使用正则表达式忽略Java中的模式

Question

I have a sentence: "we:PR show:V" . I want to match only those characters after ":" and before "\\\\s" using regex pattern matcher. I used following pattern:

Pattern pattern=Pattern.compile("^(?!.*[\\w\\d\\:]).*$");

But it did not work. What is the best pattern to get the output?

Answer 1

For a situation such as this, if you are using java, it may be easier to do something with substrings:

String input = "we:PR show:V";
String colon = ":";
String space = " ";
List<String> results = new ArrayList<String>();
int spaceLocation = -1;
int colonLocation = input.indexOf(colon);
while (colonLocation != -1) {
    spaceLocation = input.indexOf(space);
    spaceLocation = (spaceLocation == -1 ? input.size() : spaceLocation);
    results.add(input.substring(colonLocation+1,spaceLocation);

    if(spaceLocation != input.size()) {
        input = input.substring(spaceLocation+1, input.size());
    } else {
        input = new String(); //reached the end of the string
    }
}
return results;

This will be faster than trying to match on regex.

Answer 2

The following regex assumes that any non-whitespace characters following a colon (in turn preceded by non-colon characters) are a valid match:

[^:]+:(\S+)(?:\s+|$)

Use like:

String input = "we:PR show:V";
Pattern pattern = Pattern.compile("[^:]+:(\\S+)(?:\\s+|$)");
Matcher matcher = pattern.matcher(input);
int start = 0;
while (matcher.find(start)) {
    String match = matcher.group(1); // = "PR" then "V"
    // Do stuff with match
    start = matcher.end( );
}

The pattern matches, in order:

1. At least one character that isn't a colon.
2. A colon.
3. At least non-whitespace character (our match).
4. At least one whitespace character, or the end of input.

The loop continues as long as the regex matches an item in the string, beginning at the index start , which is always adjusted to point to after the end of the current match.