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')
]
]
]
);
Related
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 ?
I'm new with Guzzle and I'm trying to make a REST request to sign PDF file. The provider says :
you need to use BASIC authentication
the request must be a POST request
the mimetype should be multipart/form-data
the file sent must be application/octet-stream and its name should be "file"
the data sent must be application/json and its name should be "data"
The system returns a response which contains the signed PDF file and type is application/octet-stream
This is the code I tested with Guzzle, but the provider says that the type mime sent in application/pdf. How can I "force" the mimetype for the PDF file ?
$client = new Client([
'auth' => ['login', 'password'],
'debug' => true,
'curl' => [
CURLOPT_PROXY => '192.168.1.232',
CURLOPT_PROXYPORT => '8080',
CURLOPT_PROXYUSERPWD => 'username:password',
],
]);
$boundary = 'my_custom_boundary';
$multipart = [
[
'name' => 'data',
'contents' => "{'nomDocument':'documentTest.pdf','externalid':'123456'}",
'Content-Type' => 'application/json'
],
[
'name' => 'file',
'contents' => fopen('documentTest.pdf', 'r'),
'Content-Type' => 'application/octet-stream'
],
];
$params = [
'headers' => [
'Connection' => 'close',
'Content-Type' => 'multipart/form-data; boundary='.$boundary,
],
'body' => new GuzzleHttp\Psr7\MultipartStream($multipart, $boundary),
];
try{
$response = $client->request('POST', 'https://server.com/api/sendDocument', $params);
} catch (RequestException $e) {
echo Psr7\str($e->getRequest());
if ($e->hasResponse()) {
echo Psr7\str($e->getResponse());
}
}
Thank you for your help.
You have to pass the Content-Type in headers
$multipart = [
[
'name' => 'data',
'contents' => "{'nomDocument':'documentTest.pdf','externalid':'123456'}",
'headers' => [ 'Content-Type' => 'application/json']
],
[
'name' => 'file',
'contents' => fopen('documentTest.pdf', 'r'),
'headers' => [ 'Content-Type' => 'application/octet-stream']
],
];
in Guzzle Documentation say that you can specify headers for every multipart data.
If you not set header Guzzle put a Content-Type for you based on file.
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
]
);