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Laravel 传递的Guzzle 6异步请求返回参数1必须是数组类型,字符串为,_Laravel_Guzzle_Guzzle6 - Fatal编程技术网

Laravel 传递的Guzzle 6异步请求返回参数1必须是数组类型,字符串为,

Laravel 传递的Guzzle 6异步请求返回参数1必须是数组类型,字符串为,,laravel,guzzle,guzzle6,Laravel,Guzzle,Guzzle6,我正在尝试进行异步POST调用,但guzzle调用返回以下错误: “Request.php第220行中的ErrorException: 传递给Symfony\Component\HttpFoundation\Request::\uu construct()的参数1必须是数组类型,字符串给定,在第87行的C:\Program Files…\app\Http\Controllers\confirAccountController.php中调用并定义 在ConfirmAccountController

我正在尝试进行异步POST调用,但guzzle调用返回以下错误:

Request.php第220行中的ErrorException: 传递给Symfony\Component\HttpFoundation\Request::\uu construct()的参数1必须是数组类型,字符串给定,在第87行的C:\Program Files…\app\Http\Controllers\confirAccountController.php中调用并定义

在ConfirmAccountController中:

$client = new Client(['base_uri' => 'correct_api_address']);

$request = new Request('POST', 'testpromoboiler/updateUser',  [
        'query' => ['token' => $user->sdg_token ,
                    'address' => $user->address ,
                  ]
    ]);

$promise = $client->sendAsync($request)->then(function ($response) {
    echo 'I completed! ' . $response->getBody();
});

$promise->wait();
为什么我会犯这个错误

我做了另外两个同步调用,一切都很顺利


感谢

代码实例化了Laravel框架
请求

改用guzzles请求方法:

$request = $client->request('POST', 'testpromoboiler/updateUser',  [
        'query' => ['token' => $user->sdg_token ,
                    'address' => $user->address ,
                  ]
    ]);

Symfony\Component\HttpFoundation\Request
中,前两个参数是数组。第一个是GET参数,第二个是POST参数。不确定
Client
对象在代码中引用了什么,但您可能应该在那里添加方法和URL,而不是在请求对象中。