找不到类'TusPhp\Tus\Client‘

Question

我正在使用PHP中的vimeo-api库，当我上传文件时，它给我这个错误。我不知道为什么这个错误会出现在我身上，请看看这个，让我知道为什么这个错误是常见的错误:类'TusPhp\Tus\Client‘找不到这里是我的代码

<?php
use Vimeo\Vimeo;
use Vimeo\Exceptions\VimeoUploadException;
/**
 *   Copyright 2013 Vimeo
 *
 *   Licensed under the Apache License, Version 2.0 (the "License");
 *   you may not use this file except in compliance with the License.
 *   You may obtain a copy of the License at
 *
 *       http://www.apache.org/licenses/LICENSE-2.0
 *
 *   Unless required by applicable law or agreed to in writing, software
 *   distributed under the License is distributed on an "AS IS" BASIS,
 *   WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
 *   See the License for the specific language governing permissions and
 *   limitations under the License.
 */
$config = require(__DIR__ . '/init.php');
if (empty($config['access_token'])) {
    throw new Exception(
        'You can not upload a file without an access token. You can find this token on your app page, or generate ' .
        'one using `auth.php`.'
    );
}
// Instantiate the library with your client id, secret and access token (pulled from dev site)
$lib = new Vimeo($config['client_id'], $config['client_secret'], $config['access_token']);
// Create a variable with a hard coded path to your file system
$file_name = "C:\Users\BASIT\Downloads\SampleVideo_1280x720_1mb.mp4";
echo 'Uploading: ' . $file_name . "\n";
try {
    // Upload the file and include the video title and description.
    $uri = $lib->upload($file_name, array(
        'name' => 'Vimeo API SDK test upload',
        'description' => "This video was uploaded through the Vimeo API's PHP SDK."
    ));
    // Get the metadata response from the upload and log out the Vimeo.com url
    $video_data = $lib->request($uri . '?fields=link');
    echo '"' . $file_name . ' has been uploaded to ' . $video_data['body']['link'] . "\n";
    // Make an API call to edit the title and description of the video.
    $lib->request($uri, array(
        'name' => 'Vimeo API SDK test edit',
        'description' => "This video was edited through the Vimeo API's PHP SDK.",
    ), 'PATCH');
    echo 'The title and description for ' . $uri . ' has been edited.' . "\n";
    // Make an API call to see if the video is finished transcoding.
    $video_data = $lib->request($uri . '?fields=transcode.status');
    echo 'The transcode status for ' . $uri . ' is: ' . $video_data['body']['transcode']['status'] . "\n";
} catch (VimeoUploadException $e) {
    // We may have had an error. We can't resolve it here necessarily, so report it to the user.
    echo 'Error uploading ' . $file_name . "\n";
    echo 'Server reported: ' . $e->getMessage() . "\n";
} catch (VimeoRequestException $e) {
    echo 'There was an error making the request.' . "\n";
    echo 'Server reported: ' . $e->getMessage() . "\n";
}

Answer 1

您好，如果您使用框架php，如codeigniter，那么您还可以在自动加载文件中声明，并且您的路径$file_name = "C:\Users\BASIT\Downloads\SampleVideo_1280x720_1mb.mp4"；不工作，只需应用单引号，则它可以正常工作。