regular expression to extract string from java code [duplicate]

Question

The sample source code to match is

String string="welcome";
String k="a\"welcome";

I am using "(\"[^(\")]*\")" regex in java.

But this extracts

0:"welcome"
0:"a\"

Expected output is

0:"welcome"
0:"a\"welcome"

What change should i make in regex to get the expected output ?

Java source :

private static String pattern1="(\"[^(\")]*\")";
public void getStrings(){
    Pattern r = Pattern.compile(pattern1);
    Matcher m = r.matcher("String string=\"welcome\";\n" +
            "String k=\"a\\\"welcome\";");


    while(m.find()){
        System.out.println("0:"+m.group(0));
    }
}

You should tell us what pattern is not matched, or better what's the principle of matching or not. — Tony
– Tony, Commented Sep 6, 2014 at 14:07

Avinash Raj · Accepted Answer · 2014-09-06 14:31:58Z

1

Just use lookahead and lookbehind in your regex,,

(?<==)(".*?")(?=;)

Get the value from group index 1.

DEMO

Pattern r = Pattern.compile("(?<==)(\".*?\")(?=;)");
Matcher m = r.matcher("String string=\"welcome\";\n" +
            "String k=\"a\\\"welcome\";");
while(m.find()){
        System.out.println("0:"+m.group(1));
}

Output:

0:"welcome"
0:"a\"welcome"

OR

Use the greediness of *,

Pattern r = Pattern.compile("(\".*\")");

OR

It skips the double quotes which are preceded by a backslash,

Pattern r = Pattern.compile("(\\\".*?(?<=[^\\\\])\\\")");

edited Sep 6, 2014 at 14:31

answered Sep 6, 2014 at 13:56

Avinash Raj

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2 Comments

Avinash Raj Over a year ago

you could try this also regex101.com/r/vN2vU4/4 . It just skips the double quotes which are proceeded by a backslash.

Avinash Raj Over a year ago

or this, (\".*?[^\\]\")

plalx · Accepted Answer · 2014-09-06 14:07:21Z

0

Why do you even bother with variable assignment. You know that everything within "" is a string.

"(.+)"\s*; should do it just fine.

edited Sep 6, 2014 at 14:07

answered Sep 6, 2014 at 13:58

plalx

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