Looping over JavaScript object and adding string to end if not the last item











up vote
1
down vote

favorite












{
field_country: ["England", "Netherlands", "India", "Italy"],
field_continent: ["Europe"],
field_group: ["Building", "People", "Landscape"
}


I want to loop over each item and return the key and the array together with ending 'OR' for example:
field_country: "England" OR field_country: "Netherlands"
The last item should not end with 'OR' in the loop. I am not sure what the best process is for this using vanilla JS. So far my code is as follows:



Object.keys(facets).forEach(function(facetKey) {
if (facets[facetKey].length > 1) {
facetResults = facets[facetKey];
for (var i = 0; i < facetResults.length; i ++) {
if (i == 1) {
filter = "'" + facetKey + "'" + ":'" + facetResults[i] + " OR";
return filter;
} else {
filter = "'" + facetKey + "'" + ":'" + facetResults[i];
}
}
} else {
filter = "'" + facetKey + "'" + ": " + facets[facetKey] + "'";
return filter;
}
});


I would be very grateful for any assistance.
Thanks in advance.










share|improve this question






















  • append the or at the beginning of the string when index > 0 instead of adding it at the end when index > -1
    – blaze_125
    Nov 22 at 0:19












  • The classic Fence Post problem. The basic algorithm is to start at your first post (element at index 0 in this case) and post-pend your fence (the " OR ") up until the second to last element, then add your last post. As others have pointed out, the array join function does this for you.
    – Drew Reese
    Nov 22 at 0:53















up vote
1
down vote

favorite












{
field_country: ["England", "Netherlands", "India", "Italy"],
field_continent: ["Europe"],
field_group: ["Building", "People", "Landscape"
}


I want to loop over each item and return the key and the array together with ending 'OR' for example:
field_country: "England" OR field_country: "Netherlands"
The last item should not end with 'OR' in the loop. I am not sure what the best process is for this using vanilla JS. So far my code is as follows:



Object.keys(facets).forEach(function(facetKey) {
if (facets[facetKey].length > 1) {
facetResults = facets[facetKey];
for (var i = 0; i < facetResults.length; i ++) {
if (i == 1) {
filter = "'" + facetKey + "'" + ":'" + facetResults[i] + " OR";
return filter;
} else {
filter = "'" + facetKey + "'" + ":'" + facetResults[i];
}
}
} else {
filter = "'" + facetKey + "'" + ": " + facets[facetKey] + "'";
return filter;
}
});


I would be very grateful for any assistance.
Thanks in advance.










share|improve this question






















  • append the or at the beginning of the string when index > 0 instead of adding it at the end when index > -1
    – blaze_125
    Nov 22 at 0:19












  • The classic Fence Post problem. The basic algorithm is to start at your first post (element at index 0 in this case) and post-pend your fence (the " OR ") up until the second to last element, then add your last post. As others have pointed out, the array join function does this for you.
    – Drew Reese
    Nov 22 at 0:53













up vote
1
down vote

favorite









up vote
1
down vote

favorite











{
field_country: ["England", "Netherlands", "India", "Italy"],
field_continent: ["Europe"],
field_group: ["Building", "People", "Landscape"
}


I want to loop over each item and return the key and the array together with ending 'OR' for example:
field_country: "England" OR field_country: "Netherlands"
The last item should not end with 'OR' in the loop. I am not sure what the best process is for this using vanilla JS. So far my code is as follows:



Object.keys(facets).forEach(function(facetKey) {
if (facets[facetKey].length > 1) {
facetResults = facets[facetKey];
for (var i = 0; i < facetResults.length; i ++) {
if (i == 1) {
filter = "'" + facetKey + "'" + ":'" + facetResults[i] + " OR";
return filter;
} else {
filter = "'" + facetKey + "'" + ":'" + facetResults[i];
}
}
} else {
filter = "'" + facetKey + "'" + ": " + facets[facetKey] + "'";
return filter;
}
});


I would be very grateful for any assistance.
Thanks in advance.










share|improve this question













{
field_country: ["England", "Netherlands", "India", "Italy"],
field_continent: ["Europe"],
field_group: ["Building", "People", "Landscape"
}


I want to loop over each item and return the key and the array together with ending 'OR' for example:
field_country: "England" OR field_country: "Netherlands"
The last item should not end with 'OR' in the loop. I am not sure what the best process is for this using vanilla JS. So far my code is as follows:



Object.keys(facets).forEach(function(facetKey) {
if (facets[facetKey].length > 1) {
facetResults = facets[facetKey];
for (var i = 0; i < facetResults.length; i ++) {
if (i == 1) {
filter = "'" + facetKey + "'" + ":'" + facetResults[i] + " OR";
return filter;
} else {
filter = "'" + facetKey + "'" + ":'" + facetResults[i];
}
}
} else {
filter = "'" + facetKey + "'" + ": " + facets[facetKey] + "'";
return filter;
}
});


I would be very grateful for any assistance.
Thanks in advance.







javascript arrays string foreach






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 21 at 23:57









Peter Ayello Wright

315




315












  • append the or at the beginning of the string when index > 0 instead of adding it at the end when index > -1
    – blaze_125
    Nov 22 at 0:19












  • The classic Fence Post problem. The basic algorithm is to start at your first post (element at index 0 in this case) and post-pend your fence (the " OR ") up until the second to last element, then add your last post. As others have pointed out, the array join function does this for you.
    – Drew Reese
    Nov 22 at 0:53


















  • append the or at the beginning of the string when index > 0 instead of adding it at the end when index > -1
    – blaze_125
    Nov 22 at 0:19












  • The classic Fence Post problem. The basic algorithm is to start at your first post (element at index 0 in this case) and post-pend your fence (the " OR ") up until the second to last element, then add your last post. As others have pointed out, the array join function does this for you.
    – Drew Reese
    Nov 22 at 0:53
















append the or at the beginning of the string when index > 0 instead of adding it at the end when index > -1
– blaze_125
Nov 22 at 0:19






append the or at the beginning of the string when index > 0 instead of adding it at the end when index > -1
– blaze_125
Nov 22 at 0:19














The classic Fence Post problem. The basic algorithm is to start at your first post (element at index 0 in this case) and post-pend your fence (the " OR ") up until the second to last element, then add your last post. As others have pointed out, the array join function does this for you.
– Drew Reese
Nov 22 at 0:53




The classic Fence Post problem. The basic algorithm is to start at your first post (element at index 0 in this case) and post-pend your fence (the " OR ") up until the second to last element, then add your last post. As others have pointed out, the array join function does this for you.
– Drew Reese
Nov 22 at 0:53












3 Answers
3






active

oldest

votes

















up vote
1
down vote



accepted










Your conditional expression if (i == 1) would only trigger on the second iteration of the loop since i will only equal 1 one time.



Try something like:



if (i < (facetResults.length - 1)) {
// only add OR if this isn't the last element of the array
filter = "'" + facetKey + "'" + ":'" + facetResults[i] + " OR";
return filter;
}


Here's your updated code:



Object.keys(facets).forEach(function(facetKey) {
if (facets[facetKey].length > 1) {
facetResults = facets[facetKey];
for (var i = 0; i < facetResults.length; i ++) {
if (i < (facetResults.length - 1)) {
filter = "'" + facetKey + "'" + ":'" + facetResults[i] + " OR";
return filter;
} else {
filter = "'" + facetKey + "'" + ":'" + facetResults[i];
}
}
} else {
filter = "'" + facetKey + "'" + ": " + facets[facetKey] + "'";
return filter;
}
});





share|improve this answer





















  • Thank you @Nunchy. That worked nicely
    – Peter Ayello Wright
    Nov 22 at 10:48


















up vote
2
down vote













You can do something like this with Object.entries and Array.reduce if you would like to get the final result in the form of an object:






const data = { field_country: ["England", "Netherlands", "India", "Italy"], field_continent: ["Europe"], field_group: ["Building", "People", "Landscape"] }

const result = Object.entries(data).reduce((r, [k, v]) => {
r[k] = v.join(' OR ')
return r
}, {})

console.log(result)





It is somewhat unclear what is the final format you need to result in but that should help you to get the idea. If ES6 is not an option you can convert this to:



const result = Object.entries(data).reduce(function(r, [k, v]) {
r[k] = v.join(' OR ')
return r
}, {})


So there are is no arrow function etc.



The idea is to get the arrays into the arrays of strings and use the Array.join to do the "replacement" for you via join(' OR ')






share|improve this answer



















  • 1




    Instead of v.toString().replace(/,/g, ' OR ') you can use join: v.join(' OR ').
    – slider
    Nov 22 at 0:29










  • Sure. Good call. Thanks.
    – Akrion
    Nov 22 at 0:30


















up vote
1
down vote













Here's the idea. In your code you are appending " or " at the end of your strings starting at index 0. I suggest you append it at the the beginning starting at index 1.



var somewords = ["ORANGE", "GREEN", "BLUE", "WHITE" ];
var retval = somewords[0];
for(var i = 1; i< somewords.length; i++)
{
retval += " or " + somewords[i];
}
console.log(retval);

//result is: ORANGE or GREEN or BLUE or WHITE





share|improve this answer



















  • 1




    Just set retval = some words[0];, start at index 1, and do just your else branch.
    – Drew Reese
    Nov 22 at 0:50










  • @DrewReese, it took me a moment to understand but you are so right! Changed it and I'll remember this comment!
    – blaze_125
    Nov 22 at 2:19











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


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53422104%2flooping-over-javascript-object-and-adding-string-to-end-if-not-the-last-item%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










Your conditional expression if (i == 1) would only trigger on the second iteration of the loop since i will only equal 1 one time.



Try something like:



if (i < (facetResults.length - 1)) {
// only add OR if this isn't the last element of the array
filter = "'" + facetKey + "'" + ":'" + facetResults[i] + " OR";
return filter;
}


Here's your updated code:



Object.keys(facets).forEach(function(facetKey) {
if (facets[facetKey].length > 1) {
facetResults = facets[facetKey];
for (var i = 0; i < facetResults.length; i ++) {
if (i < (facetResults.length - 1)) {
filter = "'" + facetKey + "'" + ":'" + facetResults[i] + " OR";
return filter;
} else {
filter = "'" + facetKey + "'" + ":'" + facetResults[i];
}
}
} else {
filter = "'" + facetKey + "'" + ": " + facets[facetKey] + "'";
return filter;
}
});





share|improve this answer





















  • Thank you @Nunchy. That worked nicely
    – Peter Ayello Wright
    Nov 22 at 10:48















up vote
1
down vote



accepted










Your conditional expression if (i == 1) would only trigger on the second iteration of the loop since i will only equal 1 one time.



Try something like:



if (i < (facetResults.length - 1)) {
// only add OR if this isn't the last element of the array
filter = "'" + facetKey + "'" + ":'" + facetResults[i] + " OR";
return filter;
}


Here's your updated code:



Object.keys(facets).forEach(function(facetKey) {
if (facets[facetKey].length > 1) {
facetResults = facets[facetKey];
for (var i = 0; i < facetResults.length; i ++) {
if (i < (facetResults.length - 1)) {
filter = "'" + facetKey + "'" + ":'" + facetResults[i] + " OR";
return filter;
} else {
filter = "'" + facetKey + "'" + ":'" + facetResults[i];
}
}
} else {
filter = "'" + facetKey + "'" + ": " + facets[facetKey] + "'";
return filter;
}
});





share|improve this answer





















  • Thank you @Nunchy. That worked nicely
    – Peter Ayello Wright
    Nov 22 at 10:48













up vote
1
down vote



accepted







up vote
1
down vote



accepted






Your conditional expression if (i == 1) would only trigger on the second iteration of the loop since i will only equal 1 one time.



Try something like:



if (i < (facetResults.length - 1)) {
// only add OR if this isn't the last element of the array
filter = "'" + facetKey + "'" + ":'" + facetResults[i] + " OR";
return filter;
}


Here's your updated code:



Object.keys(facets).forEach(function(facetKey) {
if (facets[facetKey].length > 1) {
facetResults = facets[facetKey];
for (var i = 0; i < facetResults.length; i ++) {
if (i < (facetResults.length - 1)) {
filter = "'" + facetKey + "'" + ":'" + facetResults[i] + " OR";
return filter;
} else {
filter = "'" + facetKey + "'" + ":'" + facetResults[i];
}
}
} else {
filter = "'" + facetKey + "'" + ": " + facets[facetKey] + "'";
return filter;
}
});





share|improve this answer












Your conditional expression if (i == 1) would only trigger on the second iteration of the loop since i will only equal 1 one time.



Try something like:



if (i < (facetResults.length - 1)) {
// only add OR if this isn't the last element of the array
filter = "'" + facetKey + "'" + ":'" + facetResults[i] + " OR";
return filter;
}


Here's your updated code:



Object.keys(facets).forEach(function(facetKey) {
if (facets[facetKey].length > 1) {
facetResults = facets[facetKey];
for (var i = 0; i < facetResults.length; i ++) {
if (i < (facetResults.length - 1)) {
filter = "'" + facetKey + "'" + ":'" + facetResults[i] + " OR";
return filter;
} else {
filter = "'" + facetKey + "'" + ":'" + facetResults[i];
}
}
} else {
filter = "'" + facetKey + "'" + ": " + facets[facetKey] + "'";
return filter;
}
});






share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 22 at 0:07









Nunchy

838410




838410












  • Thank you @Nunchy. That worked nicely
    – Peter Ayello Wright
    Nov 22 at 10:48


















  • Thank you @Nunchy. That worked nicely
    – Peter Ayello Wright
    Nov 22 at 10:48
















Thank you @Nunchy. That worked nicely
– Peter Ayello Wright
Nov 22 at 10:48




Thank you @Nunchy. That worked nicely
– Peter Ayello Wright
Nov 22 at 10:48












up vote
2
down vote













You can do something like this with Object.entries and Array.reduce if you would like to get the final result in the form of an object:






const data = { field_country: ["England", "Netherlands", "India", "Italy"], field_continent: ["Europe"], field_group: ["Building", "People", "Landscape"] }

const result = Object.entries(data).reduce((r, [k, v]) => {
r[k] = v.join(' OR ')
return r
}, {})

console.log(result)





It is somewhat unclear what is the final format you need to result in but that should help you to get the idea. If ES6 is not an option you can convert this to:



const result = Object.entries(data).reduce(function(r, [k, v]) {
r[k] = v.join(' OR ')
return r
}, {})


So there are is no arrow function etc.



The idea is to get the arrays into the arrays of strings and use the Array.join to do the "replacement" for you via join(' OR ')






share|improve this answer



















  • 1




    Instead of v.toString().replace(/,/g, ' OR ') you can use join: v.join(' OR ').
    – slider
    Nov 22 at 0:29










  • Sure. Good call. Thanks.
    – Akrion
    Nov 22 at 0:30















up vote
2
down vote













You can do something like this with Object.entries and Array.reduce if you would like to get the final result in the form of an object:






const data = { field_country: ["England", "Netherlands", "India", "Italy"], field_continent: ["Europe"], field_group: ["Building", "People", "Landscape"] }

const result = Object.entries(data).reduce((r, [k, v]) => {
r[k] = v.join(' OR ')
return r
}, {})

console.log(result)





It is somewhat unclear what is the final format you need to result in but that should help you to get the idea. If ES6 is not an option you can convert this to:



const result = Object.entries(data).reduce(function(r, [k, v]) {
r[k] = v.join(' OR ')
return r
}, {})


So there are is no arrow function etc.



The idea is to get the arrays into the arrays of strings and use the Array.join to do the "replacement" for you via join(' OR ')






share|improve this answer



















  • 1




    Instead of v.toString().replace(/,/g, ' OR ') you can use join: v.join(' OR ').
    – slider
    Nov 22 at 0:29










  • Sure. Good call. Thanks.
    – Akrion
    Nov 22 at 0:30













up vote
2
down vote










up vote
2
down vote









You can do something like this with Object.entries and Array.reduce if you would like to get the final result in the form of an object:






const data = { field_country: ["England", "Netherlands", "India", "Italy"], field_continent: ["Europe"], field_group: ["Building", "People", "Landscape"] }

const result = Object.entries(data).reduce((r, [k, v]) => {
r[k] = v.join(' OR ')
return r
}, {})

console.log(result)





It is somewhat unclear what is the final format you need to result in but that should help you to get the idea. If ES6 is not an option you can convert this to:



const result = Object.entries(data).reduce(function(r, [k, v]) {
r[k] = v.join(' OR ')
return r
}, {})


So there are is no arrow function etc.



The idea is to get the arrays into the arrays of strings and use the Array.join to do the "replacement" for you via join(' OR ')






share|improve this answer














You can do something like this with Object.entries and Array.reduce if you would like to get the final result in the form of an object:






const data = { field_country: ["England", "Netherlands", "India", "Italy"], field_continent: ["Europe"], field_group: ["Building", "People", "Landscape"] }

const result = Object.entries(data).reduce((r, [k, v]) => {
r[k] = v.join(' OR ')
return r
}, {})

console.log(result)





It is somewhat unclear what is the final format you need to result in but that should help you to get the idea. If ES6 is not an option you can convert this to:



const result = Object.entries(data).reduce(function(r, [k, v]) {
r[k] = v.join(' OR ')
return r
}, {})


So there are is no arrow function etc.



The idea is to get the arrays into the arrays of strings and use the Array.join to do the "replacement" for you via join(' OR ')






const data = { field_country: ["England", "Netherlands", "India", "Italy"], field_continent: ["Europe"], field_group: ["Building", "People", "Landscape"] }

const result = Object.entries(data).reduce((r, [k, v]) => {
r[k] = v.join(' OR ')
return r
}, {})

console.log(result)





const data = { field_country: ["England", "Netherlands", "India", "Italy"], field_continent: ["Europe"], field_group: ["Building", "People", "Landscape"] }

const result = Object.entries(data).reduce((r, [k, v]) => {
r[k] = v.join(' OR ')
return r
}, {})

console.log(result)






share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 22 at 0:36

























answered Nov 22 at 0:07









Akrion

7,52611222




7,52611222








  • 1




    Instead of v.toString().replace(/,/g, ' OR ') you can use join: v.join(' OR ').
    – slider
    Nov 22 at 0:29










  • Sure. Good call. Thanks.
    – Akrion
    Nov 22 at 0:30














  • 1




    Instead of v.toString().replace(/,/g, ' OR ') you can use join: v.join(' OR ').
    – slider
    Nov 22 at 0:29










  • Sure. Good call. Thanks.
    – Akrion
    Nov 22 at 0:30








1




1




Instead of v.toString().replace(/,/g, ' OR ') you can use join: v.join(' OR ').
– slider
Nov 22 at 0:29




Instead of v.toString().replace(/,/g, ' OR ') you can use join: v.join(' OR ').
– slider
Nov 22 at 0:29












Sure. Good call. Thanks.
– Akrion
Nov 22 at 0:30




Sure. Good call. Thanks.
– Akrion
Nov 22 at 0:30










up vote
1
down vote













Here's the idea. In your code you are appending " or " at the end of your strings starting at index 0. I suggest you append it at the the beginning starting at index 1.



var somewords = ["ORANGE", "GREEN", "BLUE", "WHITE" ];
var retval = somewords[0];
for(var i = 1; i< somewords.length; i++)
{
retval += " or " + somewords[i];
}
console.log(retval);

//result is: ORANGE or GREEN or BLUE or WHITE





share|improve this answer



















  • 1




    Just set retval = some words[0];, start at index 1, and do just your else branch.
    – Drew Reese
    Nov 22 at 0:50










  • @DrewReese, it took me a moment to understand but you are so right! Changed it and I'll remember this comment!
    – blaze_125
    Nov 22 at 2:19















up vote
1
down vote













Here's the idea. In your code you are appending " or " at the end of your strings starting at index 0. I suggest you append it at the the beginning starting at index 1.



var somewords = ["ORANGE", "GREEN", "BLUE", "WHITE" ];
var retval = somewords[0];
for(var i = 1; i< somewords.length; i++)
{
retval += " or " + somewords[i];
}
console.log(retval);

//result is: ORANGE or GREEN or BLUE or WHITE





share|improve this answer



















  • 1




    Just set retval = some words[0];, start at index 1, and do just your else branch.
    – Drew Reese
    Nov 22 at 0:50










  • @DrewReese, it took me a moment to understand but you are so right! Changed it and I'll remember this comment!
    – blaze_125
    Nov 22 at 2:19













up vote
1
down vote










up vote
1
down vote









Here's the idea. In your code you are appending " or " at the end of your strings starting at index 0. I suggest you append it at the the beginning starting at index 1.



var somewords = ["ORANGE", "GREEN", "BLUE", "WHITE" ];
var retval = somewords[0];
for(var i = 1; i< somewords.length; i++)
{
retval += " or " + somewords[i];
}
console.log(retval);

//result is: ORANGE or GREEN or BLUE or WHITE





share|improve this answer














Here's the idea. In your code you are appending " or " at the end of your strings starting at index 0. I suggest you append it at the the beginning starting at index 1.



var somewords = ["ORANGE", "GREEN", "BLUE", "WHITE" ];
var retval = somewords[0];
for(var i = 1; i< somewords.length; i++)
{
retval += " or " + somewords[i];
}
console.log(retval);

//result is: ORANGE or GREEN or BLUE or WHITE






share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 22 at 2:17

























answered Nov 22 at 0:44









blaze_125

1,8771416




1,8771416








  • 1




    Just set retval = some words[0];, start at index 1, and do just your else branch.
    – Drew Reese
    Nov 22 at 0:50










  • @DrewReese, it took me a moment to understand but you are so right! Changed it and I'll remember this comment!
    – blaze_125
    Nov 22 at 2:19














  • 1




    Just set retval = some words[0];, start at index 1, and do just your else branch.
    – Drew Reese
    Nov 22 at 0:50










  • @DrewReese, it took me a moment to understand but you are so right! Changed it and I'll remember this comment!
    – blaze_125
    Nov 22 at 2:19








1




1




Just set retval = some words[0];, start at index 1, and do just your else branch.
– Drew Reese
Nov 22 at 0:50




Just set retval = some words[0];, start at index 1, and do just your else branch.
– Drew Reese
Nov 22 at 0:50












@DrewReese, it took me a moment to understand but you are so right! Changed it and I'll remember this comment!
– blaze_125
Nov 22 at 2:19




@DrewReese, it took me a moment to understand but you are so right! Changed it and I'll remember this comment!
– blaze_125
Nov 22 at 2:19


















 

draft saved


draft discarded



















































 


draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53422104%2flooping-over-javascript-object-and-adding-string-to-end-if-not-the-last-item%23new-answer', 'question_page');
}
);

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







Popular posts from this blog

What visual should I use to simply compare current year value vs last year in Power BI desktop

How to ignore python UserWarning in pytest?

Alexandru Averescu