PHP裁剪在子文件夹中找到的图像以显示为拇指

Question

I did found this script somewhere and it seems to be working. The only problem is that it outputs a thumb exactly the same size ratio than the original even though they are supposed to be cropped to a square...

$dir = "*/";
$images = glob($dir."main.jpg" );
echo '<div class="projects-container">'; 
foreach( $images as $image ) {
$dn = dirname($image);
$thumbsDir = $dn; // path to the thumbnails destination directory

    $imageName = "main.jpg"; // returns "cheeta.jpg"
    $thumbnail = $thumbsDir.$imageName; // thumbnail full path and name, i.e "./gallery/thumbs/cheeta.jpg"
    // for each image, get width and height
    $imageSize = getimagesize( $image ); // image size 
    $imageWidth = $imageSize[0];  // extract image width 
    $imageHeight = $imageSize[1]; // extract image height
    // set the thumb size
    if( $imageHeight > $imageWidth ){
        // images is portrait so set thumbnail width to 100px and calculate height keeping aspect ratio
        $thumbWidth = 200;
        $thumbHeight = floor( $imageHeight * ( 200 / imageWidth ) );           
        $thumbPosition  = "margin-top: -" . floor( ( $thumbHeight - 200 ) / 2 ) . "px; margin-left: 0";
    } else {
        // image is landscape so set thumbnail height to 100px and calculate width keeping aspect ratio
        $thumbHeight = 200;
        $thumbWidth = floor( $imageWidth * ( 200 / $imageHeight ) ); 
        $thumbPosition  = "margin-top: 0; margin-left: -" . floor( ( $thumbWidth - 200 ) / 2 ) . "px";
    } // END else if
    // verify if thumbnail exists, otherwise create it
    if ( !file_exists( $thumbnail ) ){
        $createFromjpeg = imagecreatefromjpeg( $image );
        $thumb_temp = imagecreatetruecolor( $thumbWidth, $thumbHeight );
        imagecopyresized( $thumb_temp, $createFromjpeg, 0, 0, 0, 0, $thumbWidth, $thumbHeight, $imageWidth, $imageHeight );
        imagejpeg( $thumb_temp, $thumbnail );
    } // END if()

echo '<div class="projects">';
echo '<div class="projects-img-container">';
echo" <a href='".$dn."'><img class='img-projet' src='". '/projects/' . $thumbnail . "'/></a>";
echo '</div>';
echo '</div>';
}
echo '</div>';
?> 

Any idea of what can be wrong?

Thank you!

Answer 1

You have absolutely NO error handling and simply assuming nothing could ever go wrong, so these two lines:

$dir = "*/";
$image2 = imagecreatefromjpeg($dir."main.jpg");

will be exactly equivalent to

$image2 = imagecreatefromjpeg("*/main.jpg");

icfj() does NOT accept wildcards, period.

Since you don't check for errors, you never see the boolean FALSE that icfj() returned, signifying failure.

Answer 2

Sorry for the misunderstanding of the situation... I've forgot to add some CSS style in the IMG tag so that the position of the image corresponds to $thumbPosition and to the projects-img-container div to have a width and height of 200px.

Everything works as supposed now!