cURL命令到Guzzle

Question

I need to convert a cURL command like this into Guzzle-way:

curl -XPOST "https://api/v1.0/endpoint" -F "file=@img.jpg"

This is what I'm trying so far:

$httpClient = new \GuzzleHttp\Client;
$req = $httpClient->createRequest('POST', $url), []);
$postBody = $req->getBody();
$postBody->addFile(new \GuzzleHttp\Post\PostFile('photo', fopen(storage_path() . '/' . $filename, 'r')));
$response = $httpClient->send($req);

But I'm not getting the same response as with the previous command, which is something like:

{"id":5378678,"url":"ui/54/68/97/24/img.jpg"}

I'm getting a GuzzleHttp\\Message\\Response object, but I'm not being able to find the id and url attributes in there.

Any help will be appreciated!

Answer 1

只是注意到我可以做$response->json()来获得我想要的东西。