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我正在从另一个现有图像中裁剪图像。它适用于 jpg、jpeg、gif 正常工作。但它不适用于 png 图像文件。给出以下错误。

Warning (2): imagecreatefrompng() [function.imagecreatefrompng]: '/var/www/shareme/app/webroot//documents/users/MTI5NzkyMjQzMmZmLWxvZ28tYmlnLnBuZw.png' is not a valid PNG file [APP/controllers/components/jq_imgcrop.php, line 80]

这是代码。

function resizeThumbnailImage($thumb_image_name, $image, $width, $height, $start_width, $start_height, $scale){
        $newImageWidth = ceil($width * $scale);
        $newImageHeight = ceil($height * $scale);

        $ext = strtolower(substr(basename($image), strrpos(basename($image), ".") + 1));
        $source = "";
        if($ext == "png"){
            $source = imagecreatefrompng($image);
        }elseif($ext == "jpg" || $ext == "jpeg"){
            $source = imagecreatefromjpeg($image);
        }elseif($ext == "gif"){
            $source = imagecreatefromgif($image);
        }
        $newImage = imagecreatetruecolor($newImageWidth,$newImageHeight);

        imagecopyresampled($newImage,$source,0,0,$start_width,$start_height,$newImageWidth,$newImageHeight,$width,$height);
        imagejpeg($newImage,$thumb_image_name,90);
        chmod($thumb_image_name, 0777);
        return $thumb_image_name;
    }

你对这个问题有任何想法吗？

4

1 回答 1

3

也许您还想检查 mime 类型而不是潜在的错误扩展。

$handle = finfo_open(FILEINFO_MIME); 
$mime_type = finfo_file($handle, $src);
$mime_type = mime_content_type($src);
switch(strtolower($mime_type)) {
    case 'image/gif':
        $img = imageCreateFromGIF($src);
        break;
    case 'image/png':
        $img = imageCreateFromPNG($src);
        break;
}
于 2011-02-17T08:13:48.607 回答