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.
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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.