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I am trying to split the string on "." Except when a part of the string is in [ ] then I just want to return what is inside the brackets.

I have the following line of code:

"blah.blah[http://blah.blah.com/blah/blah#]".split(("(\\.|\\[(?=.*\\]))")

This returns:

 [ "blah", "blah", "http:blah", "blah", "com/blah/blah#]" ]

If instead I try:

"blah.blah[http://blah.blah.com/blah/blah#]".split(("(\\.|\\[(?:.*\\]))")

I get:

["blah", "blah"]

I'm not sure how I need to define my non capturing group so that it will split on the first [ but not capture anything after up to and including the ]

Just to clarify the array I am expecting back is

["blah", "blah", "http://blah.blah.com/blah/blah#"]
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1 Answer 1

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2

To do that, the best option is to use the "find" method instead of split with this pattern:

(?<=\\[)[^\\]]*(?=\\])|[^\\][.]+

Note that the order of the alternatives is important because the first win. So (?<=\\[)[^\\]]*(?=\\]) must be before [^\\][.]+

demo

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7 Comments

The find here would match the "." and return them. The goal is to get the strings between them.
@tiger13cubed: refresh your browser.
Thanks for your help Casmir :) Not the solution I had in mind, but it does the trick. Any way you can explain what I was doing wrong with my \[(?=.*\]) regex?
@tiger13cubed: It's simple, a lookahead doesn't consume characters, it's only a check. So all the dots between square brackets can be matched. A pattern you can use with the split method is \\.(?![^[]*])|[\\]\\[] but it is far from an efficient way.
Where can I find the "find" method?
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