How to replace part of a string using regex

Question

i need to replace a part of a string in Javascript

The following example should clarify what i mean

var str = "asd[595442/A][30327][0]";
var strToReplace = "30333";

var strDesiredResult = "asd[595442/A][30333][0]";

Basically it means the second area within the brackets should get replaced with another string

How to do that?

What i did so far is something like this :

var str = "asd[595442/A][30327][0]";
var regex = /asd\[(.*)\]\[(.*)\]\[(.*)\]/;
var arrMatches = regex.exec(str);

The string appears in arrMatches[2] correctly, and i could replace this. But what happens if in arrMatches[1] is the same string ?

Because it should only replace the value in the second bracket area.

and again a downvote on a question of mine in a short time - i've asked a total of 5 questions here on this platform and everyone of them got downvoted in the last couple days, at least provide an explanation ... — Atural
– Atural, Commented Mar 5, 2018 at 7:30
I think you were downvoted because your question is not very clear. — Antoine Colson
– Antoine Colson, Commented May 6, 2018 at 19:34

Wiktor Stribiżew · Accepted Answer · 2017-04-28 09:23:33Z

22

You may use a regex that will match the first [....] followed with [ and capture that part into a group (that you will be able to refer to via a backreference), and then match 1+ chars other than ] to replace them with your replacement:

var str = "asd[595442/A][30327][0]";
var strToReplace = "30333";
console.log(str.replace(/(\[[^\]]*]\[)[^\]]*/, "$1" + strToReplace));

var strDesiredResult = "asd[595442/A][30333][0]";
console.log(strDesiredResult);

The /(\[[^\]]*]\[)[^\]]*/ has no gmodifier, it will be looking for one match only.

Since regex engine searches a string for a match from left to right, you will get the first match from the left.

The \[[^\]]*]\[ matches [, then any 0+ chars other than ] and then ][. The (...) forms a capturing group #1, it will remember the value that you will be able to get into the replacement with $1 backreference. [^\]]* matches 0+ chars other than ] and this will be replaced.

Details:

( - a capturing group start
- \[ - a literal [ symbol (if unescaped, it starts a character class)
- [^\]]* - a negated character class that matches zero or more (due to the * quantifier)
- ] - a literal ] (outside a character class, it does not have to be escaped)
- \[ - a literal [
) - end of capturing group #1 (its value can be accessed with $1 backreference from the replacement pattern)
[^\]]* - 0+ (as the * quantifier matches zero or more occurrences, replace with + if you need to only match where there is 1 or more occurrences) chars other than ] (inside a character class in JS regex, ] must be escaped in any position).

edited Apr 28, 2017 at 9:23

answered Apr 28, 2017 at 8:50

Wiktor Stribiżew

631k41 gold badges502 silver badges632 bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

Atural Over a year ago

this is really sick ;)it looks like it works and beside your explanation i've no idea why this works, but i'll try to understand this for future things - thx dude

Wiktor Stribiżew Over a year ago

Please let me know what is unclear. The point I was making is that you need to rely on a [^...] construct, a negated character class that matches any char but those defined in the class. If you use .* or even .*?, the . can match [ and ], and overmatch, or result in an incorrect match.

Atural Over a year ago

the reason why i said i've no idea whats going on is because i'm absolutely inexperienced in this kind of regex world - but i'm certainly willing to study this masterpiece ;)

Wiktor Stribiżew Over a year ago

I will add a full pattern explanation in a couple of minutes.

iboss · Accepted Answer · 2017-04-28 08:45:13Z

1

There are many ways to do this. One possible pattern is

str.replace(/^(.+)(\[.+\])(\[.+\])(\[.+\])$/, `$1$2[${strToReplace}]$4`)

You can see that $<number> is referred to captured string from regex (string groups in parentheses). We can refer to those and rearrange it however we want.

answered Apr 28, 2017 at 8:45

iboss

4966 silver badges18 bronze badges

2 Comments

Wiktor Stribiżew Over a year ago

This won't help if there are more than 3 [...] substrings in the input string.

iboss Over a year ago

Then you can use this pattern instead /^(.+)(\[.+\])(\[.+\])(.*)$/. That's why I said there are many ways. It depends on the context.

yivo · Accepted Answer · 2017-04-28 08:58:46Z

1

Use this pattern:

'asd[595442/A][30327][0]'.replace(/^(asd\[[^\[\]]+\]\[)([^\[\]]+)(\]\[0\])$/, '$130333$3')

Test here

^            - match beginning of string
first group  - match "asd[", any chars except [ and ], "]["
second group - match any chars except [ and ]
third group  - match exactly: "][0]"
$            - match end of string

edited Apr 28, 2017 at 8:58

answered Apr 28, 2017 at 8:52

yivo

3,6421 gold badge20 silver badges24 bronze badges

Comments

Aruna · Accepted Answer · 2017-04-28 08:49:59Z

0

You can use Regular Expression like this /\[[0-9]+\]/ as below.

var str = "asd[595442/A][30327][0]";
var strToReplace = "30333";

var strDesiredResult = str.replace(/\[[0-9]+\]/, '[' + strToReplace + ']'); 

console.log(strDesiredResult);  //"asd[595442/A][30333][0]";

answered Apr 28, 2017 at 8:49

Aruna

12k3 gold badges31 silver badges42 bronze badges

Collectives™ on Stack Overflow

How to replace part of a string using regex

4 Answers 4

4 Comments

2 Comments

Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

4 Comments

2 Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related