当有空格时，preg_match_all正则表达式失败

Question

I'm trying to get the image urls from html source code using the following regex, but it fails when the image url has spaces in it. For example this url:

<img src="http://a57.foxnews.com/global.fncstatic.com/static/managed/img/Entertainment/876/493/kazantsev pink bikini reuters.jpg?ve=1&amp;tl=1" alt="kazantsev pink bikini reuters.jpg" itemprop="image">

$image_regex_src_url = '/<img[^>]*'.'src=[\"|\'](.*)[\"|\']/Ui';
preg_match_all($image_regex_src_url, $string, $out, PREG_PATTERN_ORDER);

This gives me back the following.
http://a57.foxnews.com/global.fncstatic.com/static/managed/img/Entertainment/876/493/kazantsev

Is there a way to match any character including whitespace? Or is it something I have to set in the php configuration?

Answer 1

You have several issues with your regular expression.

First, you are trying to use the concatenation operator ('.') to join both parts of your expression together ( this is not necessary ). Secondly, you don't need to use the alternation operator | inside of your character classes.

The dot . will match any character except newline sequence. It is a possibility that these tags could possibly include line breaks since they are located in HTML source. You could use the s (dotall) modifier which forces the dot to match any character including line breaks or use a negated character class meaning match any character except.

Using the s (dotall) modifier:

$image_regex_src_url = '/<img[^>]*src=(["\'])(.*?)\1/si';

Using a negated character class [^ ]

$image_regex_src_url = '/<img[^>]*src=(["\'])([^"\']*)\1/i';

Although, it is much easier to use a parser such as DOM to grab the results.

$doc = new DOMDocument;
@$doc->loadHTML($html); // load the HTML

foreach($doc->getElementsByTagName('img') as $node) {
   $urls[] = $node->getAttribute('src');
}

print_r($urls);