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javaregex
JavaUser
JavaUser

Reputation: 26364

Java regex to parse a string with special chars and numbers

I need a java regex to split the words from the following pattern.

The sentence will start with @ symbol and continue with 6 numerals and then @ symbol. Then 9 numerals and then [ symbol Then 6 numerals and @ symbol Then 2 numerals

There may be space in between the special character and number

For example :

    @123456@ 123456789[ 123456@ 12

Here my output should be like :
     string1 = 123456
     string2 = 123456789
     string3 = 123456
     string4 = 12

The below regex is not spliting the words properly . Anyone help on this?

@[0-9]{6}@[0-9]{9}[[0-9]{6}@[0-9]{2}

Thanks.

Upvotes: 1

Views: 185

Answers (7)

slavik
slavik

Reputation: 1303

You have an error in your pattern

    String s = "@123456@ 123456789[ 123456@ 12";
    Pattern p = Pattern.compile("@(\\d{6})@\\s?(\\d{9})\\[\\s?(\\d{6})@\\s?(\\d{2})");
    Matcher m = p.matcher(s);
    if (m.matches()) {
        System.out.println(m.group(1));
        System.out.println(m.group(2));
        System.out.println(m.group(3));
        System.out.println(m.group(4));
    }

Upvotes: 1

zx81
zx81

Reputation: 41838

Capture Groups

You can match into capture groups:

Pattern regex = Pattern.compile("^@(\\d{6})@ *(\\d{9})\\[ (\\d{6})@ (\\d{2})");
Matcher regexMatcher = regex.matcher(subjectString);
if (regexMatcher.find()) {
    String firstGroup  = regexMatcher.group(1);
    // 123456
    String secondGroup = regexMatcher.group(2);
    // 123456789
    String thirdGroup  = regexMatcher.group(3);
    // 123456
    String fourthGroup = regexMatcher.group(4);
    // 12
}

Upvotes: 2

achingfingers
achingfingers

Reputation: 1916

The regex-pattern by which you want to split is: [@\[]\s*. This overview will help to understand the pattern. In Java it looks like follows:

String string = "@123456@ 123456789[ 123456@ 12";
String[] strings = string.split("[@\\[]\\s*");

The strings array will contain the desired output [123456,123456789,123456,12]

Upvotes: 1

Domenico Monaco
Domenico Monaco

Reputation: 1236

You must use (...) to incapsulate "capturing".

@([0-9]{6})@([0-9]{9})\[([0-9]{6})@([0-9]{2})

Upvotes: 1

Unihedron
Unihedron

Reputation: 11051

You can simply use String.split()!

{
    String[] strArr = "@123456@ 123456789[ 123456@ 12".split("[@\\[] ?");
    string1 = strArr[0];
    string2 = strArr[1];
    string3 = strArr[2];
    string4 = strArr[3];
}

Online Demo STDOUT:
123456
123456789
123456
12

Upvotes: 1

Korashen
Korashen

Reputation: 2191

If you are not sure about the spaces, add them with the * quantity to have 0 to unlimited. And make sure to escape the [

@[0-9]{6}@ *[0-9]{9}\[ *[0-9]{6}@ *[0-9]{2}

Works fine in the RegEx-Tester http://erik.eae.net/playground/regexp/regexp.html

Upvotes: 0

Maroun
Maroun

Reputation: 95978

Try this one:

@[0-9]{6}@\s[0-9]{9}\[\s[0-9]{6}@ [0-9]{2}

You should quote the actual [ and add the spaces you want.

See this working example.

Upvotes: 0

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