GuzzleHttp\Client 400 bad request on Laravel 5.5 - php

Question

I am sending a POST request to an API, Curl returns 200 and the correct response.
When Implementing with GuzzleHttp\Client, I get a 400 Bad request, what is wrong with my formatting.
here is my code using Laravel Returns 400 Bad Request:
$client = new Client();
$URI = 'http://api.example.com';
$params['headers'] = ['Content-Type' => 'application/json',
'apikey' => config('app._api_key'),
'debug' => true
];
$params['form_params'] = [
'sender' => 'Test_sender',
'recipient' => config('app.test_recipient'),
'message_body' => 'Test body'
];
return $response = $client->post($URI, $params);
Curl (Returns 200):
curl -X POST --header 'Content-Type: application/json' --header 'Accept: application/json' --header 'apikey: 212121212’ -d '{ "message_body": "test","sender": "2018","recipient": “4453424141” }' 'http://api.example.com'

Answer 1

Try the below code:
$client = new \GuzzleHttp\Client(['headers' => ['Content-Type' => 'application/json',
'apikey'=> config('app._api_key'),
'debug' => true
]
]);
$URI = 'http://api.example.com';
$body['sender']='Test_sender';
$body['recipient']=config('app.test_recipient');
$body['message_body']='Test body';
$body=json_encode($body);
$URI_Response = $client->request('POST',$URI,['body'=>$body]);
$URI_Response =json_decode($URI_Response->getBody(), true);
return $URI_Response;
Note: I would suggest you to handle error please refer GuzzleDocumentation

Answer 2

That is proper error handling:
use GuzzleHttp\Client;
use GuzzleHttp\Exception\RequestException;
try {
$response = $client->get(YOUR_URL, [
'connect_timeout' => 10
]);
// Here the code for successful request
} catch (RequestException $e) {
// Catch all 4XX errors
// To catch exactly error 400 use
if ($e->getResponse()->getStatusCode() == '400') {
echo "Got response 400";
}
// You can check for whatever error status code you need
} catch (\Exception $e) {
// There was another exception.
}
Implementation: http://guzzle.readthedocs.org/en/latest/quickstart.html

Answer 3

You can handle errors like this
use GuzzleHttp\Client;
use GuzzleHttp\Exception\ClientException;
use GuzzleHttp\Exception\RequestException;
use Exception;
try{
$client = new Client();
$response = $client->request('POST', $url,[
'headers' => $header,
'form_params' => $form-params
]);
$body = $response->getBody();
$status = 'true';
$message = 'Data found!';
$data = json_decode($body);
}catch(ClientException $ce){
$status = 'false';
$message = $ce->getMessage();
$data = [];
}catch(RequestException $re){
$status = 'false';
$message = $re->getMessage();
$data = [];
}catch(Exception $e){
$this->status = 'false';
$this->message = $e->getMessage();
$data = [];
}
return ['status'=>$status,'message'=>$message,'data'=>$data];