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Editorial Team
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Asked: 2026-05-15T17:43:51+00:00

Here’s the code I’m using : public class splitText { public static void main(String[]

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Here’s the code I’m using :

public class splitText {
public static void main(String[] args) {
    String x = "I lost my Phone. I shouldn't drive home alone";
    String[] result = x.split(".");
    for (String i : result) {
        System.out.println(i);
    }
}
}

Compiles perfectly, but nothing happens at runtime. What am I doing wrong?

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1 Answer

  1. Editorial Team

    String.split(String regex) takes a regular-expression pattern. It just so happens that . in regex is a metacharacter that matches (almost) any character, hence why split(".") doesn’t work the way you expected.

    You can escape the . by preceding it with a backslash. As a Java string literal, this is "\\.". The \ is doubled because \ itself is a Java escape character. "\\." is a String of length 2, containing a backslash and a period.

    If you’re given an arbitrary String that is to be matched literally (or if you just don’t care to escape them yourself), you can use Pattern.quote. It’ll make a pattern to literally match a given String.

    See also

    This is provided for educational purposes only:

        String text =
        "Wait a minute... what?!? Oh yeah! This is awesome!!";

    for (String part : text.split("(?<=[.?!]) ")) {
        System.out.println(part);
    }

    This prints:

    Wait a minute...
what?!?
Oh yeah!
This is awesome!!

    References

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