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,
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.
}
$client->get('http://www.example.com/path/'.$path,
[
'allow_redirects' => true,
'timeout' => 2000,
'http_errors' => true
]);
Use http_errors => false option with the request.
来源:https://stackoverflow.com/questions/25040436/guzzle-handle-400-bad-request