我无法找到一个解决方案来访问通过AJAX传递给我的PHP脚本变量的这个简单的问题。我甚至尝试使用isset($ _ POST),但它仍然未能找到用户名和密码的变量。
下面是AJAX调用:
变种U = $(#用户名的形式).VAL();
变种P = $(#密码,形式).VAL();
//测试
的console.log('用户名:'+ U); //'约翰'
的console.log(密码:+ P); // '测试'
如果(U ==''和;&放大器,P = =!!){
$阿贾克斯(网址:{url:http://www.domain.net/php/user-login.php,
数据:{用户名:U,密码:P},
类型:'后',
异步:真正的,
数据类型:JSON,
beforeSend:函数(){
//这个回调函数数据之前将触发发送
$ .mobile.loading('秀'); //这将显示ajax的微调
},
完成:函数(){
//这个回调函数将在发送的数据触发/接收完成
$ .mobile.loading(隐藏); //这将隐藏阿贾克斯微调
},
成功:功能(数据){
//保存后返回的数据localStorage的手动按钮来回切换
//window.localStorage.setItem('topGenderDataArray',data);
的console.log(登录成功+数据);
console.dir(数据);
警报(欢迎回来+数据['用户名']);
$(#loginButton)removeAttr(禁用)。
},
错误:函数(XHR,要求,错误){
//这个回调函数将在不成功的行动触发
(出现网络错误,请重试!'+ XHR +|+要求+|+错误)警告;
}
});
下面是PHP脚本的开头:
如果(使用isset($ _ POST ['用户名'])及和放大器;使用isset($ _ POST ['密码']))
{
$数据['用户名'] = $ _ POST ['用户名'];
$数据['密码'] = $ _ POST ['密码'];
}
其他{
$数据['消息'] =抱歉,发生错误[]!;
$数据['USER_ID'] = -1;
回声json_en code($的数据);
出口();
}
解决方案
现在的问题是在你的PHP code
你做回声 json_en code($数据)
的其他
子句中,而你应该后做在如果
和其他
就像这样:
如果(使用isset($ _ POST ['用户名'])及和放大器;使用isset($ _ POST ['密码']))
{
$数据['用户名'] = $ _ POST ['用户名'];
$数据['密码'] = $ _ POST ['密码'];
}
其他
{
$数据['消息'] =抱歉,发生错误[]!;
$数据['USER_ID'] = -1;
}
回声json_en code($的数据);
I can't find a solution to this simple problem of accessing variable passed to my PHP script via AJAX. I have even tried isset($_POST) but it still fails to find the username and password variables.
Here is the AJAX call:
var u = $("#username", form).val();
var p = $("#password", form).val();
//testing
console.log('Username: '+u); // 'John'
console.log('Password: '+p); // 'test'
if(u !== '' && p!=='') {
$.ajax({url: 'http://www.domain.net/php/user-login.php',
data: {username:u,password:p},
type: 'post',
async: true,
dataType:'json',
beforeSend: function() {
// This callback function will trigger before data is sent
$.mobile.loading('show'); // This will show ajax spinner
},
complete: function() {
// This callback function will trigger on data sent/received complete
$.mobile.loading('hide'); // This will hide ajax spinner
},
success: function (data) {
//save returned data to localStorage for manual button toggling later
//window.localStorage.setItem('topGenderDataArray',data);
console.log("Login successful: "+ data);
console.dir(data);
alert("Welcome back "+data['username']);
$("#loginButton").removeAttr("disabled");
},
error: function (xhr,request,error) {
// This callback function will trigger on unsuccessful action
alert('Network error has occurred please try again! '+xhr+ " | "+request+" | "+error);
}
});
Here is the beginning of the PHP script:
if(isset($_POST['username']) && isset($_POST['password']))
{
$data['username'] = $_POST['username'];
$data['password'] = $_POST['password'];
}
else{
$data['message']= "Sorry, an error occurred! []";
$data['user_id']= -1;
echo json_encode($data);
exit();
}
解决方案
The problem is in your php code
you do the echo json_encode($data)
inside the else
clause, while you should do it after the if
and else
just like this:
if(isset($_POST['username']) && isset($_POST['password']))
{
$data['username'] = $_POST['username'];
$data['password'] = $_POST['password'];
}
else
{
$data['message']= "Sorry, an error occurred! []";
$data['user_id']= -1;
}
echo json_encode($data);