Custom regex to format number
i have this number
1234567890
this is how i want to display it
1234 567 890
im trying this
console.log('1234567899'.replace(/(d)(?=(d{3})+$)/g, '$1 '));
javascript angularjs regex
asked Nov 23 '18 at 6:32
Manav
76111
add a comment |
i have this number
1234567890
this is how i want to display it
1234 567 890
im trying this
console.log('1234567899'.replace(/(d)(?=(d{3})+$)/g, '$1 '));
javascript angularjs regex
asked Nov 23 '18 at 6:32
Manav
76111
which angular version are you using?
– Abhishek Mani
Nov 23 '18 at 6:33
Can you explain the logic you're looking for? Do you want to omit the first comma when there is only one digit in the first section, or for two as well, or for all 3, or what?
– CertainPerformance
Nov 23 '18 at 6:33
add a comment |
i have this number
1234567890
this is how i want to display it
1234 567 890
im trying this
console.log('1234567899'.replace(/(d)(?=(d{3})+$)/g, '$1 '));
javascript angularjs regex
asked Nov 23 '18 at 6:32
Manav
76111
i have this number
1234567890
this is how i want to display it
1234 567 890
im trying this
console.log('1234567899'.replace(/(d)(?=(d{3})+$)/g, '$1 '));
javascript angularjs regex
javascript angularjs regex
asked Nov 23 '18 at 6:32
Manav
76111
asked Nov 23 '18 at 6:32
Manav
76111
asked Nov 23 '18 at 6:32
Manav
76111
asked Nov 23 '18 at 6:32
Manav
76111
asked Nov 23 '18 at 6:32
Manav
76111
76111
which angular version are you using?
– Abhishek Mani
Nov 23 '18 at 6:33
Can you explain the logic you're looking for? Do you want to omit the first comma when there is only one digit in the first section, or for two as well, or for all 3, or what?
– CertainPerformance
Nov 23 '18 at 6:33
add a comment |
which angular version are you using?
– Abhishek Mani
Nov 23 '18 at 6:33
Can you explain the logic you're looking for? Do you want to omit the first comma when there is only one digit in the first section, or for two as well, or for all 3, or what?
– CertainPerformance
Nov 23 '18 at 6:33
which angular version are you using?
– Abhishek Mani
Nov 23 '18 at 6:33
which angular version are you using?
– Abhishek Mani
Nov 23 '18 at 6:33
Can you explain the logic you're looking for? Do you want to omit the first comma when there is only one digit in the first section, or for two as well, or for all 3, or what?
– CertainPerformance
Nov 23 '18 at 6:33
Can you explain the logic you're looking for? Do you want to omit the first comma when there is only one digit in the first section, or for two as well, or for all 3, or what?
– CertainPerformance
Nov 23 '18 at 6:33
add a comment |
4 Answers
4
active
oldest
votes
One option would be to have an optional group that starts at the beginning of the string and (greedily) matches the number of the leading-digits-without-commas you want to have. Then, instead of replacing with just 1, replace with 12 (the optional group plus the second captured digits):
const format = str => str.replace(
/(^(?:d{1,2}))?(d{1,3})(?=(?:d{3})+$)/g,
// ^^^ change these to change the number of unbroken leading digits
'$1$2 '
);
console.log(format('1234567899'));
console.log(format('01234567899'));
console.log(format('101234567899'));The above snippet's optional group begins with d{1,2}, which means that there will be between 3 and 5 leading digits, unbroken by commas. To change that quantity, just change the number of repetitions.
The leading group (^(?:d{1,2}))? means: optionally, the beginning of the string, followed by one or two digits.
answered Nov 23 '18 at 6:53
CertainPerformance
76.5k143862
add a comment |
- Divide your string into 2 parts.
- String contains space after 4 digits
- String contains space after 3 digits.
- Add space after n digits
Try with this
var str = "1234567890";
var temp = str.substring(0,4); // get first 4 digits
//Add space after 3 digits. You can use same logic to add space after 4 digits as well
var z = [...str.substring(4)].map((d, i) => i % 3 == 0 ? ' '+d : d).join('').trim();
//Concatenate both strings
var result = temp + ' ' + z;
//Display result
console.log(temp);
console.log(z);
console.log(result);answered Nov 23 '18 at 6:54
Prasad Telkikar
1,862419
add a comment |
Try this regex which ensures the numbers will be grouped in bunches of three except the first digits which can be grouped from four to five.
Match /(^d{4}|d{3})(?=(d{3})*$)/g and replace with $1
Here is some sample javascript code,
console.log('1234567899' + ' --> ' + '1234567899'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
console.log('12345678991' + ' --> ' + '12345678991'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
console.log('123456789912' + ' --> ' + '123456789912'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
console.log('1234567899123' + ' --> ' + '1234567899123'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));answered Nov 23 '18 at 7:38
Pushpesh Kumar Rajwanshi
5,3242827
add a comment |
You can do with this way.
let number = '1234567890';
let result = number.replace(/(d{4})(d{3})(d{3})/, "$1 $2 $3");
console.warn(result);
answered Nov 23 '18 at 6:44
tron
4318
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
One option would be to have an optional group that starts at the beginning of the string and (greedily) matches the number of the leading-digits-without-commas you want to have. Then, instead of replacing with just 1, replace with 12 (the optional group plus the second captured digits):
const format = str => str.replace(
/(^(?:d{1,2}))?(d{1,3})(?=(?:d{3})+$)/g,
// ^^^ change these to change the number of unbroken leading digits
'$1$2 '
);
console.log(format('1234567899'));
console.log(format('01234567899'));
console.log(format('101234567899'));The above snippet's optional group begins with d{1,2}, which means that there will be between 3 and 5 leading digits, unbroken by commas. To change that quantity, just change the number of repetitions.
The leading group (^(?:d{1,2}))? means: optionally, the beginning of the string, followed by one or two digits.
answered Nov 23 '18 at 6:53
CertainPerformance
76.5k143862
add a comment |
One option would be to have an optional group that starts at the beginning of the string and (greedily) matches the number of the leading-digits-without-commas you want to have. Then, instead of replacing with just 1, replace with 12 (the optional group plus the second captured digits):
const format = str => str.replace(
/(^(?:d{1,2}))?(d{1,3})(?=(?:d{3})+$)/g,
// ^^^ change these to change the number of unbroken leading digits
'$1$2 '
);
console.log(format('1234567899'));
console.log(format('01234567899'));
console.log(format('101234567899'));The above snippet's optional group begins with d{1,2}, which means that there will be between 3 and 5 leading digits, unbroken by commas. To change that quantity, just change the number of repetitions.
The leading group (^(?:d{1,2}))? means: optionally, the beginning of the string, followed by one or two digits.
answered Nov 23 '18 at 6:53
CertainPerformance
76.5k143862
add a comment |
One option would be to have an optional group that starts at the beginning of the string and (greedily) matches the number of the leading-digits-without-commas you want to have. Then, instead of replacing with just 1, replace with 12 (the optional group plus the second captured digits):
const format = str => str.replace(
/(^(?:d{1,2}))?(d{1,3})(?=(?:d{3})+$)/g,
// ^^^ change these to change the number of unbroken leading digits
'$1$2 '
);
console.log(format('1234567899'));
console.log(format('01234567899'));
console.log(format('101234567899'));The above snippet's optional group begins with d{1,2}, which means that there will be between 3 and 5 leading digits, unbroken by commas. To change that quantity, just change the number of repetitions.
The leading group (^(?:d{1,2}))? means: optionally, the beginning of the string, followed by one or two digits.
answered Nov 23 '18 at 6:53
CertainPerformance
76.5k143862
One option would be to have an optional group that starts at the beginning of the string and (greedily) matches the number of the leading-digits-without-commas you want to have. Then, instead of replacing with just 1, replace with 12 (the optional group plus the second captured digits):
const format = str => str.replace(
/(^(?:d{1,2}))?(d{1,3})(?=(?:d{3})+$)/g,
// ^^^ change these to change the number of unbroken leading digits
'$1$2 '
);
console.log(format('1234567899'));
console.log(format('01234567899'));
console.log(format('101234567899'));The above snippet's optional group begins with d{1,2}, which means that there will be between 3 and 5 leading digits, unbroken by commas. To change that quantity, just change the number of repetitions.
The leading group (^(?:d{1,2}))? means: optionally, the beginning of the string, followed by one or two digits.
const format = str => str.replace(
/(^(?:d{1,2}))?(d{1,3})(?=(?:d{3})+$)/g,
// ^^^ change these to change the number of unbroken leading digits
'$1$2 '
);
console.log(format('1234567899'));
console.log(format('01234567899'));
console.log(format('101234567899'));const format = str => str.replace(
/(^(?:d{1,2}))?(d{1,3})(?=(?:d{3})+$)/g,
// ^^^ change these to change the number of unbroken leading digits
'$1$2 '
);
console.log(format('1234567899'));
console.log(format('01234567899'));
console.log(format('101234567899'));answered Nov 23 '18 at 6:53
CertainPerformance
76.5k143862
answered Nov 23 '18 at 6:53
CertainPerformance
76.5k143862
answered Nov 23 '18 at 6:53
CertainPerformance
76.5k143862
answered Nov 23 '18 at 6:53
CertainPerformance
76.5k143862
76.5k143862
add a comment |
add a comment |
- Divide your string into 2 parts.
- String contains space after 4 digits
- String contains space after 3 digits.
- Add space after n digits
Try with this
var str = "1234567890";
var temp = str.substring(0,4); // get first 4 digits
//Add space after 3 digits. You can use same logic to add space after 4 digits as well
var z = [...str.substring(4)].map((d, i) => i % 3 == 0 ? ' '+d : d).join('').trim();
//Concatenate both strings
var result = temp + ' ' + z;
//Display result
console.log(temp);
console.log(z);
console.log(result);answered Nov 23 '18 at 6:54
Prasad Telkikar
1,862419
add a comment |
- Divide your string into 2 parts.
- String contains space after 4 digits
- String contains space after 3 digits.
- Add space after n digits
Try with this
var str = "1234567890";
var temp = str.substring(0,4); // get first 4 digits
//Add space after 3 digits. You can use same logic to add space after 4 digits as well
var z = [...str.substring(4)].map((d, i) => i % 3 == 0 ? ' '+d : d).join('').trim();
//Concatenate both strings
var result = temp + ' ' + z;
//Display result
console.log(temp);
console.log(z);
console.log(result);answered Nov 23 '18 at 6:54
Prasad Telkikar
1,862419
add a comment |
- Divide your string into 2 parts.
- String contains space after 4 digits
- String contains space after 3 digits.
- Add space after n digits
Try with this
var str = "1234567890";
var temp = str.substring(0,4); // get first 4 digits
//Add space after 3 digits. You can use same logic to add space after 4 digits as well
var z = [...str.substring(4)].map((d, i) => i % 3 == 0 ? ' '+d : d).join('').trim();
//Concatenate both strings
var result = temp + ' ' + z;
//Display result
console.log(temp);
console.log(z);
console.log(result);answered Nov 23 '18 at 6:54
Prasad Telkikar
1,862419
- Divide your string into 2 parts.
- String contains space after 4 digits
- String contains space after 3 digits.
- Add space after n digits
Try with this
var str = "1234567890";
var temp = str.substring(0,4); // get first 4 digits
//Add space after 3 digits. You can use same logic to add space after 4 digits as well
var z = [...str.substring(4)].map((d, i) => i % 3 == 0 ? ' '+d : d).join('').trim();
//Concatenate both strings
var result = temp + ' ' + z;
//Display result
console.log(temp);
console.log(z);
console.log(result);var str = "1234567890";
var temp = str.substring(0,4); // get first 4 digits
//Add space after 3 digits. You can use same logic to add space after 4 digits as well
var z = [...str.substring(4)].map((d, i) => i % 3 == 0 ? ' '+d : d).join('').trim();
//Concatenate both strings
var result = temp + ' ' + z;
//Display result
console.log(temp);
console.log(z);
console.log(result);var str = "1234567890";
var temp = str.substring(0,4); // get first 4 digits
//Add space after 3 digits. You can use same logic to add space after 4 digits as well
var z = [...str.substring(4)].map((d, i) => i % 3 == 0 ? ' '+d : d).join('').trim();
//Concatenate both strings
var result = temp + ' ' + z;
//Display result
console.log(temp);
console.log(z);
console.log(result);answered Nov 23 '18 at 6:54
Prasad Telkikar
1,862419
edited Nov 23 '18 at 7:00
answered Nov 23 '18 at 6:54
Prasad Telkikar
1,862419
answered Nov 23 '18 at 6:54
Prasad Telkikar
1,862419
answered Nov 23 '18 at 6:54
Prasad Telkikar
1,862419
1,862419
add a comment |
add a comment |
Try this regex which ensures the numbers will be grouped in bunches of three except the first digits which can be grouped from four to five.
Match /(^d{4}|d{3})(?=(d{3})*$)/g and replace with $1
Here is some sample javascript code,
console.log('1234567899' + ' --> ' + '1234567899'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
console.log('12345678991' + ' --> ' + '12345678991'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
console.log('123456789912' + ' --> ' + '123456789912'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
console.log('1234567899123' + ' --> ' + '1234567899123'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));answered Nov 23 '18 at 7:38
Pushpesh Kumar Rajwanshi
5,3242827
add a comment |
Try this regex which ensures the numbers will be grouped in bunches of three except the first digits which can be grouped from four to five.
Match /(^d{4}|d{3})(?=(d{3})*$)/g and replace with $1
Here is some sample javascript code,
console.log('1234567899' + ' --> ' + '1234567899'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
console.log('12345678991' + ' --> ' + '12345678991'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
console.log('123456789912' + ' --> ' + '123456789912'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
console.log('1234567899123' + ' --> ' + '1234567899123'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));answered Nov 23 '18 at 7:38
Pushpesh Kumar Rajwanshi
5,3242827
add a comment |
Try this regex which ensures the numbers will be grouped in bunches of three except the first digits which can be grouped from four to five.
Match /(^d{4}|d{3})(?=(d{3})*$)/g and replace with $1
Here is some sample javascript code,
console.log('1234567899' + ' --> ' + '1234567899'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
console.log('12345678991' + ' --> ' + '12345678991'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
console.log('123456789912' + ' --> ' + '123456789912'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
console.log('1234567899123' + ' --> ' + '1234567899123'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));answered Nov 23 '18 at 7:38
Pushpesh Kumar Rajwanshi
5,3242827
Try this regex which ensures the numbers will be grouped in bunches of three except the first digits which can be grouped from four to five.
Match /(^d{4}|d{3})(?=(d{3})*$)/g and replace with $1
Here is some sample javascript code,
console.log('1234567899' + ' --> ' + '1234567899'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
console.log('12345678991' + ' --> ' + '12345678991'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
console.log('123456789912' + ' --> ' + '123456789912'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
console.log('1234567899123' + ' --> ' + '1234567899123'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));console.log('1234567899' + ' --> ' + '1234567899'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
console.log('12345678991' + ' --> ' + '12345678991'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
console.log('123456789912' + ' --> ' + '123456789912'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
console.log('1234567899123' + ' --> ' + '1234567899123'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));console.log('1234567899' + ' --> ' + '1234567899'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
console.log('12345678991' + ' --> ' + '12345678991'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
console.log('123456789912' + ' --> ' + '123456789912'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));
console.log('1234567899123' + ' --> ' + '1234567899123'.replace(/(^d{4}|d{3})(?=(d{3})*$)/g, '$1 '));answered Nov 23 '18 at 7:38
Pushpesh Kumar Rajwanshi
5,3242827
edited Nov 23 '18 at 8:06
answered Nov 23 '18 at 7:38
Pushpesh Kumar Rajwanshi
5,3242827
answered Nov 23 '18 at 7:38
Pushpesh Kumar Rajwanshi
5,3242827
answered Nov 23 '18 at 7:38
Pushpesh Kumar Rajwanshi
5,3242827
5,3242827
add a comment |
add a comment |
You can do with this way.
let number = '1234567890';
let result = number.replace(/(d{4})(d{3})(d{3})/, "$1 $2 $3");
console.warn(result);
answered Nov 23 '18 at 6:44
tron
4318
add a comment |
You can do with this way.
let number = '1234567890';
let result = number.replace(/(d{4})(d{3})(d{3})/, "$1 $2 $3");
console.warn(result);
answered Nov 23 '18 at 6:44
tron
4318
add a comment |
You can do with this way.
let number = '1234567890';
let result = number.replace(/(d{4})(d{3})(d{3})/, "$1 $2 $3");
console.warn(result);
answered Nov 23 '18 at 6:44
tron
4318
You can do with this way.
let number = '1234567890';
let result = number.replace(/(d{4})(d{3})(d{3})/, "$1 $2 $3");
console.warn(result);
answered Nov 23 '18 at 6:44
tron
4318
edited Nov 23 '18 at 11:13
answered Nov 23 '18 at 6:44
tron
4318
answered Nov 23 '18 at 6:44
tron
4318
answered Nov 23 '18 at 6:44
tron
4318
4318
add a comment |
add a comment |
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which angular version are you using?
– Abhishek Mani
Nov 23 '18 at 6:33
Can you explain the logic you're looking for? Do you want to omit the first comma when there is only one digit in the first section, or for two as well, or for all 3, or what?
– CertainPerformance
Nov 23 '18 at 6:33