如何在PHP中解析JSON对象？

Question

我有一个JSON对象,我发送到PHP脚本,我在解析JSON时遇到问题.这是POST请求:

http://mywebsite.com?action=somefunction&{%22id%22:1,%22Name%22:%22Mike%22}

这是我的PHP函数,显然不起作用:

$data = $_GET['data'];
$obj = json_decode($data);
echo $obj->Name;
die();

最终目标是从URL字符串中提取名称"Mike".有什么建议？

Answer 1

试着看一下PHP输出的内容json_decode():

$data = $_GET['data'];
$obj = json_decode($data);
var_dump($obj);

您的代码本身工作正常:http://ideone.com/0jsjgT

但是你的查询字符串data=在实际的JSON之前就丢失了.这个:

http://mywebsite.com?action=somefunction&{%22id%22:1,%22Name%22:%22Mike%22}

应该这样:

http://mywebsite.com?action=somefunction&data={%22id%22:1,%22Name%22:%22Mike%22}