Guzzle: handle 400 bad request

mwafi picture mwafi · Jul 30, 2014 · Viewed 37.1k times · Source

I'm using Guzzle in Laravel 4 to return some data from another server, but I can't handle Error 400 bad request

 [status code] 400 [reason phrase] Bad Request

using:

$client->get('http://www.example.com/path/'.$path,
            [
                'allow_redirects' => true,
                'timeout' => 2000
            ]);

how to solve it? thanks,

Answer

Hpatoio picture Hpatoio · Jul 30, 2014

As written in Guzzle official documentation: http://guzzle.readthedocs.org/en/latest/quickstart.html

A GuzzleHttp\Exception\ClientException is thrown for 400 level errors if the exceptions request option is set to true

For correct error handling I would use this code:

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 ($e->hasResponse()){
        if ($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.

}