正则表达式替换两组

Question

I have pattern like : <[a-zA-Z][^>]*(?:poster|src)=(['\\"])([^'\\"]+)\\\\1[^>]*> here i want to replace the value of src or poster attributes.

It is okey for

<video src='srcVal' />

and

<video poster='posterVal' src='srcVal' />

but for

<video poster='posterVal' src='srcVal' />

only changes src value, due to matcher.group(2) returning only srcVal .

public class Test {
    public static void main(String[] args) throws Exception {
        String html = "<video poster='posterVal' src='srcVal' />";
        Pattern resourcePattern = Pattern.compile("<[a-zA-Z][^>]*(?:poster|src)=(['\"])([^'\"]+)\\1[^>]*>");
        Matcher matcher = resourcePattern.matcher(html);
        int last = 0;
        StringBuilder sb = new StringBuilder();
        while(matcher.find()) {
            String path = matcher.group(2) + "Changed";
            sb.append( html.substring(last, matcher.start(2)) + path );
            last = matcher.end(2);
        }
        sb.append(html.substring(last));
        System.out.println(sb);
        //outputs <video poster='posterVal' src='srcValChanged' />
        //expecting <video poster='posterValChanged' src='srcValChanged' />
    }
}

Does any body has an idea how to do this?

Answer 1

The basic problem is with the [^>]* near the start of your expression. Because * is greedy this will eat up as many characters as it can while still allowing the rest of the expression to match, so given

<video poster='posterVal' src='srcVal' />

the [^>]* will gobble ideo poster='posterVal' up to and including the space before src= .

I would approach it differently, rather than trying to write a regex that matches the whole tag just write one that matches the attributes you're interested in, and replace all matches of that expression

html.replaceAll("\\b((?:poster|src)=)(['\"])([^'\"]+)\\1", "$1$2$3Changed$2")

But as other posters have commented it would be much more sensible to use a proper parser that understands the language rather than trying to manipulate the textual representation with regular expressions.

Answer 2

I wouldn't do this with regex, but you can try such a thing:

<[a-zA-Z]*[^>]*(?:(poster)|src)=(['\"])([^'\"]+)\\2(?(1)[^>]*(?:src=(['\"])([^'\"]+)\\4)?[^>]*|[^>]*(?:poster=(['\"])([^'\"]+)\\6)?[^>]*)>

Though I don't have time to test it as of now, sorry.

Edit:
Less performance-oriented:

<[a-zA-Z]*(?=(?:[^>]*?poster=['\"]([^'\"]+))?)(?=(?:[^>]*?src=['\"]([^'\"]+))?)[^>]*(?:poster|src)[^>]*>

If you only want to match video tags, change it to (as it'd greatly improve it):

<video(?=(?:[^>]*?poster=['\"]([^'\"]+))?)(?=(?:[^>]*?src=['\"]([^'\"]+))?)[^>]*(?:poster|src)[^>]*>

Explanation: (as I guess it must look quite disturbing)

We're using 2 lookaheads to capture what's interesting. Lookaheads will allow us to check twice what comes ahead, therefore ignoring the order. However, those lookaheads must always work (using * and ? to make sure of that), but still being greedy, while being lazy (what?): we have to stop as soon as we see poster/src, but go far enough to catch those. .*?a? will always catch nothing. So we use here (?:.*?a)? . The behavior here is to try to catch the a with laziness, while if it fails it's not a problem.
The last part of the regex is to make sure we catch only tags with a poster or a src attribute, as our lookaheads only do that catching and certainly can't be used to do that.

Note that I removed the check for your attributes, as anyway it was useless.