Question

通过使用imagick创建一些缩略图，我发现了一个奇怪的错误。我的php脚本接收一些以base 64编码的图片，将其保存到文件夹中，然后为每个图片创建一个缩略图。这是我保存图片的代码：

$directory = getParentDirectory() . "/img/myfolder";

if (!is_dir($directory)) {
    $res = mkdir($directory, 0777, true);
}
$ifp = fopen($directory . "/back.jpg ", "wb");
fwrite($ifp, base64_decode($back));
fclose($ifp);

//dx
$ifp = fopen($directory . "/dx.jpg ", "wb");
fwrite($ifp, base64_decode($dx));
fclose($ifp);

//front
$ifp = fopen($directory . "/front.jpg ", "wb");
fwrite($ifp, base64_decode($front));
fclose($ifp);

//sx
$ifp = fopen($directory . "/sx.jpg ", "wb");
fwrite($ifp, base64_decode($sx));
fclose($ifp);

 createPreview($directory . "/back.jpg", $directory, "back");
createPreview($directory . "/dx.jpg", $directory, "dx");
createPreview($directory . "/sx.jpg", $directory, "sx");
createPreview($directory . "/front.jpg", $directory, "front");

这是crearePreview和getParentdirectyory代码：

function createPreview($image_path, $directory, $filename) {
    $img = new Imagick($image_path);
    $img->thumbnailImage(350, 350, true);
    $img->writeImage($directory . "/" . $filename . "_prev.jpg");
}


function getParentDirectory() {

    $currentwork = getcwd();
    $parent = basename(dirname($_SERVER['PHP_SELF']));
    return str_replace("/" . $parent, "", $currentwork);
}

但是当我调用createPreview时出现此错误：

11-18 10:32:28.662: I/JSON(23562): <b>Fatal error</b>:  Uncaught exception 'ImagickException' with message 'unable to open image `/web/htdocs/www.mysite.it/home/old-site/sitefolder/img/myfolder/back.jpg': No such file or directory' in /web/htdocs/www.mysite.it/home/old-site/sitefolder/scripts/load_haircut_new.php:221

但如果我检查内部/ img文件夹，所有图片（原始大小）都已保存，但没有预览。想象力为何无法找到它？

想象PHP：没有这样的文件或目录

0 个答案: