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

I believe a regular expression will do the trick here. – Sid Oct 06 '10 at 11:21 — Sid, Oct 06 '10 at 11:21

score 5 · Answer 1 · answered Oct 06 '10 at 11:16

5

Break up your problem:

parsing into tokens
converting tokens into objects
operate on objects

answered Oct 06 '10 at 11:16

duffymo

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jmj · Answer 2 · 2010-10-06T13:42:01.590

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 06 '10 at 13:42

answered Oct 06 '10 at 11:28

jmj

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@Chankey yeah it was typo. Thanks – jmj Oct 06 '10 at 13:42
Sorry for the downvote... it seems in this special case, it IS legit... but thanks for the update. +1 – helpermethod Oct 06 '10 at 15:21

score 1 · Accepted Answer · edited May 23 '17 at 11:55

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 '17 at 11:55

Community

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answered Oct 06 '10 at 11:15

Bozho

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3

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. – vstoyanov Oct 06 '10 at 11:18
@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 Oct 06 '10 at 11:20
@vstoyanov http://stackoverflow.com/questions/3684079/is-it-ok-to-try-catch-something-just-to-check-if-an-exception-was-thrown-or-not/3684093#3684093 – Bozho Oct 06 '10 at 11:22
for (String part : parts) please detail this statement what it does. – Sameek Mishra Oct 06 '10 at 11:31
this is the foreach loop introduced in Java 5. It loops all elements of the array/collection – Bozho Oct 06 '10 at 11:32
@Bozho http://commons.apache.org/lang/api-2.4/org/apache/commons/lang/math/NumberUtils.html there is a api to check , check my answer – jmj Oct 06 '10 at 11:57

score 0 · Answer 4 · answered Oct 06 '10 at 14:48

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.

How to extract number and String in java

4 Answers4