Question

我正在尝试学习如何通过CURL将PHP发布到另一个PHP页面并阅读响应。

我有2页。第一个是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;
?>

然后在页面上我从中得到数据;

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

当我在test.php上运行脚本时，我得到的是;

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

我不明白这个错误足以理解要改变什么才能使其发挥作用？

Answer 1

尝试： curl_setopt（$ curl，CURLOPT_HTTPHEADER，阵列（＆＃39;方法＆＃39; =＆GT; ＆＃39; POST＆＃39 ;, ＆＃39;头＆＃39; =＆GT; ＆＃39;内容类型：application / json＆＃39; 。＆＃34; \ r \ n＆＃34; 。＆＃39;内容长度：＆＃39; 。 strlen（$ data_string）。＆＃34; \ r \ n＆＃34 ;, ＆＃39;内容＆＃39; =＆GT; $ data_string，）;

或类似的

将PHP和CURL的POST JSON字符串发送到php页面并获得响应

1 个答案: