正则表达式在php中捕获<img> src [重复]

Question

This question already has an answer here:

• How to extract img src, title and alt from html using php? [duplicate] 10 answers

I'd like to extract the img src and using preg_match_all I have this:

$tag = '<img src="path/to/image.png" />';
preg_match_all('/(width|height|src)=("[^"]*")/i',$tag, $img[$tag]);

which returns:

Array
(
[<img src="path/to/image.png" />] => Array
    (
        [0] => Array
            (
                [0] => src="path/to/image.png"
            )

        [1] => Array
            (
                [0] => src
            )

        [2] => Array
            (
                [0] => "path/to/image.png"
            )

    )

)

How can I write the regex to return a similar result regardless of double or single quotes used in tag? I can write:

$tag = "<img src='path/to/image.png' />";
preg_match_all('/(width|height|src)=(\'[^\']*\')/i',$tag, $img[$tag]);

Which works, but I'm not familiar enough with regex to write one expression to handle either. I did try:

preg_match_all('/(width|height|src)=((\'[^\']*\')|("[^"]*"))/i',$tag, $img[$tag]);

But this seems to return extra matches in the array which I don't want.

</div>

Answer 1

You can use this:

(width|height|src)=("[^"]*"|'[^']*')

I've basically used an alternation to either match "fds" or 'fds'.