How to find and replace every nth character occurrence using regex? - javascript

How can I find and replace for example every third occurrence of 'x' character in string with regular expression?
There is similar question but it addresses every fourth character but I need every "nth" occurrence of specific character in string.
I'm using JavaScript replace() method, this below example will change every x character to y:
replace(/\x/g, 'y')
But how can I write regex for question above?
Some random text for example, and expected result:
I amx nexver axt hoxme on Sxundxaxs.
to:
I amx nexver ayt hoxme on Sxundyaxs.
I can use JS for loop with split() and join() but solution with replace() method seams more elegant.

You can use:
str = str.replace(/((?:[^x]*x){2}[^x]*)x/g, '$1y');
So if you are searching for every 3rd occurrence then use n-1 between {} in above regex.
To build it dynamically use:
let str = 'I amx nexver axt hoxme on Sxundxaxs';
let n = 3;
let ch = 'x';
let regex = new RegExp("((?:[^" +ch+ "]*" +ch+ "){" + (n-1) + "}[^" +ch+ "]*)" +ch, "g");
str = str.replace(regex, '$1y');
console.log(str);
//=> I amx nexver ayt hoxme on Sxundyaxs
RegEx Demo

Related

Getting the content between two characters

So I have this (example) string: 1234VAR239582358X
And I want to get what's in between VAR and X. I can easily replace it using .replace(/VAR.*X/, "replacement");
But, how would I get the /VAR.*X/as a variable?
I think what you are looking for might be
string.match(/VAR(.*)X/)[1]
The brackets around the .* mark a group. Those groups are returned inside the Array that match creates :)
If you want to only replace what's in between "VAR" and "X" it would be
string.replace(/VAR(.*)X/, "VAR" + "replacement" + "X");
Or more generic:
string.replace(/(VAR).*(X)/, "$1replacement$2");
You can try use the RegExp class, new RegExp(`${VAR}.*X`)
You can store it as variable like this,
const pattern = "VAR.*X";
const reg = new RegExp(pattern);
Then use,
.replace(reg, "replacement");
If you
want to get what's in between VAR and X
then using .* would do the job for the given example string.
But note that is will match until the end of the string, and then backtrack to the first occurrence of X it can match, being the last occurrence of the X char in the string and possible match too much.
If you want to match only the digits, you can match 1+ digits in a capture group using VAR(\d+)X
const regex = /VAR(\d+)X/;
const str = "1234VAR239582358X";
const m = str.match(regex);
if (m) {
let myVariable = m[1];
console.log(myVariable);
}
Or you can match until the first occurrence of an X char using a negated character class VAR([^\r\nX]+)X
const regex = /VAR([^\r\nX]+)X/;
const str = "1234VAR239582358X";
const m = str.match(regex);
if (m) {
let myVariable = m[1];
console.log(myVariable);
}

Javascript string replace certain characters

I have this string:
var s = '/channels/mtb/videos?page=2&per_page=100&fields=uri%2Cname%2Cdescription%2Cduration%2Cwidth%2Cheight%2Cprivacy%2Cpictures.sizes&sort=date&direction=asc&filter=embeddable&filter_embeddable=true'
I want to repace per_page number (in this case 100, but it can be any number from 1-100, maybe more?)
I can select first part of the string with:
var s1 = s.substr(0, s.lastIndexOf('per_page=')+9)
which give me:
/channels/mtb/videos?page=2&per_page=
but how would I select next '&' after that so I can replace number occurrence?
dont assume same order of parameters!
You can use following regex to replace the content you want.
regex:- /per_page=[\d]*/g(this is only for your requirement)
var new_no=12; //change 100 to 12
var x='/channels/mtb/videos?page=2&per_page=100&fields=uri%2Cname%2Cdescription%2Cduration%2Cwidth%2Cheight%2Cprivacy%2Cpictures.sizes&sort=date&direction=asc&filter=embeddable&filter_embeddable=true';
var y=x.replace(/per_page=[\d]*/g,'per_page='+new_no);
console.log(y);
Explanation:-
/per_page=[\d]*/g
/ ----> is for regex pattern(it inform that from next character onward whatever it encounter will be regex pattern)
per_page= ----> try to find 'per_page=' in string
[\d]* ----> match 0 or more digit (it match until non digit encounter)
/g ---->/ to indicate end of regex pattern and 'g' is for global means find in all string(not only first occurrence)
Use replace with a regular expression to find the numbers after the text per_page=. Like this:
s.replace(/per_page=\d+/,"per_page=" + 33)
Replace the 33 with the number you want.
Result:
"/channels/mtb/videos?page=2&per_page=33&fields=uri%2Cname%2Cdescription%2Cduration%2Cwidth%2Cheight%2Cprivacy%2Cpictures.sizes&sort=date&direction=asc&filter=embeddable&filter_embeddable=true"
Start with the index from the lastIndexOf-per_page instead of 0.
Get the index of the first & and create a substr s2 to the end.
Then concat s1 + nr + s2.
I would not use regex, because it is much slower for this simple stuff.
With Array.filter you can do this, where one split the text into key/value pairs, and filter out the one that starts with per_page=.
Stack snippet
var s = '/channels/mtb/videos?page=2&per_page=100&fields=uri%2Cname%2Cdescription%2Cduration%2Cwidth%2Cheight%2Cprivacy%2Cpictures.sizes&sort=date&direction=asc&filter=embeddable&filter_embeddable=true'
var kv_pairs = s.split('&');
var s2 = s.replace((kv_pairs.filter(w => w.startsWith('per_page=')))[0],'per_page=' + 123);
//console.log(s2);
var matches = /(.*\bper_page=)(\d+)(.*)/;
if (matches) {
s = matches[0] + newValue + matches[2];
}

Regex Match Punctuation Space but Retain Punctuation

I have a large paragraph string which I'm trying to split into sentences using JavaScript's .split() method. I need a regex that will match a period or a question-mark [?.] followed by a space. However, I need to retain the period/question-mark in the resulting array. How can I do this without positive lookbehinds in JS?
Edit: Example input:
"This is sentence 1. This is sentence 2? This is sentence 3."
Example output:
["This is sentence 1.", "This is sentence 2?", "This is sentence 3."]
This regex will work
([^?.]+[?.])(?:\s|$)
Regex Demo
JS Demo
Ideone Demo
var str = 'This is sentence 1. This is sentence 2? This is sentence 3.';
var regex = /([^?.]+[?.])(?:\s|$)/gm;
var m;
while ((m = regex.exec(str)) !== null) {
document.writeln(m[1] + '<br>');
}
Forget about split(). You want match()
var text = "This is an example paragragh. Oh and it has a question? Ok it's followed by some other random stuff. Bye.";
var matches = text.match(/[\w\s'\";\(\)\,]+(\.|\?)(\s|$)/g);
alert(matches);
The generated matches array contains each sentence:
Array[4]
0:"This is an example paragragh. "
1:"Oh and it has a question? "
2:"Ok it's followed by some other random stuff. "
4:"Bye. "
Here is the fiddle with it for further testing: https://jsfiddle.net/uds4cww3/
Edited to match end of line too.
May be this one validates your array items
\b.*?[?\.](?=\s|$)
Debuggex Demo
This is tacky, but it works:
var breakIntoSentences = function(s) {
var l = [];
s.replace(/[^.?]+.?/g, a => l.push(a));
return l;
}
breakIntoSentences("how? who cares.")
["how?", " who cares."]
(Really how it works: the RE matches a string of not-punctuation, followed by something. Since the match is greedy, that something is either punctuation or the end-of-string.)
This will only capture the first in a series of punctuation, so breakIntoSentences("how???? who cares...") also returns ["how?", " who cares."]. If you want to capture all the punctuation, use /[^.?]+[.?]*/g as the RE instead.
Edit: Hahaha: Wavvves teaches me about match(), which is what the replace/push does. You learn something knew every goddamn day.
In its minimal form, supporting three punctuation marks, and using ES6 syntax, we get:
const breakIntoSentences = s => s.match(/[^.?,]+[.?,]*/g)
I guess .match will do it:
(?:\s?)(.*?[.?])
I.e.:
sentence = "This is sentence 1. This is sentence 2? This is sentence 3.";
result = sentence.match(/(?:\s?)(.*?[.?])/ig);
for (var i = 0; i < result.length; i++) {
document.write(result[i]+"<br>");
}

Remove (n)th space from string in JavaScript

I am trying to remove some spaces from a few dynamically generated strings. Which space I remove depends on the length of the string. The strings change all the time so in order to know how many spaces there are, I iterate over the string and increment a variable every time the iteration encounters a space. I can already remove all of a specific type of character with str.replace(' ',''); where 'str' is the name of my string, but I only need to remove a specific occurrence of a space, not all the spaces. So let's say my string is
var str = "Hello, this is a test.";
How can I remove ONLY the space after the word "is"? (Assuming that the next string will be different so I can't just write str.replace('is ','is'); because the word "is" might not be in the next string).
I checked documentation on .replace, but there are no other parameters that it accepts so I can't tell it just to replace the nth instance of a space.
If you want to go by indexes of the spaces:
var str = 'Hello, this is a test.';
function replace(str, indexes){
return str.split(' ').reduce(function(prev, curr, i){
var separator = ~indexes.indexOf(i) ? '' : ' ';
return prev + separator + curr;
});
}
console.log(replace(str, [2,3]));
http://jsfiddle.net/96Lvpcew/1/
As it is easy for you to get the index of the space (as you are iterating over the string) , you can create a new string without the space by doing:
str = str.substr(0, index)+ str.substr(index);
where index is the index of the space you want to remove.
I came up with this for unknown indices
function removeNthSpace(str, n) {
var spacelessArray = str.split(' ');
return spacelessArray
.slice(0, n - 1) // left prefix part may be '', saves spaces
.concat([spacelessArray.slice(n - 1, n + 1).join('')]) // middle part: the one without the space
.concat(spacelessArray.slice(n + 1)).join(' '); // right part, saves spaces
}
Do you know which space you want to remove because of word count or chars count?
If char count, you can Rafaels Cardoso's answer,
If word count you can split them with space and join however you want:
var wordArray = str.split(" ");
var newStr = "";
wordIndex = 3; // or whatever you want
for (i; i<wordArray.length; i++) {
newStr+=wordArray[i];
if (i!=wordIndex) {
newStr+=' ';
}
}
I think your best bet is to split the string into an array based on placement of spaces in the string, splice off the space you don't want, and rejoin the array into a string.
Check this out:
var x = "Hello, this is a test.";
var n = 3; // we want to remove the third space
var arr = x.split(/([ ])/); // copy to an array based on space placement
// arr: ["Hello,"," ","this"," ","is"," ","a"," ","test."]
arr.splice(n*2-1,1); // Remove the third space
x = arr.join("");
alert(x); // "Hello, this isa test."
Further Notes
The first thing to note is that str.replace(' ',''); will actually only replace the first instance of a space character. String.replace() also accepts a regular expression as the first parameter, which you'll want to use for more complex replacements.
To actually replace all spaces in the string, you could do str.replace(/ /g,""); and to replace all whitespace (including spaces, tabs, and newlines), you could do str.replace(/\s/g,"");
To fiddle around with different regular expressions and see what they mean, I recommend using http://www.regexr.com
A lot of the functions on the JavaScript String object that seem to take strings as parameters can also take regular expressions, including .split() and .search().

How to split the values in JavaScript

I have to split the values using JavaScript and want to find the last occuring slash / from a string and replace the contents after the last slash / For example:
var word = "www.abc/def/ghf/ijk/**default.aspx**";
should become
var word ="www.abc/def/ghf/ijk/**replacement**";
The number of slashes may vary each time.
Try using regexp:
"www.abc/def/ghf/ijk/default.aspx".replace(/\/[^\/]+$/, "/replacement");
An alternative without regular expression (I just remembered lastIndexOf() method)
var word = "www.abc/def/ghf/ijk/default.aspx";
word = word.substring(0, word.lastIndexOf("/")) + "/replacement";
You can array split on '/', then pop the last element off the array, and rejoin.
word = word.split('/');
word.pop();
word = word.join('/') + replacement;
How about the KISS principle?
var word = "www.abc/def/ghf/ijk/default.aspx";
word = word.substring(0, word.lastIndexOf("/")) + "/replacement";
What about using a combination of the join() and split() functions?
var word = "www.abc/def/ghf/ijk/default.aspx";
// split the word using a `/` as a delimiter
wordParts = word.split('/');
// replace the last element of the array
wordParts[wordParts.length-1] = 'replacement';
// join the array back to a string.
var finalWord = wordParts.join('/');
The number of slashes doesn't matter here because all that is done is to split the string at every instance of the delimiter (in this case a slash).
Here is a working demo
Use regexp or arrays, something like:
[].splice.call(word = word.split('/'), -1, 1, 'replacement');
word = word.join('/');

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