我有一个可以使用的 curl 片段，但是 Guzzle 出现 400 错误，为什么？

Question

谁能看到我在这里可能做错了什么？Curl 有效，但 Guzzle (v6.3) 出现 400 错误。

Curl 由 Postman 生成：

卷曲：

    $curl = curl_init();

    curl_setopt_array($curl, array(

        CURLOPT_URL => "https://unnamed-website.com/api2/v2/charges?chargeId=273628584&paymentGate=inovio",

        CURLOPT_RETURNTRANSFER => true,

        CURLOPT_ENCODING => "",

        CURLOPT_MAXREDIRS => 10,

        CURLOPT_TIMEOUT => 0,

        CURLOPT_FOLLOWLOCATION => true,

        CURLOPT_HTTP_VERSION => CURL_HTTP_VERSION_1_1,

        CURLOPT_CUSTOMREQUEST => "GET",

        CURLOPT_HTTPHEADER => array(

            "app-token: toktoktoktoktotkoktoktokt",

            "Content-type: application/json",

            "Authorization: Basic MDFkNWJiMTllYTdlZjVjODYzYTVjYjI3ZjUzNWY4NmM6YzFhMzRlNWJkZGYyMzljZTFmZDcwZjNiZDk0Y2Q4ZjA="

        ),

    ));

    $response = curl_exec($curl);

输出：

大嘴：

$guzzle = new Client();

$request = $guzzle->get("https://unnamed-website.com/api2/v2/charges?chargeId=273628584&paymentGate=inovio",

[

    'headers' => [

        "app-token" => "toktoktoktoktotkoktoktokt",

        "Content-type" => "application/json",

        "Authorization" => "Basic MDFkNWJiMTllYTdlZjVjODYzYTVjYjI3ZjUzNWY4NmM6YzFhMzRlNWJkZGYyMzljZTFmZDcwZjNiZDk0Y2Q4ZjA="

    ]

]);

$response = $request->send();

输出：

   Client error response

   [status code] 400

   [reason phrase] Bad Request

Answer 1

Guzzle 可以这样调用：<?php&nbsp; &nbsp; $client = new GuzzleHttp\Client();&nbsp; &nbsp; $response = $client->request('GET',"https://unnamed-website.com/api2/v2/charges?chargeId=273628584&paymentGate=inovio",[&nbsp; &nbsp; &nbsp; &nbsp; 'headers' => [&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; "app-token" => "toktoktoktoktotkoktoktokt",&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; "Content-type" => "application/json",&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; "Authorization" => "Basic MDFkNWJiMTllYTdlZjVjODYzYTVjYjI3ZjUzNWY4NmM6YzFhMzRlNWJkZGYyMzljZTFmZDcwZjNiZDk0Y2Q4ZjA="&nbsp; &nbsp; &nbsp; &nbsp; ]&nbsp; &nbsp; ]);&nbsp; &nbsp; $response = $request->send();&nbsp; &nbsp; $result = json_decode($response->getBody()->getContents());?>

Answer 2

当我在 slack 上发布它时，我的开发人员同事们对此进行了深入研究，他们发现了问题。我在&nbsp;use Guzzle\Http\Client;应该使用的时候使用use GuzzleHttp\Client;...前者是可用的，因为在一个巨大的专有框架内有许多不受管理的供应商包。