问拉威尔:我从一个API收到了歌曲文件"form-data“，然后我需要将该文件转发到另一个API。有什么想法可以转发吗？

EN

try{
$path = 'var/www/html/myproject/public/file.txt';//your file path
if (!empty($path) && file_exists($path)) {

   $guzzleResponse = $client->post($api_url, [
       'multipart' => [
                        [
                            'name' => 'file',// it is the name as specfied in the payload of api_url
                            'contents' => fopen($path, 'r')// or you can use file_get_contents()
                        ]
                    ],
                    'headers' => $headers
       ]);
}
if ($guzzleResponse->getStatusCode() == 200) {
   $response = json_decode($guzzleResponse->getBody());// whatever you want to do with response
}
}catch(RequestException $e){
    return $e;  //You can also handle specific status codes here using eg $e->getResponse()->getStatusCode() == '400'
}

[
     'Accept'                => 'application/json',
     'Authorization'         => 'Bearer '. $userToken,
]