Regular expression to get characters before brackets or comma

Question

I'm pulling my hair out a bit with this.

Say I have a string 7f8hd::;;8843fdj fls "": ] fjisla;vofje]]} fd)fds,f,f

I want to now extract this 7f8hd::;;8843fdj fls "": from the string based on the premise that the string ends with either a } or ] or , or ) but all those characters could be present I only need the first one.

I have tried without success to create a regular expression with a Matcher and Pattern class but I just can't seem to get it right.

The best I could come up with is below but my reg exp just doesn't seem to work like I think it should.

String line = "7f8hd::;;8843fdj fls "": ] fjisla;vofje]]} fd)fds,f,f";
Matcher m = Pattern.compile("(.*?)\\}|(.*?)\\]|(.*?)\\)|(.*?),").matcher(line);
while (matcher.find()) {
    System.out.println(matcher.group());
}

I'm clearly not understanding reg exp correctly. Any help would be great.

Answer 1

^[^\]}),]*

matches from the start of the string until (but excluding) the first ] , } , ) or , .

In Java:

Pattern regex = Pattern.compile("^[^\\]}),]*");
Matcher regexMatcher = regex.matcher(line);
if (regexMatcher.find()) {
    System.out.println(regexMatcher.group());
}

(You can actually remove the backslashes ( [^]}),] ), but I like to keep them there for clarity and for compatibility since not all regex engines recognize that idiom.)

Explanation:

^         # Match the start of the string
[^\]}),]* # Match zero or more characters except ], }, ) or ,

Answer 2

you could just cut the rest part by replaceAll :

  String newStr = yourStr.replaceAll("[\\])},].*", "");

or by split() and get the first element.

String newStr = yourStr.split("[\\])},]")[0];

Answer 3

You can use this (as java string):

"(.+?)[\\]},)].*"

here is a fiddle

Answer 4

你可以试试正则表达式(.*?)[}\\]),](.*?)我在rubular上测试它并且反对你的例子。