如果success: function(result)
错误而不是alert()
,但如果确实如此,我想重定向。任何人都可以帮我解决这个问题。
function log_me_in_now() {
var eml = $('#email').val();
var pwd = $('#password').val();
$.ajax({
url: 'http://dev/ajax/shopper_login.php',
data: { 'email': eml, 'password': pwd },
dataType: 'json',
cache: false,
failure: function(results) {
alert('99 problems...\n' + results);
},
success: function(result) {
alert(result);
//window.location = "http://dev/test/logged_in"
}
});
}
AJAX:
include_once('../includes/settings.php');
if (isset($_GET['email']) && $_GET['email'] != '' && isset($_GET['password']) && $_GET['password'] != '') {
// todo: change their connection string to become their Oracle user.
$shopper = new Shopper($_GET['email']);
$token = $shopper->login($_GET['password']);
if (isset($token) && $token != '') {
echo json_encode($token);
}
}
答案 0 :(得分:1)
success: function(result) {
if (result == "true") {
window.location = "http://dev/test/logged_in"
}
else{
alert("Login Failed -- Please try again.");
};
}
这取决于你的ajax调用实际返回的内容。从理论上讲,如果登录正确,它应该返回“true”或“false”。如果这是真正的呼叫返回的内容,那么上面的代码将适合您。
答案 1 :(得分:0)
检查一下:
function log_me_in_now(){
var eml = $('#email').val();
var pwd = $('#password').val();
$.ajax({
url: 'http://dev/ajax/shopper_login.php',
data: { 'email': eml, 'password': pwd },
dataType: 'json',
cache: false,
failure: function(results) {
alert('99 problems...\n' + results);
},
success: function(result) {
window.location.assign("http://dev/test/logged_in");
}
});
}
答案 2 :(得分:0)
检查一下! 对于默认浏览器,您的提交必须为preventdefault false!
$("#inputID" ).click(function( event ) {
event.preventDefault();
var eml = $('#email').val();
var pwd = $('#password').val();
$.ajax({
url: 'http://dev/ajax/shopper_login.php',
data: { 'email': eml, 'password': pwd },
dataType: 'json',
cache: false,
failure: function(results) {
alert('99 problems...\n' + results);
},
success: function(result) {
window.location.assign("http://dev/test/logged_in");
}
});
});