I'd like to add some data to a Guzzle Http Request. There are file name, file content and header with authorization key.
$this->request = $this->client->request('POST', 'url', [
'multipart' => [
'name' => 'image_file',
'contents' => fopen('http://localhost:8000/vendor/l5-swagger/images/logo_small.png', 'r'),
'headers' =>
['Authorization' => 'Bearer uCMvsgyuYm0idmedWFVUx8DXsN8QzYQj82XDkUTw']
]]);
but I get error
Catchable Fatal Error: Argument 2 passed to GuzzleHttp\Psr7\MultipartStream::addElement() must be of the type array, string
given, called in vendor\guzzlehttp\psr7\src\MultipartStream.php on line 70 and defined in vendor\guzzlehttp\psr7\src\MultipartStream.php line 79
In Guzzle 6 documentation is something like this: http://docs.guzzlephp.org/en/latest/request-options.html#multipart
Who knows where I made a mistake?
Here is the solution. Header with access token should be outside multipart section.
$this->request = $this->client->request('POST', 'request_url', [
'headers' => [
'Authorization' => 'Bearer access_token'
],
'multipart' => [
[
'Content-type' => 'multipart/form-data',
'name' => 'image_file',
'contents' => fopen('image_file_url', 'r')
]
]
]);
As per the docs, "The value of multipart is an array of associative arrays", so you need to nest one level deeper:
$this->request = $this->client->request('POST', 'url', [
'multipart' => [
[
'name' => 'image_file',
'contents' => fopen('http://localhost:8000/vendor/l5-swagger/images/logo_small.png', 'r'),
'headers' => ['Authorization' => 'Bearer uCMvsgyuYm0idmedWFVUx8DXsN8QzYQj82XDkUTw']
]
]
]);
try this
works for me
use GuzzleHttp\Client;
use GuzzleHttp\Psr7\Utils;
$this->client = new Client([
'base_uri' => 'https://baseurl'
]);
$body = Utils::tryFopen($tempPath . $fileName, 'r');
$res = $this->client->request(
'POST',
'url',
[
'headers' => [
...
],
'body' => $body
]
);
Related
i use Fast Excel library from https://github.com/rap2hpoutre/fast-excel, can i put return from ->download('file.xlsx') to Guzzle post request? with this code i'm not get anything
$httpClient = new \GuzzleHttp\Client([
'headers' => [
'Authorization' => mdmTokenGenerate($token),
'Accept' => 'application/json'
],
'verify' => false
]);
$response = $httpClient->post(
"{$client->domain}/api/push", [
'multipart' => [
[
'name' => 'file',
'contents' => (new FastExcel(MasterNumbers::limit(1000)->get()))->download('file.xlsx')
]
]
]
);
How can I pass the $request->file(...) to Guzzle thru POST request?
Or is it even possible? Should I upload it first?
This is my attempt:
$client = new Client([ 'base_uri' => 'http://api.domain.com']);
$response = $client->post(
'/api/dosomething',
[
'multipart' => [
[
'Content-type' => 'multipart/form-data',
'name' => 'photo_1',
'contents' => $request->file('photo_1')
]
]
]
);
This is not working. The response body is just blank.
I found the solution below (We need to specify the path of the temporary file in the temp folder with the code $file->getPathName()):
$client = new Client([ 'base_uri' => 'http://api.domain.com']);
$file = $request->file('photo_1')
$response = $client->post(
'/api/dosomething',
[
'multipart' => [
[
'Content-type' => 'multipart/form-data',
'name' => 'photo_1',
'contents' => fopen($file->getPathName(), 'r')
]
]
]
);
I am having a hard time converting cURL to Guzzle6. I'm want to send a name and reference UUID via JSON AND the contents of an XML file to process to a REST endpoint.
cURL
curl -H 'Expect:' -F
'request={"name":"test", "reference":"870e0320-021e-4c67-9169-d4b2c7e5b9c9"}'
-F 'file=#sample.xml' http://ec2-48-88-173-9.us-east-1.compute.amazonaws.com:8180/rs/process
Guzzle
$client = new Client(['debug' => true]);
$request = $client->request('POST',
'http://ec2-48-88-173-9.us-east-1.compute.amazonaws.com:8180/rs/process', [
'multipart' => [
[
'name' => 'data',
'contents' => "{'name':'test','reference':870e0320-021e-4c67-9169-d4b2c7e5b9c9}",
'headers' => ['Content-Type' => 'application/json']
],
[
'name' => 'file',
'contents' => fopen('sample.xml', 'r'),
'headers' => ['Content-Type' => 'text/xml']
],
]
]
);
$response = $request->getBody()->getContents();
Also, I'm not sure what the 'name' fields should be ('name' => 'data'), etc.
This is the Guzzle equivalent of your curl command:
$client = new Client(['debug' => true]);
$request = $client->request('POST',
'http://ec2-48-88-173-9.us-east-1.compute.amazonaws.com:8180/rs/process', [
'multipart' => [
[
'name' => 'request',
'contents' => "{'name':'test','reference':870e0320-021e-4c67-9169-d4b2c7e5b9c9}",
],
[
'name' => 'file',
'contents' => fopen('sample.xml', 'r'),
],
]
]
);
$response = $request->getBody()->getContents();
For the file Guzzle will specify the appropriate content type, as curl does. Name for the first part is request — from -F
'request={"name":"test", "reference":"870e0320-021e-4c67-9169-d4b2c7e5b9c9"}'
I am using guzzle 6.3 to post a file along with some data to my api built on laravel 5.5. When i post the data, i am not able to get the data sent to the api except the file posted.
Client Side
$client = new Client([
'headers' => [ 'Content-Type' => 'application/json' ]
]);
$response = $client->request('POST',$url, [
'multipart' => [
[
'name' => 'body',
'contents' => json_encode(['documentation' => 'First Doc', 'recipient' => ['78011951231']]),
'headers' => ['Content-Type' => 'application/json']
],
[
'name' => 'file',
'contents' => fopen('/path/public/media/aaaah.wav', 'r'),
'headers' => ['Content-Type' => 'audio/wav']
],
],
]);
echo($response->getBody()->getContents());
API Controller
if (($Key)) {
return response()->json([
'status' => 'success',
'documenation' => $request->get('documentation'),
'recipient' => $request->get('recipient'),
'file' =>$request->get('file'),
'media'=>$request->hasFile('file')
]);
}
Response
{"status":"error","documentation":null,,"recipient":null,"file":null,"media":true}
Why am i getting NULL returned for the data that i am posting? Could it be because of the file that i am posting ?
My working Guzzle5 code looks roughly as follows:
$guzzle = new \GuzzleHttp\Client();
$request = $guzzle->createRequest('POST', $url);
$request->setHeader('Authorization', 'Bearer ' . $token);
$postBody = $request->getBody();
$postBody->setField('name', 'content');//several times
if (check for file) {
$postBody->addFile(new \GuzzleHttp\Post\PostFile('name', fopen(...));
}
$response = $guzzle->send($request);
What with setting a header and maybe adding a file, I’m not sure how to do this with Guzzle6.
Here an example from the official documentation how can you set headers and adding file into your POST request with Guzzle 6:
$client = new \GuzzleHttp\Client();
$client->post('/post', [
'multipart' => [
[
'name' => 'foo',
'contents' => 'data',
'headers' => ['X-Baz' => 'bar']
],
[
'name' => 'baz',
'contents' => fopen('/path/to/file', 'r')
],
[
'name' => 'qux',
'contents' => fopen('/path/to/file', 'r'),
'filename' => 'custom_filename.txt'
],
]
]);
The multipart option sets the body of the request to a multipart/form-data form, if you don't need to work with files you can just use form_params instead of multipart option.
Any headers you can easy set with help headers option.
Additional info you can find here Guzzle Upgrade Guide (5.0 to 6.0)
Here is some code copied from one of my projects:
$client = new GuzzleHttp\Client();
$url = 'someurl.com/api';
$body = json_encode([
'variable1' => 'this',
'variable2' => 'that'
]);
$response = $client->post($url, [
'headers' => [
'header_variable1' => 'foo',
'header_variable2' => 'bar'
],
'json' => true,
'timeout' => 3,
'body' => $body
]);
$data = $response->json();