正则表达式，如何分割|并且在\之前避免分裂

Question

I have the next text

 aaa|bbbb|cccc|dddd\|eeee|ffff

and i want to split by | and excluding when | is preceded by \\ and obtain

aaa

bbbb

cccc

dddd\\|eeee

ffff

Thanks.

ps : i tried using some regexp generator (for example http://txt2re.com/ ) but frankly regexp is anything but friendly.

update: finally i give up. Regexp is not fast (i did a benchmark), neither is clear (in comparison with a function that everybody can follow), then i skip it and now i am using real code.

Answer 1

This should do it:

(?<!\\\\)\\|

If you want to allow backslash-escaped backslashes, you can use:

(?<!(?<!\\\\)\\\\)\\|

So given the string aaa|bbbb|cccc|dddd\\|eeee\\\\|ffff , the split would be:

    aaa
    bbbb
    cccc
    dddd|eeee\*
    ffff

* Or dddd\\|eeee\\\\ if you're not stripping escape-backslashes for some reason.

Edit: not familiar with Java regular expression flavor, added escapes per ratchet freak's comment.

Answer 2

Tried to add this as a comment to eyelidlessness's answer, but don't know how to format it there...

Anyhow, eyelidlessness answer looks correct to me:

    String str = "aaa|bbbb|cccc|dddd\\|eeee|ffff";
    String[] tokens = str.split("(?<!\\\\)\\|");
    System.out.println(Arrays.toString(tokens));    

which prints:

[aaa, bbbb, cccc, dddd\|eeee, ffff]

Answer 3

Don't use split() for this. (You could if Java supported indefinite repetition inside lookbehind assertions. But it doesn't.)

Better collect all the matches between | s:

List<String> matchList = new ArrayList<String>();
Pattern regex = Pattern.compile("(?:\\\\.|[^\\\\|])*");
Matcher regexMatcher = regex.matcher(subjectString);
while (regexMatcher.find()) {
    matchList.add(regexMatcher.group());
}

This correctly splits aaa|bbbb\\\\|cccc|dddd\\|eeee|ffff\\\\\\|ggg\\\\\\\\|hhhh into

aaa
bbbb\\
cccc
dddd\|eeee
ffff\\\|ggg\\\\
hhhh