php url返回json_php获取通过url的json数据，返回的是空值

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我是靠谱客的博主伶俐皮卡丘，最近开发中收集的这篇文章主要介绍php url返回json_php获取通过url的json数据，返回的是空值，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

http://php.net/api/apply.php?&apply={"apply":{"name":"4747474","site":"http://4444444444444444","icon":"47","information":"474","about":"747","weibo":"74444444444","email":"74444444444444444","moblie":"89742589633","isfree":"y","applytime":1349939108}}

这个是一个链接，我在这个页面上要怎么把那段json格式给接下出来呢？

回复讨论(解决方案)

$x = json_decode($_GET['apply']);

得出来还是空值，可能那个字符串存在错误吧...

var_dump(json_decode($_GET['apply'],true));

看?出什麽？

var_dump(json_decode($_GET['apply'],true));

看?出什麽？

NULL

var_dump(json_decode(urldecode($_GET['apply']),true));

也是不行，最后我把那些多余的去掉了，不知道是为啥啊？

print_r(json_decode('{"apply":{"name":"4747474","site":"http://4444444444444444","icon":"47","information":"474","about":"747","weibo":"74444444444","email":"74444444444444444","moblie":"89742589633","isfree":"y","applytime":1349939108}}')); stdClass Object( [apply] => stdClass Object ( [name] => 4747474 [site] => http://4444444444444444 [icon] => 47 [information] => 474 [about] => 747 [weibo] => 74444444444 [email] => 74444444444444444 [moblie] => 89742589633 [isfree] => y [applytime] => 1349939108 ))

PHP code

print_r(json_decode('{"apply":{"name":"4747474","site":"http://4444444444444444","icon":"47","information":"474","about":"747","weibo":"74444444444","email":"74444444444444444","moblie":"……

嗯，谢谢了

LZ地址中参数apply前面多了个&符号，你看会不会是这个的问题

之前是我的文本格式有问题，得出来的结果有错误

$json_string = $_GET["apply"];

if(ini_get("magic_quotes_gpc")=="1")

{

$json_string=stripslashes($json_string);

}

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