Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
749 views
in Technique[技术] by (71.8m points)

php - Guzzle: handle 400 bad request

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,

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

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

A GuzzleHttpExceptionClientException 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 GuzzleHttpClient;
use GuzzleHttpExceptionRequestException;

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.

}

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...