Home  >  Article  >  Backend Development  >  php获取通过url的json数据,返回的是空值

php获取通过url的json数据,返回的是空值

WBOY
WBOYOriginal
2016-06-23 13:57:121559browse

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']);

$x = json_decode($_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));

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);
 }

Statement:
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn