Question

我已将此变量传递给php页面

 type: 'POST',
 url:   url',
 data: "id="+id+"&nowrole="+role+"&last="+last+"&title="+title+"&devlopdes="+devlopdes

有时候，第一组值是

id=1234
role=staff
last=45
title=check page
developdes= view this
http://hibihi.com/index.php?act=view&id=1
http://hibihi.com/index.php?act=add&shipperid=1

在php页面

检索id =“1http：//hibihi.com/index.php？act = add”不检索传递'id'值

有时候，第二组值是

id=1234
role=staff
last=45
title=check page
developdes= view this
http://hibihi.com/index.php?act=view
http://hibihi.com/index.php?act=add&shipperid=1

检索正确的id = 1234值。

Answer 1

如果您正在使用对象，则可以在javascript中使用此代码：

var obj = {};
obj[id]         = "1234";
obj[role]       = "staff";
obj[last]       = "45";
obj[title]      = "check page";
obj[developdes] = "view this";
obj[http1]      = "http://hibihi.com/index.php?act=view&id=1";
obj[http2]      = http://hibihi.com/index.php?act=add&shipperid=1";

然后

type: 'POST',
url:   url',
data: JSON.stringify(obj)

对于你的PHP

$json = jsondecode($_POST[data]);
$id = $json->{'id'};

ajax发送变量，有时无法在php中检索完美值

1 个答案: