Question

Hey. I have an XML file and would like to use PHP to display the medium size only. The part of the XML looks like this:

<image size="small">/small23.png</image>
<image size="medium">/medium23.png</image>
<image size="large">/large23.png</image>

我用<代码><img src=”/>在网页上展示图像,但我不敢肯定如何只展示中的形象。在通过所有其他内容的论坛内,我尝试:

if($file->image->attributes()->size == "medium")
   echo "<img src=$file->image />";

但没有。

感谢

Answer 1

根据您的榜样,我假定你重新使用简单的XML,这样就可以:

$mediums = $file->xpath("image[@size= medium ]");
if (count($mediums)) {
    echo  <img src="  . (string) $mediums[0] .  " /> ;
}

Answer 2


$done = false;
foreach($file->image as $img) {
    foreach($img->attributes() as $key => $value) {
        if($key == "size" && $value == "medium") {
            echo "<img src = {$file->image} />";
            $done = true;
            break;
        }
    }
    if($done) { break; }
}

This might work, try it out and let me know.

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