Question

I am trying to generate some HTML code to list some images for a slide show.

I arrived at the following idea:

function galGetTopPhotos()
{
    //path to directory to scan
    $directory = SITE_ROOT_PATH."/gallery/best/";

    //get all files
    $images = glob($directory . "*.*");

    //print each file name
    $ret = "";
    $ret .= '<div id="myslides">';        

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

    $ret .= '</div>';

    return $ret;
}

The problem is that it only works when I use root path for $directory…if I use URL it will not work. And it causes the images to not load. Here is what this code generates:

<div id="myslides">
<img src="D:/xampp/htdocs/mrasti/gallery/best/1.jpg" />
<img src="D:/xampp/htdocs/mrasti/gallery/best/10.jpg" />
</div>

So the question is, how to get the list of files so it generates img sources in http://127.0.0.1/.... format?

What I mean if I use the code like this it returns no file!

$directory ="http://127.0.0.1/mrasti/gallery/best/";

Answer 1

This looks like a job for PHP function basename. This takes a file path and returns only the final element of the path – in this case the actual name of the jpeg image.

You could amend your code so that it looks something like this:

$urlPath = "http://127.0.0.1/mrasti/gallery/best/";

...
...

foreach($images as $image)
{
    $relative_path = $urlPath.basename($image);
    $ret .= '<img src="'.$relative_path.'" />';
}

The above takes the path and appends the filename “example.jpg” to your image directory url