Java pattern matching for {0}

Question

I need to replace the {0} with values from hashmap. hashmap key is 0,1,2,3 respectively. what would be the right option.i feel we can implement Pattern matcher. What is the pattern for this value?

Dear {0}, 
You are being contacted because the '{1}' named '{2}' has been changed by '{3}' in the Application. These changes may impact any existing reports you are using which depend upon this information. If you have not authorized these changes, please contact '{4}' or send a request to  IT Support to have the changes reversed

Output:

Dear abc , 
You are being contacted because the ' Attribute' named 'prod1_group' has been changed by ' Guest ' in the Application. These changes may impact any existing reports you are using which depend upon this information. If you have not authorized these changes, please contact 'Guest' or send a request to IT Support to have the changes reversed

anubhava · Accepted Answer · 2014-09-19 11:02:00Z

2

You can use this regex:

Pattern p = Pattern.compile("\\{(\\d+)}");

And use matcher.group(1) as key to your HashMap in a while (matcher.find()) {..} loop.

Where matcher is:

Matcher matcher = p.matcher( input );

RegEx Demo

edited Sep 19, 2014 at 11:02

answered Sep 19, 2014 at 9:51

anubhava

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3 Comments

RahulArackal Over a year ago

This will fail with PatternSyntaxException, you have to escape { character.

anubhava Over a year ago

Thanks @iMmo: I had \\{ and \\} earlier but then it worked without error on regex101 site so I removed it :)

RahulArackal Over a year ago

I saw it, but regex101 doesnt support java, btw dont need to escape } :)

Jothi · Accepted Answer · 2014-09-19 10:32:38Z

You need to ecsape { character otherwise you will get PatternSyntaxException and in order to capture digits \\d+. Lastly matcher.group(1) will return String , so you need to cast it to Integer Below is a sample

import java.util.HashMap;
import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class Registrar {
    public static void main(String[] args) {

        String input = "Dear {0}, \n" +
                "You are being contacted because the '{1}' named '{2}' has been changed by '{3}' in the Application. These changes may impact any existing reports you are using which depend upon this information. If you have not authorized these changes, please contact '{4}' or send a request to IT Support to have the changes reversed";


        Pattern pattern = Pattern.compile("\\{(\\d+)}");

        HashMap<Integer, String> map = new HashMap<>();
        map.put(0, "zero");
        map.put(1, "one");
        map.put(2, "two");
        map.put(3, "three");
        map.put(4, "four");

        Matcher matcher = pattern.matcher(input);

        while (matcher.find()) {
            String val = matcher.group(1);
            String group = matcher.group();
            input = input.replace(group, map.get(Integer.parseInt(val)));
        }

        System.out.println(input);
    }
}

And it outputs

Dear zero, 
You are being contacted because the 'one' named 'two' has been changed by 'three' in the Application. These changes may impact any existing reports you are using which depend upon this information. If you have not authorized these changes, please contact 'four' or send a request to IT Support to have the changes reversed

RahulArackal · Accepted Answer · 2014-09-19 10:58:18Z

0

You could do it with \\{[0-9]{1}}

    String testString = "{0}";
    String myPattern = "\\{[0-9]{1}}";
    Pattern pattern = Pattern.compile(myPattern);
    Matcher m = pattern.matcher(testString);
    if(m.matches()) {
        System.out.println("Correct Value");
    } else {
        System.out.println("Wrong Value");          
    }

To match it from a String do like the following

    String testString = "Dear {0},";
    String myPattern = ".*\\{[0-9]{1}}.*";

edited Sep 19, 2014 at 10:58

answered Sep 19, 2014 at 10:07

RahulArackal

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