从图像路径获取文件名（来自函数）

Question

I am trying to get a file name from an image path using basename and echo it inside the title tag but I can't get it to work since I am echoeing the image path from a functions file.

What I am trying to do now (in product.php):

$path = ProductBekijkPlaatje($contenti[0]['images']);

$basename = basename($path);

echo $basename;

But this breaks the function and only echoes the filepath, while the actual image is lost.

This is the function I am calling (in functions.php):

function ProductBekijkPlaatje($plaatje) {

    $path = $img->image_intro;

    $basename = basename($path);

    $img = json_decode($plaatje);

    if ($img->image_intro == '') {
        $image = '<img src="images/no-img.jpg" alt="">';
    } else {

        $image = '<img class="shorterimageimg" title="'.$basename.'" src="cms/'.$img->image_intro.'" alt="'.$img->image_intro_alt.'" >';

    }
    return $image;

}

I also try to show the image name in the title inside the function, but this isn't working either. What am I doing wrong?

Answer 1

it doesn't work in your function because you try to access the property of the object before you do json_decode():

$path = $img->image_intro;     // $img doesn't exist here
$basename = basename($path);
$img = json_decode($plaatje);  // here $img is created

Just move the json_decode() to the front:

$img = json_decode($plaatje);
$path = $img->image_intro;
$basename = basename($path);

Now, you could modify your function to return an array with different information:

function ProductBekijkPlaatje($plaatje) {
    $img = json_decode($plaatje);
    $return = array();
    $return['path'] = $img->image_intro;
    $return['basename'] = basename($return['path']);
    if ($img->image_intro == '') {
        $return['image'] = '<img src="images/no-img.jpg" alt="">';
    } else {
        $return['image'] = '<img class="shorterimageimg" title="'.$return['basename'].'" src="cms/'.$img->image_intro.'" alt="'.$img->image_intro_alt.'" >';
    }
    return $return;
}

Then you can use this array later:

$image = ProductBekijkPlaatje($contenti[0]['images']);
echo $image['image']; // contains the html output
echo $image['basename']; // contains the basename only