【发布时间】:2021-07-23 19:46:20
【问题描述】:
我无法理解如何将此 cURL 修改为 Laravel 5.8,得到 {"code":"11", "message":"invalid Request found", "status":"DECLINED"} 响应。
这是我的cURL 代码,运行良好(在邮递员和浏览器中):
$curl = curl_init();
curl_setopt_array($curl, [
CURLOPT_URL => 'https://api.hylo.biz/Api/v1.0/Payment',
CURLOPT_RETURNTRANSFER => true,
CURLOPT_ENCODING => '',
CURLOPT_MAXREDIRS => 10,
CURLOPT_TIMEOUT => 0,
CURLOPT_FOLLOWLOCATION => true,
CURLOPT_HTTP_VERSION => CURL_HTTP_VERSION_1_1,
CURLOPT_CUSTOMREQUEST => 'POST',
CURLOPT_POSTFIELDS => '{
"orderId":"KJDHKSGIU768",
"amount":"100",
"redirect_url":"google.com"
}',
CURLOPT_HTTPHEADER => [
'Authorization: Basic WUR1pYQ1hYY3U2Og2OGM6JJE5tWEDJhJDEwZReVE=',
'Content-Type: application/json',
],
]);
$response = curl_exec($curl);
curl_close($curl);
echo $response;
这是我的 laravel 代码:
public function process(Request $request)
{
// return $request['request'];
$client = new \GuzzleHttp\Client();
$url = "https://api.hylo.biz/Api/v1.0/Payment";
$response = $client->request('POST', $url, [
'headers' => [
'Authorization' => 'Basic WUR1pYQ1hYY3U2Og2OGM6JJE5tWEDJhJDEwZReVE=',
'Content-Type' => 'application/json'
],
'form_params' => $request['request']
]);
return $response = $response->getBody();
}
【问题讨论】:
-
如何在 laravel 代码中发送 POST 表单数据? ,为什么不直接在你的 laravel 应用程序中使用你的 PHP curl 代码
-
是的,但我想使用 GuzzleHttp 你能帮我解决这个问题吗@Mahdimehrabi 谢谢...