Splitting By " | " Not Between Square Brackets

I need to split a string, but ignoring stuff between square brackets. You can imagine this akin to ignoring stuff within quotes. I came across a solution for quotes; (\|)(?=(?:[^']

Solution 1:

This is the adapted version of the "" regex you posted:

(\|)(?=(?:[^\]]|\[[^\]]*\])*$)
(\|)(?=(?:[^ "]| "[^ "]* ")*$) <- "" version for comparison

You replace the 2nd " with \[ and the 1st and 3rd with \]

Working on RegExr

Solution 2:

You should better use String#match:

s='one|two[ignore|ignore]|three';
m = s.match(/[^|\[]+\[.*?\]|[^|]+/g);
//=> ["one", "two[ignore|ignore]", "three"]

Solution 3:

Yet another solution:

"one|two[ignore|ignore]|three".split( /\|(?![^\[]*\])/ )

It uses negative lookahead, unfortunately lookbehind is not supported in JS.

Solution 4:

with the split method you can use this pattern since the capturing group is displayed with results:

\|((?:[^[|]+|\[[^\]]+])*)\|