Question

我想使用Php在网站上获取Api响应（在nodejs中创建），因此我正在使用卷曲，但不起作用，我尝试使用以下代码，但对我不起作用（显示空白页），我在哪里错了？这是我的代码

$post = ['email'=> "example@xyz.com",'password'=> "testing"];
  $ch = curl_init();
  curl_setopt($ch, CURLOPT_URL,'http://35.154.149.228:8000/api/admin/login');
  curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
  curl_setopt($ch, CURLOPT_POSTFIELDS, http_build_query($post));
  $response = curl_exec($ch);
  $result = json_decode($response); 
  print_R($result);

Answer 1

更改帖子数组


  $post = array('email'=> "example@xyz.com",'password'=> "testing");
  $ch = curl_init();
  curl_setopt($ch, CURLOPT_URL,'http://35.154.149.228:8000/api/admin/login');
  curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
  curl_setopt($ch, CURLOPT_POSTFIELDS, http_build_query($post));
  $response = curl_exec($ch);
  $result = json_decode($response); 
  print_r($result);

Answer 2

更改

$result = json_decode($response);

收件人

$result = json_decode($response, true);

然后

echo '<pre>';
print_r($result);

响应：-

Array
(
 [statusCode] => 401
 [error] => Unauthorized
 [message] => Invalid username or password
 [responseType] => INVALID_USER_PASS
)

Answer 3

尝试显示错误，以防错误/警告被抑制。

在php标记后的文件顶部使用这些

ini_set("display_errors", "On");
error_reporting(E_ALL);

还尝试在json_decoding之前打印原始响应，这是因为如果您得到的响应不是有效的json，则在解码后将不会打印出任何内容。

使用此

print_r("The response is: " . $response);

总而言之，您的代码应该看起来像

ini_set("display_errors", "On");
error_reporting(E_ALL);

$post = ['email'=> "example@xyz.com",'password'=> "testing"];
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL,'http://35.154.149.228:8000/api/admin/login');
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_POSTFIELDS, http_build_query($post));
$response = curl_exec($ch);
//Printing the original response before trying to decode it
//$result = json_decode($response); 

print_r("The response from the server before decoding is: " . $response);

让我们知道您从中得到的确切答复是什么

api响应未在php中显示为curl

3 个答案: