php - 如何使用Laravel在Controller中捕获500错误

Question

我需要连接到API，所以我编写了一个函数:

try {
    $res4 = $client3->post('https://api.example.co.uk/Book', [
        'headers' => [
            'Accept' => 'application/json',
            'Content-Type' => 'application/json',
            'Authorization' => 'Bearer ajhsdbjhasdbasdbasd',
        ],
        'json' => [
            'custFirstName' => $FirstName,
            'custLastName' => $Surname,
            'custPhone' => $Mobile,
            'custEmail' => $Email,
        ]
    ]);
} catch (GuzzleHttp\Exception\ClientException $e) {
    $response = $e->getResponse();
    $result = json_decode($response->getBody()->getContents());
    $item->update(['status' => 'Problems at step3']);
    Mail::raw('Problem at STEP 3', function ($message) use ($serial) {
        $message->from('asd.asd@gmail.com', 'asd.asd@gmail.com');
        $message->subject('We got a problem etc.');
        $message->to('john.smith@gmail.com');
    });
}

如您所见，我需要调用API，但是在API中断的情况下，我会编写catch函数。

但是现在，当API关闭并且API返回“500 Internal Error”时，此函数就崩溃了...

我的问题是为什么不捕获它？

我如何处理错误-当API中断或请求错误时...为什么catch {}不起作用？

更新:这是我的laravel.log

[2018-10-25 14:51:04] local.ERROR: GuzzleHttp\Exception\ServerException: Server error: `POST https://api.example.co.uk/Book` resulted in a `500 Internal Server Error` response:
{"message":"An error has occured. Please contact support."}
 in /home/public_html/vendor/guzzlehttp/guzzle/src/Exception/RequestException.php:107
Stack trace:
#0 /home/public_html/vendor/guzzlehttp/guzzle/src/Middleware.php(65): GuzzleHttp\Exception\RequestException::create(Object(GuzzleHttp\Psr7\Request), Object(GuzzleHttp\Psr7\Response))
#1 /home/public_html/vendor/guzzlehttp/promises/src/Promise.php(203): GuzzleHttp\Middleware::GuzzleHttp\{closure}(Object(GuzzleHttp\Psr7\Response))

Answer 1

问题是这里的 namespace ，而不是:

} catch (GuzzleHttp\Exception\ClientException $e) {

您应该使用:

} catch (\GuzzleHttp\Exception\ClientException $e) {

否则，PHP会假定该类在当前的 namespace 中，因此实际上，当您使用 GuzzleHttp\Exception\ClientException时，实际上您可能使用了 App\Http\Controllers\GuzzleHttp\Exception\ClientException，并且此类异常显然不会被Guzzle抛出。