Java regex only bashslash(\\) not working

Question

I am incorporating a pattern with has a backslash(\) with an escape sequence once.But that is not working at all.I am getting result as no match.

package com.test;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class TestClassRegex {
    private static final String VALIDATION = "^[0-9\\-]+$";
    public static void main(String[] args) {
        String line = "1234\56";
        Pattern r = Pattern.compile(VALIDATION);
        Matcher m = r.matcher(line);
        if (m.matches()) {
         System.out.println("match");
        }
        else {
            System.out.println("no match !!");
        }

    }
}

How can I write a pattern which can recognize backslash literally.

I have actually seen another post : Java regular expression value.split("\\."), "the back slash dot" divides by character? which doesn't answer my question completely.Hence needs some heads up here.

I think you need four total backslashes to get what you want. `"^[0-9\\\\-]+$"` — markspace, Jun 26 '16 at 06:05
You probably need `String line = "1234\\56";`. But it would be helpful if you told us what your code does - does it compile? What does it print? — Duncan Jones, Jun 26 '16 at 06:09
@markspace, is there a simple and more accurate way to do it. — Ajay, Jun 26 '16 at 06:23
A string created using the string literal `"1234\56"` does not contain a backslash; `\5` is an octal escape sequence for `U+0005`, a control character. I assume your real string is coming from somewhere else, and really has a backslash in it, but the literal your test code should contain two backslashes, as @Duncan said. And the string literal representing the regex should have four backslashes, as @markspace said. — Alan Moore, Jun 26 '16 at 06:32
Ducan didn't intended to change your data, but in Java, this `"1234\56" means "123456" because you espace the character '5' which gives '5'. So to have a backslash in a variable, you need to double it, as he said. Oups, just saw the comment of Alan Moore, which already explains it. — Master DJon, Jun 26 '16 at 06:54
Also, that other question you linked to doesn't relate to your problem. Here;s a better one: http://stackoverflow.com/questions/2242417/how-to-remove-the-backslash-in-string-using-regex-in-java/2242486#2242486 — Alan Moore, Jun 26 '16 at 06:56
Doh! It's actually `\56` (`U+002E`, or `.`), as Yassin points out below. But the takeaway is the same: your regex isn't trying to match a backslash, and your test string doesn't contain any. — Alan Moore, Jun 26 '16 at 07:24

score 1 · Accepted Answer · answered Jun 26 '16 at 07:00

"1234\56" will not produce "123456" but instead "1234."

Why?

The \ in a String is used to refer to the octal value of a character in the ASCII table. Here, you're calling \056 which is the character number 46 in the ASCII table and is represented by .

That's exactly the reason why you're not getting a match here.

Solution

You should first of all change your regex to ^[0-9\\\\-]+$ because in Java you need to escape the \ in a String. Even if your initial RegEx does not do it.

Your input needs to look like 1234\\56 for the same reason as above.

Java regex only bashslash(\\) not working

1 Answers1

Why?

Solution