validate digit in java, error using length

Question

I tried to validate digit and make this function.

public int digitCheck(String string)
    {
        int flag=0;

        for(int i=0;i<string.length();i++)
        {
            if(!Character.isDigit(string.charAt(i)))    
                flag++; 
        }       
        return flag;
    }

But it's always said an error on length on for

for(int i=0;i<string.length();i++)

Does anyone know why there's an error on length?

I assume you mean exception? What is the exception? is the string being passed in null, since that's the only error I can imagine happening calling `string.length()`? — Krease, Dec 11 '13 at 04:23

score 3 · Accepted Answer · answered Dec 11 '13 at 04:25

You forgot to check for null, and I think you want a boolean function like this.

public static boolean digitCheck(String string) {
  if (string != null) { // <--- Add This
    for (int i = 0; i < string.length(); i++) {
      if (!Character.isDigit(string.charAt(i)))
        return false;
    }
    return true; // we only got here if there were characters, and they're all digit(s).
  }
  return false; 
}

score 2 · Answer 2 · edited Dec 11 '13 at 16:05

2

Checking for null and using Java's Pattern class is another solution.

Here is a code sample:

public static final Pattern DIGIT_PATTERN = java.util.regex.Pattern.compile("[0-9]*");

public boolean digitCheck(String str) {
    if ( str == null ) {
        return true; // Assume Null is valid digit string;
    }
    return DIGIT_PATTERN.matcher(str).matches();
}

edited Dec 11 '13 at 16:05

Krease

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answered Dec 11 '13 at 04:28

LHA

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Can you clarify what you mean by 'use java pattern'? – Krease Dec 11 '13 at 07:11
Check this: http://docs.oracle.com/javase/7/docs/api/java/util/regex/Pattern.html – LHA Dec 11 '13 at 14:36

validate digit in java, error using length

2 Answers2