php - 使用 Guzzle 6 将文件上传到 API 端点

Question

我可以使用 Postman 将文件上传到 API 端点。

我正在尝试将其转换为从表单上传文件，使用 Laravel 上传文件并使用 Guzzle 6 发布到端点。

它在 Postman 中的截图(我故意省略了 POST URL)

下面是当您在 POSTMAN 中单击“生成代码”链接时它生成的文本:

POST /api/file-submissions HTTP/1.1
Host: strippedhostname.com
Authorization: Basic 340r9iu34ontoeioir
Cache-Control: no-cache
Postman-Token: 6e0c3123-c07c-ce54-8ba1-0a1a402b53f1
Content-Type: multipart/form-data; boundary=----WebKitFormBoundary7MA4YWxkTrZu0gW

----WebKitFormBoundary7MA4YWxkTrZu0gW
Content-Disposition: form-data; name="FileContents"; filename=""
Content-Type: 


----WebKitFormBoundary7MA4YWxkTrZu0gW
Content-Disposition: form-data; name="FileInfo"

{ "name": "_aaaa.txt", "clientNumber": "102425", "type": "Writeoff" }
----WebKitFormBoundary7MA4YWxkTrZu0gW

下面是用于保存文件和其他信息的 Controller 函数。文件上传正确，我可以获取文件信息。

我认为我遇到的问题是使用正确的数据设置多部分和标题数组。

public function fileUploadPost(Request $request)
{
    $data_posted = $request->input();
    $endpoint = "/file-submissions";
    $response = array();
    $file = $request->file('filename');
    $name = time() . '_' . $file->getClientOriginalName();
    $path = base_path() .'/public_html/documents/';

    $resource = fopen($file,"r") or die("File upload Problems");

    $file->move($path, $name);

    // { "name": "test_upload.txt", "clientNumber": "102425", "type": "Writeoff" }
    $fileinfo = array(
        'name'          =>  $name,
        'clientNumber'  =>  "102425",
        'type'          =>  'Writeoff',
    );

    $client = new \GuzzleHttp\Client();

    $res = $client->request('POST', $this->base_api . $endpoint, [
        'auth' => [env('API_USERNAME'), env('API_PASSWORD')],
        'multipart' => [
            [
                'name'  =>  $name,
                'FileContents'  => fopen($path . $name, 'r'),
                'contents'      => fopen($path . $name, 'r'),
                'FileInfo'      => json_encode($fileinfo),
                'headers'       =>  [
                    'Content-Type' => 'text/plain',
                    'Content-Disposition'   => 'form-data; name="FileContents"; filename="'. $name .'"',
                ],
                // 'contents' => $resource,
            ]
        ],
    ]);

    if($res->getStatusCode() != 200) exit("Something happened, could not retrieve data");

    $response = json_decode($res->getBody());

    var_dump($response);
    exit();
}

我收到的错误，它是如何使用 Laravel 的调试 View 显示的屏幕截图:

Answer 1

您发布数据的方式有误，因此收到的数据格式不正确。

Guzzle docs :

The value of multipart is an array of associative arrays, each containing the following key value pairs:

name: (string, required) the form field name

contents:(StreamInterface/resource/string, required) The data to use in the form element.

headers: (array) Optional associative array of custom headers to use with the form element.

filename: (string) Optional string to send as the filename in the part.

使用上述列表之外的键并设置不必要的 header 而不将每个字段分隔到一个数组中将导致发出错误的请求。

$res = $client->request('POST', $this->base_api . $endpoint, [
    'auth'      => [ env('API_USERNAME'), env('API_PASSWORD') ],
    'multipart' => [
        [
            'name'     => 'FileContents',
            'contents' => file_get_contents($path . $name),
            'filename' => $name
        ],
        [
            'name'     => 'FileInfo',
            'contents' => json_encode($fileinfo)
        ]
    ],
]);