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Java parser code to extract 2 distincts values from an String array

Posted on 2014-10-30
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Last Modified: 2014-10-30
Need to parse 2 values (tn and id) from a response String similar to this:

tn='447795584165' ptr {id=0}
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Question by:yvan_vallee
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3 Comments
 
LVL 35

Accepted Solution

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mccarl earned 500 total points
ID: 40414790
Do you know regular expressions? If so, the below should hopefully make sense. If not, try it and see if it works, and if there are different format strings that need to be catered for, I can modify it for you.

package ee.yvanVallee;

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

public class ExtractFields {
    
    public static void main(String[] args) {
        String testString = "tn='447795584165' ptr {id=0}";
        
        
        
        Pattern pattern = Pattern.compile("tn='([^']*)' ptr \\{id=([^}]*)\\}");
        
        Matcher matcher = pattern.matcher(testString);
        if (matcher.matches() && matcher.groupCount() >= 2) {
            String tn = matcher.group(1);
            String id = matcher.group(2);
            
            
            System.out.println("tn: " + tn);
            System.out.println("id: " + id);
        }
    }
    
}

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Or if the values of both tn and id are always numeric (longs) only, you could do this...

package ee.yvanVallee;

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

public class ExtractFields {
    
    public static void main(String[] args) {
        String testString = "tn='447795584165' ptr {id=0}";
        
        
        
        Pattern pattern = Pattern.compile("tn='(\\d*)' ptr \\{id=(\\d*)\\}");
        
        Matcher matcher = pattern.matcher(testString);
        if (matcher.matches() && matcher.groupCount() >= 2) {
            long tn = new Long(matcher.group(1));
            long id = new Long(matcher.group(2));
            
            
            System.out.println("tn: " + tn);
            System.out.println("id: " + id);
        }
    }
    
}

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Author Closing Comment

by:yvan_vallee
ID: 40414907
Hi mccarl,,
This is exactly what I need, already tested and working fine.

Thank you,
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LVL 35

Expert Comment

by:mccarl
ID: 40414927
You're welcome, glad to help! :)
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