Question

I’m trying to display the biggest image url returned from an xml result. So far the largest returned is 400 high so I hardcoded 400 in. If possible I would like to select just the largest in case in the future I get results that don’t have a 400 height image in them.

I’ve tried

$x = file_get_contents($url);
$xml = simplexml_load_string($x);

$imageURL=$xml->categories->category->items->product->images->image[@height='400']->sourceURL;

Which gives me “syntax error, unexpected ‘=’, expecting ‘]'”.

And I also tried:

$imageURL= $xml->xpath("/categories/category/items/producct/images/image[@height='400']/sourceURL");

But got a bad link.
Here is the XML:

 <images>
        <image available="true" height="100" width="100">
            <sourceURL>
                Someurl.com
            </sourceURL>
        </image>
        <image available="true" height="200" width="200">
            <sourceURL>
                Someurl.com
            </sourceURL>
        </image>
        <image available="true" height="300" width="300">
            <sourceURL>
                Someurl.com
            </sourceURL>
        </image>
        <image available="true" height="400" width="400">
            <sourceURL>
                Someurl.com
            </sourceURL>
        </image>
        <image available="true" height="399" width="400">
            <sourceURL>
                Someurl.com
            </sourceURL>
        </image>
    </images>

Any ideas?

Answer 1

->image[@height='400'] is a direct PHP array reference. This’d be interpreted as supressing errors (@) on a defined() constant (height), and trying to set its value via an assignment ='400'.

For your xpath version, remember that an xpath query returns a DOMNodeList, not an actual DOMElement. To get the URLs you need from the query results, you have to ierate over the node list:

$nodes = $xpath->query(...) {
foreach($nodes as $node) {
   $url = $node->nodeValue;
}