Dan*_*l H 2 javascript php ajax jquery post
我正在使用这个功能:
function end_incident() {
var dataString = 'name=Daniel&phone=01234123456';
$.ajax({
type: "POST",
url: "http://www.example.co.uk/erc/end_incident.php",
data: dataString,
success: function(msg){
alert('Success!'+dataString);
}
});
};
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发送信息end_incident.php,但我无法访问$_POST变量.我试过这样做:
$name = $_POST['name'];
$phone = $_POST['phone'];
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难道我做错了什么?
谢谢你的帮助
Ric*_*ton 13
尝试将数据作为对象发送:
function end_incident() {
$.ajax({
type: "POST",
url: "http://www.example.co.uk/erc/end_incident.php",
data: { name: "Daniel", phone: "01234123456" },
success: function(msg){
alert('Success!');
}
});
};
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