使用正则表达式从html代码中提取第一个图像源？

Question

I would like to know how this can be achieved.

Assume: That there's a lot of html code containing tables, divs, images, etc.

Problem: How can I get matches of all occurances. More over, to be specific, how can I get the img tag source (src = ?).

example:

<img src="http://example.com/g.jpg" alt="" />

How can I print out http://example.com/g.jpg in this case. I want to assume that there are also other tags in the html code as i mentioned, and possibly more than one image. Would it be possible to have an array of all images sources in html code?

I know this can be achieved way or another with regular expressions, but I can't get the hang of it.

Any help is greatly appreciated.

Answer 1

While regular expressions can be good for a large variety of tasks, I find it usually falls short when parsing HTML DOM. The problem with HTML is that the structure of your document is so variable that it is hard to accurately (and by accurately I mean 100% success rate with no false positive) extract a tag.

What I recommend you do is use a DOM parser such as SimpleHTML and use it as such:

function get_first_image($html) {
    require_once('SimpleHTML.class.php')

    $post_html = str_get_html($html);

    $first_img = $post_html->find('img', 0);

    if($first_img !== null) {
        return $first_img->src;
    }

    return null;
}

Some may think this is overkill, but in the end, it will be easier to maintain and also allows for more extensibility. For example, using the DOM parser, I can also get the alt attribute.

A regular expression could be devised to achieve the same goal but would be limited in such way that it would force the alt attribute to be after the src or the opposite, and to overcome this limitation would add more complexity to the regular expression.

Also, consider the following. To properly match an <img> tag using regular expressions and to get only the src attribute (captured in group 2), you need the following regular expression:

<\s*?img\s+[^>]*?\s*src\s*=\s*(["'])((\\?+.)*?)\1[^>]*?>

And then again, the above can fail if:

• The attribute or tag name is in capital and the i modifier is not used.
• Quotes are not used around the src attribute.
• Another attribute then src uses the > character somewhere in their value.
• Some other reason I have not foreseen.

So again, simply don't use regular expressions to parse a dom document.

---

EDIT: If you want all the images:

function get_images($html){
    require_once('SimpleHTML.class.php')

    $post_dom = str_get_dom($html);

    $img_tags = $post_dom->find('img');

    $images = array();

    foreach($img_tags as $image) {
        $images[] = $image->src;
    }

    return $images;
}

Answer 2

Use this, is more effective:

preg_match_all('/<img [^>]*src=["|\']([^"|\']+)/i', $html, $matches);
foreach ($matches[1] as $key=>$value) {
    echo $value."<br>";
}

Example:

$html = '
<ul>     
  <li><a target="_new" href="http://www.manfromuranus.com">Man from Uranus</a></li>       
  <li><a target="_new" href="http://www.thevichygovernment.com/">The Vichy Government</a></li>      
  <li><a target="_new" href="http://www.cambridgepoetry.org/">Cambridge Poetry</a></li>      
  <img width="190" height="197" border="0" align="right" alt="upload.jpg" title="upload.jpg" class="noborder" src="value1.jpg" />
  <li><a href="http://www.verot.net/pretty/">Electronaut Records</a></li>      
  <img width="190" height="197" border="0" align="right" alt="upload.jpg" title="upload.jpg" class="noborder" src="value2.jpg" />
  <li><a target="_new" href="http://www.catseye-crew.com">Catseye Productions</a></li>     
  <img width="190" height="197" border="0" align="right" alt="upload.jpg" title="upload.jpg" class="noborder" src="value3.jpg" />
</ul>
<img width="190" height="197" border="0" align="right" alt="upload.jpg" title="upload.jpg" class="noborder" src="res/upload.jpg" />
  <li><a target="_new" href="http://www.manfromuranus.com">Man from Uranus</a></li>       
  <li><a target="_new" href="http://www.thevichygovernment.com/">The Vichy Government</a></li>      
  <li><a target="_new" href="http://www.cambridgepoetry.org/">Cambridge Poetry</a></li>      
  <img width="190" height="197" border="0" align="right" alt="upload.jpg" title="upload.jpg" class="noborder" src="value4.jpg" />
  <li><a href="http://www.verot.net/pretty/">Electronaut Records</a></li>      
  <img src="value5.jpg" />
  <li><a target="_new" href="http://www.catseye-crew.com">Catseye Productions</a></li>     
  <img width="190" height="197" border="0" align="right" alt="upload.jpg" title="upload.jpg" class="noborder" src="value6.jpg" />
';   
preg_match_all('/<img .*src=["|\']([^"|\']+)/i', $html, $matches);
foreach ($matches[1] as $key=>$value) {
    echo $value."<br>";
} 

Output:

value1.jpg
value2.jpg
value3.jpg
res/upload.jpg
value4.jpg
value5.jpg
value6.jpg

Answer 3

This works for me:

preg_match('@<img.+src="(.*)".*>@Uims', $html, $matches);
$src = $matches[1];

Answer 4

i assume all your src= have " around the url

<img[^>]+src=\"([^\"]+)\"

the other answers posted here make other assumsions about your code

Answer 5

I agree with Andrew Moore. Using the DOM is much, much better. The HTML DOM images collection will return to you a reference to all image objects.

Let's say in your header you have,

<script type="text/javascript">
    function getFirstImageSource()
    {
        var img = document.images[0].src;
        return img;
    }
</script>

and then in your body you have,

<script type="text/javascript">
  alert(getFirstImageSource());
</script>

This will return the 1st image source. You can also loop through them along the lines of, (in head section)

function getAllImageSources()
    {
        var returnString = "";
        for (var i = 0; i < document.images.length; i++)
        {
            returnString += document.images[i].src + "\n"
        }
        return returnString;
    }

(in body)

<script type="text/javascript">
  alert(getAllImageSources());
</script>

If you're using JavaScript to do this, remember that you can't run your function looping through the images collection in your header. In other words, you can't do something like this,

<script type="text/javascript">
    function getFirstImageSource()
    {
        var img = document.images[0].src;
        return img;
    }
    window.onload = getFirstImageSource;  //bad function

</script>

because this won't work. The images haven't loaded when the header is executed and thus you'll get a null result.

Hopefully this can help in some way. If possible, I'd make use of the DOM. You'll find that a good deal of your work is already done for you.

Answer 6

I don't know if you MUST use regex to get your results. If not, you could try out simpleXML and XPath, which would be much more reliable for your goal:

First, import the HTML into a DOM Document Object. If you get errors, turn errors off for this part and be sure to turn them back on afterward:

 $dom = new DOMDocument();
 $dom -> loadHTMLFile("filename.html");

Next, import the DOM into a simpleXML object, like so:

 $xml = simplexml_import_dom($dom);

Now you can use a few methods to get all of your image elements (and their attributes) into an array. XPath is the one I prefer, because I've had better luck with traversing the DOM with it:

 $images = $xml -> xpath('//img/@src');

This variable now can treated like an array of your image URLs:

 foreach($images as $image) {
    echo '<img src="$image" /><br />
    ';
  }

Presto, all of your images, none of the fat.

Here's the non-annotated version of the above:

---

 $dom = new DOMDocument();
 $dom -> loadHTMLFile("filename.html");

 $xml = simplexml_import_dom($dom);

 $images = $xml -> xpath('//img/@src');

 foreach($images as $image) {
    echo '<img src="$image" /><br />
    ';
  }

Answer 7

I really think you can not predict all the cases with on regular expression.

The best way is to use the DOM with the PHP5 class DOMDocument and xpath. It's the cleanest way to do what you want.

$dom = new DOMDocument();
$dom->loadHTML( $htmlContent );
$xml = simplexml_import_dom($dom);
$images = $xml -> xpath('//img/@src');

Answer 8

You can try this:

preg_match_all("/<img\s+src=\"(.+)\"/i", $html, $matches);
foreach ($matches as $key=>$value) {
    echo $key . ", " . $value . "<br>";
}

Answer 9

since you're not worrying about validating the HTML, you might try using strip_tags() on the text first to clear out most of the cruft.

Then you can search for an expression like

"/\<img .+ \/\>/i"

The backslashes escape special characters like <,>,/. .+ insists that there be 1 or more of any character inside the img tag You can capture part of the expression by putting parentheses around it. eg (.+) captures the middle part of the img tag.

When you decide what part of the middle you wish specifically to capture, you can modify the (.+) to something more specific.

Answer 10

<?php    
/* PHP Simple HTML DOM Parser @ http://simplehtmldom.sourceforge.net */

require_once('simple_html_dom.php');

$html = file_get_html('http://example.com');
$image = $html->find('img')[0]->src;

echo "<img src='{$image}'/>"; // BOOM!

PHP Simple HTML DOM Parser will do the job in few lines of code.