將帶有PHP和CURL的JSON字符串發布到php頁面並獲取響應

Question

我正在嘗試學習如何通過CURL將PHP發布到另一個PHP頁面並閱讀響應。

我有2頁。 第一個是test.php

<?php
$array['User'] = array();
$array['User']['AppId'] = 'sdfgfd9-sdfgsdf-sdfgfdgfgff';
$array['User']['UserName'] = 'me@example.co.uk';
$array['User']['Token'] = 'fsdgf5-455g-223ee-bggg-asdsadsda';
$array['User']['Timestamp'] = '2018-05-30BST16:28:293600';

$url = "https://www.myurl.co.uk/api/index.php";    
$content = json_encode($array['User']);

$curl = curl_init($url);
curl_setopt($curl, CURLOPT_HEADER, false);
curl_setopt($curl, CURLOPT_RETURNTRANSFER, true);
curl_setopt($curl, CURLOPT_HTTPHEADER,
array("Content-type: application/json"));
curl_setopt($curl, CURLOPT_POST, true);
curl_setopt($curl, CURLOPT_POSTFIELDS, $content);

$json_response = curl_exec($curl);

$status = curl_getinfo($curl, CURLINFO_HTTP_CODE);

if ( $status != 201 ) 
{
   die("Error: call to URL $url failed with status $status, response 
  $json_response, curl_error " . curl_error($curl) . ", curl_errno " . 
  curl_errno($curl));
}

curl_close($curl);

$response = json_decode($json_response, true);
echo $response;
?>

然后在頁面上，我只是從中獲取數據。

<?php
$userArray = array();   
$userArray = trim(file_get_contents("php://input"));
echo json_decode($userArray, true);
?>

當我在test.php上運行腳本時，得到的響應是：

Error: call to URL https://www.myurl.co.uk/api/index.php failed with status 200, response Array, curl_error , curl_errno 0

我對錯誤的理解不足，無法理解要進行哪些更改才能使此工作正常進行？

Answer 1

試試：curl_setopt（$ curl，CURLOPT_HTTPHEADER，array（'method'=>'POST'，'header'=>'Content-Type：application / json'。“ \\ r \\ n”。'Content-Length：'。strlen （$ data_string）。“ \\ r \\ n”，'content'=> $ data_string，）;

或類似