Guzzle: handle 400 bad request

问题:

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,

回答1:

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->getResponse()->getStatusCode() == '400') {
            echo "Got response 400";
    }

    // You can check for whatever error status code you need 

} catch (\Exception $e) {

    // There was another exception.

}

回答2:

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

Use http_errors => false option with the request.