How to find a string using pattern in java

Question

please refer below code

    String line = "abc_dfgb_tf";

    String pattern1 = "(\\w+)([+-])(\\d+)(\\w+)";
    Pattern r1 = Pattern.compile(pattern1);
    Matcher m1 = r1.matcher(line);

    if (m1.find( )) 
    {
       System.out.println("Found value: " + m1.group(1) );
       System.out.println("Found value: " + m1.group(2) );
       System.out.println("Found value: " + m1.group(3) );
       System.out.println("Found value: " + m1.group(4) );
    }

in case of "abc_dfgb_tf" string m1.find( ) is coming false.

please suggest the pattern which will use for both type of string "abc_dfgb_tf" and "abc_dfgb_tf+1cbv"

help

Just make the [+-] part optional: [+-]?.

Keppil
– Keppil

2014-12-22 20:32:02 +00:00
Commented Dec 22, 2014 at 20:32 — Keppil
– Keppil, Commented Dec 22, 2014 at 20:32

John Bollinger · Accepted Answer · 2014-12-22 20:30:52Z

5

You appear to want something like this:

String pattern1 = "(\\w+)(?:([+-])(\\d+)(\\w+))?";

That makes the optional tail in fact optional.

answered Dec 22, 2014 at 20:30

John Bollinger

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1 Comment

Rajeev Over a year ago

got it .. Thnaks :) John Bollinger

Elliott Frisch · Accepted Answer · 2014-12-22 20:32:52Z

2

You appear to want optional everything except the first group. Something like

String line = "abc_dfgb_tf";
String pattern1 = "(\\w+)([+-]*)(\\d*)(\\w*)";
Pattern r1 = Pattern.compile(pattern1);
Matcher m1 = r1.matcher(line);

if (m1.find()) {
    System.out.println("Found value: " + m1.group(1));
    System.out.println("Found value: " + m1.group(2));
    System.out.println("Found value: " + m1.group(3));
    System.out.println("Found value: " + m1.group(4));
}

Output is

Found value: abc_dfgb_tf
Found value: 
Found value: 
Found value:

and if I change line to String line = "abc_dfgb_tf+1cbv"; output is

Found value: abc_dfgb_tf
Found value: +
Found value: 1
Found value: cbv

answered Dec 22, 2014 at 20:32

Elliott Frisch

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1 Comment

Rajeev Over a year ago

got it .. Thnaks :) Elliott Frisch

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