number out of string in java

问题

I have something like "ali123hgj". i want to have 123 in integer. how can i make it in java?

回答1:

Use the following RegExp (see http://java.sun.com/docs/books/tutorial/essential/regex/):

\d+

By:

final Pattern pattern = Pattern.compile("\\d+"); // the regex
final Matcher matcher = pattern.matcher("ali123hgj"); // your string

final ArrayList<Integer> ints = new ArrayList<Integer>(); // results

while (matcher.find()) { // for each match
    ints.add(Integer.parseInt(matcher.group())); // convert to int
}

回答2:

int i = Integer.parseInt("blah123yeah4yeah".replaceAll("\\D", ""));
// i == 1234

Note how this will "merge" digits from different parts of the strings together into one number. If you only have one number anyway, then this still works. If you only want the first number, then you can do something like this:

int i = Integer.parseInt("x-42x100x".replaceAll("^\\D*?(-?\\d+).*$", "$1"));
// i == -42

The regex is a bit more complicated, but it basically replaces the whole string with the first sequence of digits that it contains (with optional minus sign), before using Integer.parseInt to parse into integer.

回答3:

This is the Google Guava #CharMatcher Way.

String alphanumeric = "12ABC34def";

String digits = CharMatcher.JAVA_DIGIT.retainFrom(alphanumeric); // 1234

String letters = CharMatcher.JAVA_LETTER.retainFrom(alphanumeric); // ABCdef

If you only care to match ASCII digits, use

String digits = CharMatcher.inRange('0', '9').retainFrom(alphanumeric); // 1234

If you only care to match letters of the Latin alphabet, use

String letters = CharMatcher.inRange('a', 'z')
                         .or(inRange('A', 'Z')).retainFrom(alphanumeric); // ABCdef

回答4:

You could probably do it along these lines:

Pattern pattern = Pattern.compile("[^0-9]*([0-9]*)[^0-9]*");
Matcher matcher = pattern.matcher("ali123hgj");
boolean matchFound = matcher.find();
if (matchFound) {
    System.out.println(Integer.parseInt(matcher.group(0)));
}

It's easily adaptable to multiple number group as well. The code is just for orientation: it hasn't been tested.

回答5:

int index = -1;
for (int i = 0; i < str.length(); i++) {
   if (Character.isDigit(str.charAt(i)) {
      index = i; // found a digit
      break;
   }
}
if (index >= 0) {
   int value = String.parseInt(str.substring(index)); // parseInt ignores anything after the number
} else {
   // doesn't contain int...
}

回答6:

public static final List<Integer> scanIntegers2(final String source) {
    final ArrayList<Integer> result = new ArrayList<Integer>(); 
    // in real life define this as a static member of the class.
    // defining integers -123, 12 etc as matches.
    final Pattern integerPattern = Pattern.compile("(\\-?\\d+)");
    final Matcher matched = integerPattern.matcher(source);
    while (matched.find()) {
     result.add(Integer.valueOf(matched.group()));
    }
    return result;

Input "asg123d ddhd-2222-33sds --- ---222 ss---33dd 234" results in this ouput [123, -2222, -33, -222, -33, 234]

来源：https://stackoverflow.com/questions/2574604/number-out-of-string-in-java

标签

java

string

numbers