Splitting by “ | ” not between square brackets

Question

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;

(\\|)(?=(?:[^"]|"[^"]*")*$)

So;

one|two"ignore|ignore"|three

would be;

one
two"ignore|ignore"
three

However, I am struggling to adapt this to ignore [] instead of ". Partly its down to there being two characters, not one. Partly its down to [] needing to be escaped, and partly its down to me being pretty absolutely awful at regexs.

My goal is to get;

So;

one|two[ignore|ignore]|three

split to;

 one
 two[ignore|ignore]
 three

I tried figuring it out myself but I got into a right mess; (\\|)(?=(?:[^\\[|\\]]|\\|][^\\[|\\]]*\\[|\\])*$) I am seeing brackets and lines everywhere now.

Help!

Answer 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

Answer 2

使用split方法，您可以使用此模式，因为捕获组显示结果：

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

Answer 3

You should better use String#match :

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

Answer 4

Yet another solution:

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

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