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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,

mwafi
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2 Answers2

73

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.

}
userbloom
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Hpatoio
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23
$client->get('http://www.example.com/path/'.$path,
            [
                'allow_redirects' => true,
                'timeout' => 2000,
                'http_errors' => true
            ]);

Use http_errors => false option with the request.

adam
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  • This is the solution that worked for me. In my case, I have no idea why is the request will not go to exception. setting the http_errors to false and get the status code to handle exception is what i did. thx for this solution – Dave Cruise Jun 22 '19 at 05:54