Api响应在php中不随curl显示

Question

我想在网站中使用Php获取Api响应（在nodejs中创建），所以为此我使用了curl，但它不起作用，我尝试了以下代码，但不起作用（显示空白页面），我错在哪里？这是我的密码

$post = ['email'=> "[e-mail]",'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

尝试显示错误，以防错误/警告被抑制。

在文件的顶部使用这些，就在php标签之后

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

在json_解码之前，还要尝试打印原始响应，这是因为如果您得到的响应无效，那么解码后将不会打印任何内容。

用这个

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

在摘要中，您的代码应该如下所示

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

$post = ['email'=> "[e-mail]",'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响应是什么

Answer 2

更改您的帖子数组

  $post = array('email'=> "[e-mail]",'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 3

改变

$result = json_decode($response);

到

$result = json_decode($response, true);

然后

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

回应：-

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