CODEWITHSUNDEEP
javaregexparsing
mickthompson
mickthompson

Reputation: 5512

Regex - extract info from a String in Java

I have a problem in applying a regex in my Java code.
My text string is like this (String myString)

        name: Abc Def;
        blah: 1 2 3;
        second name: Ghi;

I need to extract the name information (Abc Def). Name can contain a string of 1+ words. All the properties (name, blah, second name) are spaced with some whitespaces at the beginning

The code I'm trying to use is

String patternStr = "\\s{2,}name:\\s([\\w ]*\\w+);";
Matcher matcher = Pattern.compile(patternStr).matcher(myString);
if (matcher.matches()) 
    System.out.println(matcher.group(1));

My regex seems working fine with online tools (regex: \s{2,}name:\s([\w ]*\w+);) but when I port it to java code it simply doesn't work. Any idea of what I'm missing?

Thanks a lot

Edit: If I use matcher.find() it works fine.

Upvotes: 1

Views: 960

Answers (2)

rsp
rsp

Reputation: 23373

The problem is probably that .match( ) tries to match the whole input line. So either you can add a .+ at the end of the pattern to eat the rest of the line or use .contains( ). (with .match( ) the ^ and $ at the start and end of the pattern are implicit)

Note that you need to provide a flag when you need to match against multi-line input: .compile(patternStr, Perl5Compiler.MULTILINE_MASK)

Upvotes: 2

genio
genio

Reputation: 882

Here ya go:

^\s*name:\s*([^;]+);
  • Start at the beginning of the line.
  • Eat as much whitespace as there is.
  • Find "name:"
  • Eat as much whitespace as there is.
  • Capture anything before the next semicolon
  • Make sure there's a semicolon

Upvotes: 5

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