Http中传输对象，最好的表现形式莫过于 JSON 字符串了，但是作为参数的接收方，又是需要被告知传过来的是JSON！
其实这不能，只需要发送一个 http 头信息即可，即 “ Content-Type: application/json; charset=utf-8 ”，参考代码如下：
<?php
/**
* PHP发送Json对象数据
*
* @param $url 请求url
* @param $jsonStr 发送的json字符串
* @return array
*/
function http_post_json($url, $jsonStr)
{
$ch = curl_init();
curl_setopt($ch, CURLOPT_POST, 1);
curl_setopt($ch, CURLOPT_URL, $url);
curl_setopt($ch, CURLOPT_POSTFIELDS, $jsonStr);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($ch, CURLOPT_HTTPHEADER, array(
'Content-Type: application/json; charset=utf-8',
'Content-Length: ' . strlen($jsonStr)
)
);
$response = curl_exec($ch);
$httpCode = curl_getinfo($ch, CURLINFO_HTTP_CODE);
return array($httpCode, $response);
}
$url = "http://blog.snsgou.com";
$jsonStr = json_encode(array('a' => 1, 'b' => 2));
list($returnCode, $returnContent) = http_post_json($url, $jsonStr);原文 http://blog.snsgou.com/post-866.html 最佳答案
