Javascript Regex matching [closed]

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Trying to match x occurance of a string using javascript regex :

var regex = new RegExp("(?:\[([0-9]+\]{1}$)",'g')
string.replace(regex, '[3]');

String source :

sometext[2][markers][2][title]

Expected results :

sometext[2][markers][3][title]

What is wrong with the regex?

Edit : Changed the regex to : regex = new RegExp("(?:[([0-9]+]{" + Filter + "})",'g');

The results, javascript is changing all the [x]: sometext[3][markers][3][title]

When you build regex with the RegExp constructor you need to double escape special characters. See the MDN developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/… — elclanrs
– elclanrs, Commented Apr 14, 2014 at 2:03
possible duplicate of JavaScript Regex, where to use escape characters? — elclanrs
– elclanrs, Commented Apr 14, 2014 at 2:05
There are multiple issues with your regex, not just backslash escape issues. For example, the parentheses don't match. — Qantas 94 Heavy
– Qantas 94 Heavy, Commented Apr 14, 2014 at 2:05

assembly_wizard · Accepted Answer · 2014-04-14 22:57:27Z

1

The RegEx: "\[\d+\]"

\[    First, an open square bracket (escaped)
\d    This means a digit (this is the same as [0-9])
+     The plus means there can be as many digits as you wish
\]    Ending it up with an escaped close bracket.

Note: We can use /\[\d+\]/g as a replacement for new RegEx(regex) and another benefit is that we don't have to double escape the characters (escaping the character in RegEx, and also escaping the backslash so JS will treat it as a normal backslash).

Change replaceThe to replace the nth occurrence.

var string = 'sometext[2][markers][2][title]';
var replaceThe = 2; // Replace the -second- occurrence
var i = 0;
string = string.replace(/\[\d+\]/g,
             function(m) { return (replaceThe == (++i) ? '[3]' : m); })

The last line uses an overload of the string.replace function which accepts a function. This function will be called for each match and the m parameter will be set to the match itself. The function is expected to return a replacement for the match. This way we can wait until the nth occurrence until we actually replace the match, otherwise we return the match we received, and that way it won't be replaced.

edited Apr 14, 2014 at 22:57

answered Apr 14, 2014 at 2:12

assembly_wizard

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4 Comments

Alan Moore Over a year ago

Good, but be aware that you're creating the regex using a regex literal. If you write it as a string literal like the OP did, you should escape the backslashes as per the comment above.

assembly_wizard Over a year ago

Actually, this is exactly why I chose to create it literally

Alan Moore Over a year ago

So I assumed, but I originally thought the "\[\d+\]" represented a JavaScript string literal, and you were making the same mistake as the OP. Now I see you were using it to illustrate the error. you should add some text to explain that. An answer isn't really an answer if doesn't explain (in words, not in source code) what the problem was.

assembly_wizard Over a year ago

I agree, I was in a rush when I gave the answer. Added some explaining.

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