java - Regular expression to get characters before brackets or comma -

i'm pulling hair out bit this.

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

i want extract 7f8hd::;;8843fdj fls "": string based on premise string ends either } or ] or , or ) characters present need first one.

i have tried without success create regular expression matcher , pattern class can't seem right.

the best come below reg exp doesn't seem work think 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 not understanding reg exp correctly. great.

^[^\]}),]* 

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

in java:

pattern regex = pattern.compile("^[^\\]}),]*"); matcher regexmatcher = regex.matcher(line); if (regexmatcher.find()) {     system.out.println(regexmatcher.group()); } 

(you can remove backslashes ([^]}),]), keep them there clarity , compatibility since not regex engines recognize idiom.)

explanation:

^         # match start of string [^\]}),]* # match 0 or more characters except ], }, ) or ,