ajax提交param 后台接受是对象

<script type="text/javascript">
function funSubmit()
{
var param = {
LoginName: $("#loginName").val(),
PassWord: $("#passWord").val(),
NewPassWord: $("#newPassWord").val(),
ConfirmPassword: $("#confirmPassword").val()
};
$.ajax({
type: "POST",
url: "/Login/ChangePassword",
datatype: "JSON",
data: JSON.stringify(param),
contentType: 'application/json; charset=utf-8',
success: function (data) {
if (data.Status == "Ok") {
alert(data.Message);
window.location = "/Login/Index";
} else {
alert(data.Message);
}
}
});
}
</script>

 

转载于:https://www.cnblogs.com/hanran/p/8889119.html