Guzzle 不抛出异常

Question

我正在围绕 themoviedb.org api 构建一个类包装器。我在请求中使用 guzzle 7，但它似乎没有抛出任何异常。

namespace App\Classes;

use App\Models\Movie;
use App\Models\Series;
use GuzzleHttp\Client;

use GuzzleHttp\Exception\GuzzleException;
use GuzzleHttp\Handler\CurlHandler;
use GuzzleHttp\HandlerStack;
use GuzzleHttp\Middleware;
use GuzzleHttp\Psr7\Uri;

use Psr\Http\Message\RequestInterface;

class TMDBScraper{

private string $apiKey;
private string $language;
private Client $client;
private const API_URL = "http://api.themoviedb.org/3/";
private const IMAGE_URL = "http://image.tmdb.org/t/p/";
private const POSTER_PATH_SIZE = "w500";
private const BACKDROP_PATH_SIZE = "original";


public function __construct(string $apiKey = "default_api_key") {
    $this->apiKey = $apiKey;
    $this->language = app()->getLocale();

    $handlerStack = new HandlerStack(new CurlHandler());
    $handlerStack->unshift(Middleware::mapRequest(function (RequestInterface $request) {
        return $request->withUri(Uri::withQueryValues($request->getUri(), [
            'api_key' => $this->apiKey,
            'language' => $this->language
        ]));
    }));

    $this->client = new Client([
        'base_uri' => self::API_URL,
        'handler' => $handlerStack
    ]);

}

public function search($screenplayType, $query): ?array {

    try {

        $response = json_decode($this->client->get('search/' . $screenplayType, [
            'query' => compact('query')
        ])->getBody());

        return $this->toModel($response, $screenplayType);

    } catch (GuzzleException $e) {
        echo $e->getMessage();
        return null;
    }

}

... more code }

我尝试使用错误的 api 密钥，但没有抛出客户端异常。我也尝试将http_errors设置为true，应该默认设置，但也没有用。

Answer 1

你可以试试这个代码：

$handler = new CurlHandler();
$stack = HandlerStack::create($handler);