How can I put [] (square brackets) in RegExp javascript?

I'm trying this:

str = "bla [bla]";
str = str.replace(/\\[\\]/g,"");
console.log(str);

And the replace doesn't work, what am I doing wrong?

UPDATE: I'm trying to remove any square brackets in the string, what's weird is that if I do

replace(/\[/g, '')
replace(/\]/g, '')

it works, but
replace(/\[\]/g, '');
doesn't.

Answers:

Answer

Two backslashes produces a single backslash, so you're searching for "a backslash, followed by a character class consisting of a 1 or a right bracket, and then you're missing an closing bracket.

Try

str.replace(/\[1\]/g, '');
Answer

What exactly are you trying to match?

If you don't escape the brackets, they are considered character classes. This:

/[1\\]/

Matches either a 1 or a backslash. You may want to escape them with one backslash only:

/\[1\]/

But this won't match either, as you don't have a [1] in your string.

Answer

I stumbled on this question while dealing with square bracket escaping within a character class that was designed for use with password validation requiring the presence of special characters.

Note the double escaping:

var regex = new RegExp('[\\]]');

As @rudu mentions, this expression is within a string so it must be double escaped. Note that the quoting type (single/double) is not relevant here.

Here is an example of using square brackets in a character class that tests for all the special characters found on my keyboard:

var regex = new RegExp('[-,_,\',",;,:,!,@,#,$,%,^,&,*,(,),[,\\],\?,{,},|,+,=,<,>,~,`,\\\\,\,,\/,.]', 'g')
Answer

It should be:

str = str.replace(/\[.*?\]/g,"");

You don't need double backslashes (\) because it's not a string but a regex statement, if you build the regex from a string you do need the double backslashes ;).

It was also literally interpreting the 1 (which wasn't matching). Using .* says any value between the square brackets.

The new RegExp string build version would be:

str=str.replace(new RegExp("\\[.*?\\]","g"),"");

UPDATE: To remove square brackets only:

str = str.replace(/\[(.*?)\]/g,"$1");

Your above code isn't working, because it's trying to match "[]" (sequentially without anything allowed between). We can get around this by non-greedy group-matching ((.*?)) what's between the square brackets, and using a backreference ($1) for the replacement.

UPDATE 2: To remove multiple square brackets

str = str.replace(/\[+(.*?)\]+/g,"$1");
// bla [bla] [[blaa]] -> bla bla blaa
// bla [bla] [[[blaa] -> bla bla blaa

Note this doesn't match open/close quantities, simply removes all sequential opens and closes. Also if the sequential brackets have separators (spaces etc) it won't match.

Answer

You have to escape the bracket, like \[ and \]. Check out http://regexpal.com/. It's pretty useful :)

To replace all brackets in a string, this should do the job:

str.replace(/\[|\]/g,'');

I hope this helps.
Hristo

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