Java how to use pattern matcher using regular expressions to find certain string

Question

I am not familiar with Patterns & matchers, and I am pretty stuck with this problem.

I have a string that I would like to manipulate in Java. I understand that I have to use

Pattern p = Pattern.compile("\\[(.*?)\\]");
Matcher m = p.matcher(in);

while(m.find()) {
     String found = m.group(1).toString();
}

But in my string, let's say I have the following examples:

String in = "test anything goes here [A#1234|567]"; //OR
String in = "test anything goes here [B#1234|567]"; //OR
String in = "test anything goes here [C#1234|567]";

I want to find [A# | ] or [B# | ] or [C# | ] in the string, how do I use the regex to find the expression?

Regex you are using will find these values for you. What exactly is problem you are facing? — Pshemo
– Pshemo, Commented Mar 10, 2014 at 13:28
@Ryven, yes, but the numbers should be "dynamic". I can have things like [A#656464566|874834637647], numbers can vary. — prog rice bowl
– prog rice bowl, Commented Mar 10, 2014 at 13:28

Sabuj Hassan · Accepted Answer · 2014-03-10 13:24:23Z

1

Use [ABC]# in your regex to match your expression.

Pattern p = Pattern.compile("(\\[[ABC]#.*?\\])");

If the fields are digit then you can safely use \d+

Pattern p = Pattern.compile("(\\[[ABC]#\\d+\\|\\d+\\])");

answered Mar 10, 2014 at 13:24

Sabuj Hassan

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2 Comments

prog rice bowl Over a year ago

How about the "|"? That would need to be present too... Is it possible to get that in?

Sabuj Hassan Over a year ago

@progricebowl I already updated the answer. Please have a look.

Mena · Accepted Answer · 2014-03-10 13:29:25Z

I'd use a simple Pattern as in the following example:

String[] in = { "test anything goes here [A#1234|567]",
            "test anything goes here [B#1234|567]",
            "test anything goes here [C#1234|567]" };

    Pattern p = Pattern.compile("\\[[A-Z]#\\d+\\|\\d+\\]");
    for (String s: in) {
        Matcher m = p.matcher(s);
        while (m.find()) {
            System.out.println("Found: " + m.group());
        }
    }
}

Output

Found: [A#1234|567]
Found: [B#1234|567]
Found: [C#1234|567]

I'm assuming here that your Pattern has specific restrictions:

Starts with [
Followed by one upper-case non-accented letter
Followed by #
Followed by any number of digits
Followed by |
Followed by any number of digits
Followed by ]

Rhys · Accepted Answer · 2014-03-10 13:27:37Z

0

Try:

Pattern p = Pattern.compile("(\\[[A-Z]#.*\\])");

If you want to match any capital A through Z. Unclear if you want all the data between [] though.

answered Mar 10, 2014 at 13:27

Rhys

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Comments

Amit Joki · Accepted Answer · 2014-03-10 13:46:55Z

My solution:

import java.util.*;
import java.lang.*;
import java.io.*;
import java.util.regex.Matcher;
import java.util.regex.Pattern;

class Some {
    public static void main(String[] args) throws java.lang.Exception {
        String[] in = {
            "test anything goes here [A#1234|567]",
                "test anything goes here [B#1234|567]",
                "test anything goes here [C#1234|567]"
        };

        Pattern p = Pattern.compile("\\[(.*?)\\]");
        for (String s: in ) {
            Matcher m = p.matcher(s);
            while (m.find()) {
                System.out.println("Found: " + m.group().replaceAll("\\d", ""));
            }
        }

    }
}

This uses your original regex.

Demo: http://ideone.com/4Z5oYD

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