Java Regex Matcher works, but String.split() doesn't

Question

I am looking to use a one line String.split() to pull out the 'name' in a query I am writting..

The Pattern + Matcher works like expected, but I am pulling my hair out trying to figure out why String.split() doesn't return a match!

public static void main(String[] asdf)
{
    final String queryText = "id <equals> `1` <AND> name <equals> `some name`";
    final String regex = "^(.*name <equals> `)([\\S\\s]*)(`.*)$";

    System.out.println("Splitting...");
    final String[] split = queryText.split(regex);
    for (int i = 0; i < split.length; i++)
    {
        System.out.println(split[i]);
    }

    System.out.println("Matching...");
    final Pattern pattern = Pattern.compile(regex);
    final Matcher matcher = pattern.matcher(queryText);

    if (matcher.find())
    {
        for (int i = 0; i < matcher.groupCount(); i++)
        {
            System.out.println(matcher.group(i + 1));
        }
    }
}

Prints the output

Splitting...
Matching...
id <equals> `1` <AND> name <equals> `
some name
`

Your match covers the entire string. Hence, if you split by the match, the two resulting strings are empty. And trailing empty strings are omitted by default. Use the [overload with an integer argument](docs.oracle.com/javase/1.4.2/docs/api/java/lang/…, int)) and supply a negative integer to get the empty strings, if that is what you want. In Java, split does not return the matched groups in addition to the splits (as it works in some other languages). — Martin Ender
– Martin Ender, Commented Aug 19, 2013 at 10:35
The regex [\\S\\s] is insane - it means "any character that is a whitespace or not a whitespace. It's the same as . — Bohemian
– Bohemian ♦, Commented Aug 19, 2013 at 10:53
Thanks for the input Bohemian.. I had started with . and forgot to put it back. — Thomas Beauvais
– Thomas Beauvais, Commented Aug 19, 2013 at 11:50

sp00m · Accepted Answer · 2013-08-19 12:24:34Z

2

You regex matches the whole string. Thus, when splitting, the whole string gets removed. It is exactly the same as "a".split("a"), which returns an empty array.

What you could use instead is:

queryText.replaceAll(".*name <equals> `([^`]+)`.*", "$1")

which returns some name.

answered Aug 19, 2013 at 12:24

sp00m

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CodePi · Accepted Answer · 2013-08-19 10:43:15Z

0

Your regex expression covers all the string. Try to remove "^" and "$". Split method wants the expressions to remove not the expressions that must be found to detect a particular substring.

answered Aug 19, 2013 at 10:43

CodePi

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Sedecimdies · Accepted Answer · 2013-08-19 11:10:56Z

0

Try using the regex

((?<=`)[^`]*(?=`$))

as this gives you the result "some name" from

id <equals> `1` <AND> name <equals> `some name`

or if the "some name" does not Always appears at the end you can use

((?<=name <equals> `)[^`]*(?=`))

with the same result

edited Aug 19, 2013 at 11:10

answered Aug 19, 2013 at 11:03

Sedecimdies

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Thomas Beauvais · Accepted Answer · 2013-08-19 11:57:59Z

0

Thanks for all of the input. [m.buettner] was correct. I was merely misunderstanding what String.split() is supposed to do. Per the Java documentation..

This method works as if by invoking the two-argument split method with the given expression and a limit argument of zero. Trailing empty strings are therefore not included in the resulting array.

This isn't what I wanted. I will merely create a util method with my Pattern + Matcher.

Thanks again everyone.

answered Aug 19, 2013 at 11:57

Thomas Beauvais

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Community · Accepted Answer · 2017-05-23 11:49:51Z

-1

Your regex variable is not properly formed for the String split() method... Try breaking the regex as shown in the answer of this thread...

java - split string using regular expression

edited May 23, 2017 at 11:49

CommunityBot

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answered Aug 19, 2013 at 11:00

Sid

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