16 '13 at 13:29
zzzzBov
129k33261305
edited May 23 '17 at 11:55
Community♦
11
edited May 23 '17 at 11:55
Community♦
11
edited May 23 '17 at 11:55
Community♦
11
11
answered Jan 16 '13 at 13:29
zzzzBov
129k33261305
answered Jan 16 '13 at 13:29
zzzzBov
129k33261305
answered Jan 16 '13 at 13:29
zzzzBov
129k33261305
129k33261305
The only way this worked for me was if I deletedReGeXtext. So I used:return new RegExp(input, flags);
– Michael Rader
Dec 30 '16 at 1:47
1
@MichaelRader, yes, the "ReGeX" text was part of the question and used as an example, not as something that's required to make the code work.
– zzzzBov
Dec 30 '16 at 3:49
lol just realized that was just your placeholder, but as a noob this took me awhile to figure out. Thanks.
– Michael Rader
Dec 30 '16 at 4:43
add a comment |
The only way this worked for me was if I deletedReGeXtext. So I used:return new RegExp(input, flags);
– Michael Rader
Dec 30 '16 at 1:47
1
@MichaelRader, yes, the "ReGeX" text was part of the question and used as an example, not as something that's required to make the code work.
– zzzzBov
Dec 30 '16 at 3:49
lol just realized that was just your placeholder, but as a noob this took me awhile to figure out. Thanks.
– Michael Rader
Dec 30 '16 at 4:43
The only way this worked for me was if I deleted
ReGeX text. So I used: return new RegExp(input, flags);– Michael Rader
Dec 30 '16 at 1:47
The only way this worked for me was if I deleted
ReGeX text. So I used: return new RegExp(input, flags);– Michael Rader
Dec 30 '16 at 1:47
1
1
@MichaelRader, yes, the "ReGeX" text was part of the question and used as an example, not as something that's required to make the code work.
– zzzzBov
Dec 30 '16 at 3:49
@MichaelRader, yes, the "ReGeX" text was part of the question and used as an example, not as something that's required to make the code work.
– zzzzBov
Dec 30 '16 at 3:49
lol just realized that was just your placeholder, but as a noob this took me awhile to figure out. Thanks.
– Michael Rader
Dec 30 '16 at 4:43
lol just realized that was just your placeholder, but as a noob this took me awhile to figure out. Thanks.
– Michael Rader
Dec 30 '16 at 4:43
add a comment |
up vote
6
down vote
You can always give regular expression as string, i.e. "ReGeX" + testVar + "ReGeX". You'll possibly have to escape some characters inside your string (e.g., double quote), but for most cases it's equivalent.
You can also use RegExp constructor to pass flags in (see the docs).
answered Oct 27 '10 at 0:33
Nikita Rybak
56.2k18137163
add a comment |
up vote
6
down vote
You can always give regular expression as string, i.e. "ReGeX" + testVar + "ReGeX". You'll possibly have to escape some characters inside your string (e.g., double quote), but for most cases it's equivalent.
You can also use RegExp constructor to pass flags in (see the docs).
answered Oct 27 '10 at 0:33
Nikita Rybak
56.2k18137163
add a comment |
up vote
6
down vote
up vote
6
down vote
You can always give regular expression as string, i.e. "ReGeX" + testVar + "ReGeX". You'll possibly have to escape some characters inside your string (e.g., double quote), but for most cases it's equivalent.
You can also use RegExp constructor to pass flags in (see the docs).
answered Oct 27 '10 at 0:33
Nikita Rybak
56.2k18137163
You can always give regular expression as string, i.e. "ReGeX" + testVar + "ReGeX". You'll possibly have to escape some characters inside your string (e.g., double quote), but for most cases it's equivalent.
You can also use RegExp constructor to pass flags in (see the docs).
answered Oct 27 '10 at 0:33
Nikita Rybak
56.2k18137163
answered Oct 27 '10 at 0:33
Nikita Rybak
56.2k18137163
answered Oct 27 '10 at 0:33
Nikita Rybak
56.2k18137163
answered Oct 27 '10 at 0:33
Nikita Rybak
56.2k18137163
56.2k18137163
add a comment |
add a comment |
up vote
3
down vote
if you're using es6 template literals are an option...
string.replace(new RegExp(`ReGeX${testVar}ReGeX`), "replacement")
answered Apr 5 at 10:19
shunryu111
2,72531513
add a comment |
up vote
3
down vote
if you're using es6 template literals are an option...
string.replace(new RegExp(`ReGeX${testVar}ReGeX`), "replacement")
answered Apr 5 at 10:19
shunryu111
2,72531513
add a comment |
up vote
3
down vote
up vote
3
down vote
if you're using es6 template literals are an option...
string.replace(new RegExp(`ReGeX${testVar}ReGeX`), "replacement")
answered Apr 5 at 10:19
shunryu111
2,72531513
if you're using es6 template literals are an option...
string.replace(new RegExp(`ReGeX${testVar}ReGeX`), "replacement")
answered Apr 5 at 10:19
shunryu111
2,72531513
edited Apr 5 at 10:28
answered Apr 5 at 10:19
shunryu111
2,72531513
answered Apr 5 at 10:19
shunryu111
2,72531513
answered Apr 5 at 10:19
shunryu111
2,72531513
2,72531513
add a comment |
add a comment |
up vote
2
down vote
accepted answer doesn't work for me and doesn't follow MDN examples
see the 'Description' section in above link
I'd go with the following it's working for me:
let stringThatIsGoingToChange = 'findMe';
let flagsYouWant = 'gi' //simple string with flags
let dynamicRegExp = new RegExp(`${stringThatIsGoingToChange}`, flagsYouWant)
// that makes dynamicRegExp = /findMe/gi
answered Jun 7 at 22:52
Timothy Mitchell
213
It still works. The issue is that 'ReGeX' + testVar +' ReGeX' makes the solution confusing. Adding an actually example would clear things up in my opinion. var testVar = '321'; var regex = new RegExp( "[" + testVar + "]{2}", "g" ); // above equivalent to /[321]{2}/g
– HelloWorldPeace
Jun 23 at 19:38
flagsYouWant is a string, not a variable, so that won't work
– macasas
Sep 14 at 11:24
@macasas right you are i think i fixed thanks!
– Timothy Mitchell
Oct 4 at 15:40
add a comment |
up vote
2
down vote
accepted answer doesn't work for me and doesn't follow MDN examples
see the 'Description' section in above link
I'd go with the following it's working for me:
let stringThatIsGoingToChange = 'findMe';
let flagsYouWant = 'gi' //simple string with flags
let dynamicRegExp = new RegExp(`${stringThatIsGoingToChange}`, flagsYouWant)
// that makes dynamicRegExp = /findMe/gi
answered Jun 7 at 22:52
Timothy Mitchell
213
It still works. The issue is that 'ReGeX' + testVar +' ReGeX' makes the solution confusing. Adding an actually example would clear things up in my opinion. var testVar = '321'; var regex = new RegExp( "[" + testVar + "]{2}", "g" ); // above equivalent to /[321]{2}/g
– HelloWorldPeace
Jun 23 at 19:38
flagsYouWant is a string, not a variable, so that won't work
– macasas
Sep 14 at 11:24
@macasas right you are i think i fixed thanks!
– Timothy Mitchell
Oct 4 at 15:40
add a comment |
up vote
2
down vote
up vote
2
down vote
accepted answer doesn't work for me and doesn't follow MDN examples
see the 'Description' section in above link
I'd go with the following it's working for me:
let stringThatIsGoingToChange = 'findMe';
let flagsYouWant = 'gi' //simple string with flags
let dynamicRegExp = new RegExp(`${stringThatIsGoingToChange}`, flagsYouWant)
// that makes dynamicRegExp = /findMe/gi
answered Jun 7 at 22:52
Timothy Mitchell
213
accepted answer doesn't work for me and doesn't follow MDN examples
see the 'Description' section in above link
I'd go with the following it's working for me:
let stringThatIsGoingToChange = 'findMe';
let flagsYouWant = 'gi' //simple string with flags
let dynamicRegExp = new RegExp(`${stringThatIsGoingToChange}`, flagsYouWant)
// that makes dynamicRegExp = /findMe/gi
answered Jun 7 at 22:52
Timothy Mitchell
213
edited Oct 4 at 15:39
answered Jun 7 at 22:52
Timothy Mitchell
213
answered Jun 7 at 22:52
Timothy Mitchell
213
answered Jun 7 at 22:52
Timothy Mitchell
213
213
It still works. The issue is that 'ReGeX' + testVar +' ReGeX' makes the solution confusing. Adding an actually example would clear things up in my opinion. var testVar = '321'; var regex = new RegExp( "[" + testVar + "]{2}", "g" ); // above equivalent to /[321]{2}/g
– HelloWorldPeace
Jun 23 at 19:38
flagsYouWant is a string, not a variable, so that won't work
– macasas
Sep 14 at 11:24
@macasas right you are i think i fixed thanks!
– Timothy Mitchell
Oct 4 at 15:40
add a comment |
It still works. The issue is that 'ReGeX' + testVar +' ReGeX' makes the solution confusing. Adding an actually example would clear things up in my opinion. var testVar = '321'; var regex = new RegExp( "[" + testVar + "]{2}", "g" ); // above equivalent to /[321]{2}/g
– HelloWorldPeace
Jun 23 at 19:38
flagsYouWant is a string, not a variable, so that won't work
– macasas
Sep 14 at 11:24
@macasas right you are i think i fixed thanks!
– Timothy Mitchell
Oct 4 at 15:40
It still works. The issue is that 'ReGeX' + testVar +' ReGeX' makes the solution confusing. Adding an actually example would clear things up in my opinion. var testVar = '321'; var regex = new RegExp( "[" + testVar + "]{2}", "g" ); // above equivalent to /[321]{2}/g
– HelloWorldPeace
Jun 23 at 19:38
It still works. The issue is that 'ReGeX' + testVar +' ReGeX' makes the solution confusing. Adding an actually example would clear things up in my opinion. var testVar = '321'; var regex = new RegExp( "[" + testVar + "]{2}", "g" ); // above equivalent to /[321]{2}/g
– HelloWorldPeace
Jun 23 at 19:38
flagsYouWant is a string, not a variable, so that won't work
– macasas
Sep 14 at 11:24
flagsYouWant is a string, not a variable, so that won't work
– macasas
Sep 14 at 11:24
@macasas right you are i think i fixed thanks!
– Timothy Mitchell
Oct 4 at 15:40
@macasas right you are i think i fixed thanks!
– Timothy Mitchell
Oct 4 at 15:40
add a comment |
up vote
1
down vote
Here's an pretty useless function that return values wrapped by specific characters. :)
jsfiddle: https://jsfiddle.net/squadjot/43agwo6x/
function getValsWrappedIn(str,c1,c2){
var rg = new RegExp("(?<=\"+c1+")(.*?)(?=\"+c2+")","g");
return str.match(rg);
}
var exampleStr = "Something (5) or some time (19) or maybe a (thingy)";
var results = getValsWrappedIn(exampleStr,"(",")")
// Will return array ["5","19","thingy"]
console.log(results)
answered Feb 17 at 1:20
Jakob Sternberg
1,2171010
add a comment |
up vote
1
down vote
Here's an pretty useless function that return values wrapped by specific characters. :)
jsfiddle: https://jsfiddle.net/squadjot/43agwo6x/
function getValsWrappedIn(str,c1,c2){
var rg = new RegExp("(?<=\"+c1+")(.*?)(?=\"+c2+")","g");
return str.match(rg);
}
var exampleStr = "Something (5) or some time (19) or maybe a (thingy)";
var results = getValsWrappedIn(exampleStr,"(",")")
// Will return array ["5","19","thingy"]
console.log(results)
answered Feb 17 at 1:20
Jakob Sternberg
1,2171010
add a comment |
up vote
1
down vote
up vote
1
down vote
Here's an pretty useless function that return values wrapped by specific characters. :)
jsfiddle: https://jsfiddle.net/squadjot/43agwo6x/
function getValsWrappedIn(str,c1,c2){
var rg = new RegExp("(?<=\"+c1+")(.*?)(?=\"+c2+")","g");
return str.match(rg);
}
var exampleStr = "Something (5) or some time (19) or maybe a (thingy)";
var results = getValsWrappedIn(exampleStr,"(",")")
// Will return array ["5","19","thingy"]
console.log(results)
answered Feb 17 at 1:20
Jakob Sternberg
1,2171010
Here's an pretty useless function that return values wrapped by specific characters. :)
jsfiddle: https://jsfiddle.net/squadjot/43agwo6x/
function getValsWrappedIn(str,c1,c2){
var rg = new RegExp("(?<=\"+c1+")(.*?)(?=\"+c2+")","g");
return str.match(rg);
}
var exampleStr = "Something (5) or some time (19) or maybe a (thingy)";
var results = getValsWrappedIn(exampleStr,"(",")")
// Will return array ["5","19","thingy"]
console.log(results)
answered Feb 17 at 1:20
Jakob Sternberg
1,2171010
edited Aug 7 at 22:14
answered Feb 17 at 1:20
Jakob Sternberg
1,2171010
answered Feb 17 at 1:20
Jakob Sternberg
1,2171010
answered Feb 17 at 1:20
Jakob Sternberg
1,2171010
1,2171010
add a comment |
add a comment |
draft saved
draft discarded
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f4029109%2fjavascript-regex-how-to-put-a-variable-inside-a-regular-expression%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
9
Be aware that if you let the user supply this variable, it's easy for a malicious user to crash your application via catastrophic backtracking.
– Tim Pietzcker
Oct 27 '10 at 6:23
1
what is "ReGeX"? is it a variable name? it makes answers confusing by implying that it is a part of RegExp construction (in case a reader did not see those weird characters in the question).
– Eduard
Oct 17 '17 at 13:12