How to extract number and String in java

Question

i want to extract number and add these numbers using Java and String remain same.

String as-

String msg="1,2,hello,world,3,4";

output should come like- 10,hello,world

Thanks

all are saperated by commma ?

Jigar Joshi
– Jigar Joshi

2010-10-06 11:13:42 +00:00
Commented Oct 6, 2010 at 11:13 — Jigar Joshi
– Jigar Joshi, Commented Oct 6, 2010 at 11:13
I believe a regular expression will do the trick here.

Sid
– Sid

2010-10-06 11:21:04 +00:00
Commented Oct 6, 2010 at 11:21 — Sid
– Sid, Commented Oct 6, 2010 at 11:21

duffymo · Accepted Answer · 2010-10-06 11:16:02Z

5

Break up your problem:

parsing into tokens
converting tokens into objects
operate on objects

answered Oct 6, 2010 at 11:16

duffymo

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Jigar Joshi · Accepted Answer · 2010-10-06 13:42:01Z

5

String pieces[] = msg.split(",");  
int sum=0;
StringBuffer sb = new StringBuffer();
for(int i=0;i < pieces.length;i++){

      if(org.apache.commons.lang.math.NumberUtils.isNumber(pieces[i])){
             sb.appendpieces[i]();
      }else{
             int i = Integer.parseInt(pieces[i]));
             sum+=i;    
      }

 }
 System.out.println(sum+","+sb.);
 }

edited Oct 6, 2010 at 13:42

answered Oct 6, 2010 at 11:28

Jigar Joshi

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2 Comments

Jigar Joshi Over a year ago

@Chankey yeah it was typo. Thanks

helpermethod Over a year ago

Sorry for the downvote... it seems in this special case, it IS legit... but thanks for the update. +1

Community · Accepted Answer · 2017-05-23 11:55:22Z

1

String[] parts = msg.split(",");
int sum = 0;
StringBuilder stringParts = new StringBuilder();
for (String part : parts) {
    try {
        sum += Integer.parseInt(part);
    } catch (NumberFormatException ex) {
        stringParts.append("," + part);
    }
}
stringParts.insert(0, String.valueOf(sum));

System.out.println(stringParts.toString()); // the final result

Note that the above practice of using exceptions as control flow should be avoided almost always. This concrete case is I believe an exception, because there is no method that verifies the "parsability" of the string. If there was Integer.isNumber(string), then that would be the way to go. Actually, you can create such an utility method. Check this question.

edited May 23, 2017 at 11:55

CommunityBot

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answered Oct 6, 2010 at 11:15

Bozho

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6 Comments

vstoyanov Over a year ago

Isn't relying on exceptions to control program flow a bad practice? Otherwise it looks like a pretty nice simplistic solution with close to no accidental complexity.

Bozho Over a year ago

@vstoyanov there was a discussion about that specific case (with parsing numbers/date), let me find it. Generally, yes, it's a bad practice to rely on exceptions, but since there is no method to verify the parsability..

Bozho Over a year ago

@vstoyanov stackoverflow.com/questions/3684079/…

Sameek Mishra Over a year ago

for (String part : parts) please detail this statement what it does.

Bozho Over a year ago

this is the foreach loop introduced in Java 5. It loops all elements of the array/collection

|

Sean Patrick Floyd · Accepted Answer · 2010-10-06 14:48:01Z

Here's a very simple regex version:

/**
 * Use a constant pattern to skip expensive recompilation.
 */
private static final Pattern INT_PATTERN = Pattern.compile("\\d+",
    Pattern.DOTALL);

public static int addAllIntegerOccurrences(final String input){
    int result = 0;
    if(input != null){
        final Matcher matcher = INT_PATTERN.matcher(input);
        while(matcher.find()){
            result += Integer.parseInt(matcher.group());
        }
    }
    return result;

}

Test code:

public static void main(final String[] args){
    System.out.println(addAllIntegerOccurrences("1,2,hello,world,3,4"));
}

Output:

10

Caveats:

This will not work if the numbers add up to anything larger than Integer.Max_VALUE, obviously.

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