我不确定如何去做但想要验证用户登录:
<form onsubmit="return validateLogIn()" > and
<script>
function validateLogIn()
{
$.ajax({
url: 'login.php', //checking the login in
data: "",
dataType: 'json',
success: function(data)
{
continue to crud.html
}
else do an alert("please lgo in again");
});
}
</script>
我会有类似上面的内容吗?
答案 0 :(得分:1)
您可以使用此代码
<script>
function validateLogIn()
{
var username = $("#IdOfYourUserFiled").val();
var password = $("#IdOfYourPasswordFiled").val();
$.ajax({
url: 'login.php', //checking the login in
data: {username:username,password:password}, //You have to pass user inputs to next page to validate in DB
type: "POST", //Method by which data being transmitted
dataType: 'json',
success: function(data)
{
continue to crud.html
}
else do an alert("please lgo in again");
});
}
</script>
在login.php上访问用户名和密码的值
使用
$_POST['username'] and $_POST['password']
并编写查询。
答案 1 :(得分:0)
我个人在发送数据之前使用beforeSend()
进行验证。
$.ajax({
url:'login.php',
data:{username:user,password:pass},
dataType:'json',
beforeSend:function(xhr){
if(user.length<7||pass.length<7){
xhr.abort();//stop process.
//validation failed etc
}
},
success:function(result){
}
});
答案 2 :(得分:-1)
尝试以下编码,
function validateLogIn()
{
username=$("#name").val();
password=$("#word").val();
$.ajax({
url: 'login.php', //checking the login in
data: "name="+username+"&pwd="+password,
dataType: 'json',
success: function(data)
{
if(data === 0){
alert("Username or Password is incorrect");
} else if (data == 1){
window.open('index.php', '_self');
}
}
});
}