在 PHP 中转换 URL 会导致凌空错误

Question

嘿，我有这样的代码将 url 目录保存到数据库并将文件传输到目录

   $app->post('/gambar', function() use($app) {
            // check for required params

        verifyRequiredParams(array('directory'));
        if (isset($_FILES['image'])) {
            if(isset($_POST['directory'])){
                $directory = $app->request->post('directory');
                $full_directory_path = '../' . $directory;
                //Pengecekan folder, sudah tersedia atau belum
                if(!is_dir($full_directory_path)){
                    //Pembuatan folder baru
                    mkdir($full_directory_path, 0777, true);
                }
                //Menentukan tempat file akan disimpan
                $target_path = $full_directory_path . '/' . basename($_FILES['image']['name']);
                if (!move_uploaded_file($_FILES['image']['tmp_name'], $target_path)) {
                    //File gagal dipindahkan ke server, biasanya karena folder yang dituju tidak tersedia
                    $response['kode'] = 1;
                    $response['pesan'] = "File tidak dapat dipindahkan ke server";
                    echo json_encode($response);
                }else{
                    // File berhasil diupload
                    $response['kode'] = 2;
                    $response['gili'] = $target_path;
                    $response['pesan'] = "File berhasil diupload";
                    echo json_encode($response);
                }
            }else{
            }
        } else {
            //Jika file tidak terkirim dari android
            $response['kode'] = 0;
            $response['pesan'] = 'File tidak terkirim ke server';
            echo json_encode($response);
        }
        $user_id = 1;           
        // $gambar = $app->request->put('gambar');

        $db = new DbHandler();
        $res = array();
        // updating task
        $result = $db->updateGambar($target_path,$user_id);
        if ($result) {
            // task updated successfully
            $res["error"] = false;
            $res["message"] = "Task updated successfully";
        } else {
            // task failed to update
            $res["error"] = true;
            $res["message"] = "Task failed to update. Please try again!";
        }
        echoRespnse(200, $res);
    });

我想更改为此内容，

以便它保存完整的URL，而不仅仅是此"../'

$full_directory_path = 'http://192.168.0.13/task_manager/' . $directory;

但是当我用完整的 url 更改它时，它会在 Colley Android 中给我错误。 谁能知道这段代码出了什么问题？

日志中的错误只是这个

03-22 16:01:37.148 32391-19996/com.anakacara.anakacara E/Volley: [773] BasicNetwork.performRequest: Unexpected response code 500 for http://192.168.0.13/task_manager/v1/gambar

Answer 1

如果你想改变网址，为什么不做一个新变量：

$example = 'http://192.168.0.13/task_manager/'.$directory. '/' . basename($_FILES['image']['name']);

并将此变量发送到您的数据库。

我想在你的凌空抽射中需要'../' 这要代码工作，所以它给你 500 错误