Delete the substring in the string?

Question

String view:

var str = "some text [[[[test 1]] test 2]] other text";

How to remove everything in square brackets, including the brackets themselves?

My decision:

 new_str = str.replace(/\[.*\]\].*\]\]/gi,"");

Doesn't quite fit, what's wrong?

updated as advised:

 var str = "some text [[[[test 1]] test 2]] other text"; new_str = str.replace(/\[\[(.)+\[\[(.)+\]\] \]\]/i,''); alert(new_str);

As a result, the entire line returns.

Accepted Answer · 2012-12-27T18:28:51

Well, you have only one open bracket on the left, and there are two.

 REGEX: #\[\[(.)+\[\[(.)+\]\] \]\]#i

Should work.

Or + replace with * if there may be no characters.

For the first example it works, and this is for the second:

 /\[\[\[\[(.)+\]\](.)+\]\]/gi

Your code:

 var str = "some text [[[[test 1]] test 2]] other text"; new_str = str.replace(/\[\[\[\[(.)+\]\](.)+\]\]/gi,''); alert(new_str);

You can even like this:

 /\[.*\]/gi

Add the g modifier and the face characters at the beginning and at the end of the regular expression.

Answer 2 · 2012-12-27T18:42:16

It depends on what you need. for example

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

As @GenchiK correctly noted in the comments, you may just need

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

if the brackets are not necessarily double.

To avoid problems with the translation of lines, note that the dot does not match the line break . So what you need is this :

 str = str.replace(/\[[\s\S]*\]/gm, "");

Since the vehicle does not say what it is he needs, perhaps a more specific expression is more correct.

2 answers 2