Java data structure read numeric content from a string [duplicate]

This question already has answers here: 开发者_如何学C Closed 11 years ago.

Possible Duplicate:

Using Regular Expressions to Extract a Value in Java

For example, the input string is AB100FF10. I need to read 100 and 10 from the string. Is there any classes/objects in Java that I can use?

---

try this

String[] nums = "AB100FF10".split("\\D+");
for (String num : nums) {
    System.out.println(num);
}

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Other than that, you could try passing the string to a class like Scanner

Scanner scan = new Scanner("AB100FF10").useDelimiter("\\D+");
while (scan.hasNextInt()) {
    System.out.println(scan.nextInt());
}

Edit: using \\D instead of \\w as a delimiter, as Bohemian suggested in his answer and comments.

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Just use split(<non-digits>), like this:

String[] numbers = "AB100FF10CCC".replaceAll("(^\\D*|\\D*$)", "").split("\\D+"); // "[100, 10]"

Now numbers contains an array of Strings that are all guaranteed to be numeric. You can use Integer.parseInt() to get ints.

The replaceAll("(^\\D*|\\D*$)", "") is used to trim non-digits from the front and back of the input string, otherwise split() will give you a blank string as the first/last element. It just makes to code simpler, rather than having to test the first/last specially.

As a method, it would look like this:

public static int[] parseInts(String input) {        
    String[] numbers = input.replaceAll("(^\\D*|\\D*$)", "").split("\\D+");
    int[] result = new int[numbers.length];
    for (int i = 0; i < numbers.length; i++) {
        result[i] = Integer.parseInt(numbers[i]);
    }
    return result;
}

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If you want to get only integers you can do the following:

ArrayList<Integer> numbers = new ArrayList<Integer>();
char[] characters = "AB100FF10".toCharArray();
StringBuffer buf = new StringBuffer();
for (int i = 0; i < characters.length; i++) {
     if (Character.isDigit(characters[i])) 
        buf.append(characters[i]);
     else if (buf.length() != 0) {
         numbers.add(Integer.parseInt(buf.toString()));
         buf = new StringBuffer();
     }
}

After that you will have an arrayList of numbers

---

I think pattern might be a better solution to because pattern object is more powerful comparing to simple String split() method.

for example, following code can resolve the same problem without any exception

Pattern p = Pattern.compile("\\d+");
Matcher m = p.matcher(test[0]);
while(m.find()){
  int x = Integer.parseInt( m.group() );
}

But if I use String.split(), there is one NumberFormatException is hard to dealt with.For example, below code can't escape NumberFormatException

for(int i = 0 ; i < test.length; i++){
  String[] numstr= test[i].split("\\D+");
  try{              
    for(int j=0; j<numstr.length;j++){
      if( numstr[j] == null || numstr[j] == ""){
    System.out.println("empty string \n");
      }
      else
       Integer.parseInt(numstr[j]);                 
  }catch(NumberFormatException ie){
      ie.printStackTrace();
  }
}