ajax 如何使 dropdownlist 无刷新,Jquery实现无刷新DropDownList联动实现代码
先看HTML,我們引用Jquery,放兩個DropDownList:
#ddlEmployeeCars
{
display:none;
position:absolute;
top:50px;
left:9px;
}
接著寫核心的Script:
$(function() {
var $ddl = $("select[name$=ddlEmployee]");
var $ddlCars = $("select[name$=ddlEmployeeCars]");
$ddl.focus();
$ddl.bind("change keyup", function() {
if ($(this).val() != "0") {
loadEmployeeCars($("select option:selected").val());
$ddlCars.fadeIn("slow");
} else {
$ddlCars.fadeOut("slow");
}
});
});
function loadEmployeeCars(selectedItem) {
$.ajax({
type: "POST",
url: "Default.aspx/FetchEmployeeCars",
data: "{id: " + selectedItem + "}",
contentType: "application/json; charset=utf-8",
dataType: "json",
async: true,
success: function Success(data) {
printEmployeeCars(data.d);
}
});
}
function printEmployeeCars(data) {
$("select[name$=ddlEmployeeCars] > option").remove();
for (var i = 0; i < data.length; i++) {
$("select[name$=ddlEmployeeCars]").append(
$("").val(data[i].Id).html(data[i].Car)
);
}
}
非常簡單，檢查值是不是０，然后ajax傳值到server，成功后remove掉原來的option,append新的option.
看下WebPage的code:
public partial class _Default : System.Web.UI.Page
{
[WebMethod]
public static List FetchEmployeeCars(int id)
{
var emp = new EmployeeCar();
return emp.FetchEmployeeCars(id);
}
protected void Page_Load(object sender, EventArgs e)
{
if (!IsPostBack)
{
var employees = new Employee();
ddlEmployee.DataSource = employees.FetchEmployees();
ddlEmployee.DataTextField = "Surname";
ddlEmployee.DataValueField = "Id";
ddlEmployee.DataBind();
}
}
}
我們的Datasource class:
public class EmployeeCar
{
public int Id { get; set; }
public string Car { get; set; }
private static List LoadData()
{
return new List
{
new EmployeeCar {Id = 1, Car = "Ford"},
new EmployeeCar {Id = 1, Car = "Holden"},
new EmployeeCar {Id = 1, Car = "Honda"},
new EmployeeCar {Id = 2, Car = "Toyota"},
new EmployeeCar {Id = 2, Car = "General Motors"},
new EmployeeCar {Id = 2, Car = "Volvo"},
new EmployeeCar {Id = 3, Car = "Ferrari"},
new EmployeeCar {Id = 3, Car = "Porsche"},
new EmployeeCar {Id = 3, Car = "Ford2"}
};
}
public List FetchEmployeeCars(int id)
{
return (from p in LoadData()
where p.Id == id
select p).ToList();
}
}
public class Employee
{
public int Id { get; set; }
public string GivenName { get; set; }
public string Surname { get; set; }
public List FetchEmployees()
{
return new List
{
new Employee {Id = 1, GivenName = "Tom", Surname = "Hanks"},
new Employee {Id = 2, GivenName = "Hugh", Surname = "Jackman"},
new Employee {Id = 3, GivenName = "Petter", Surname = "Liu"}
};
}
public Employee FetchEmployee(int id)
{
var employees = FetchEmployees();
return (from p in employees
where p.Id == id
select p).First();
}
}
完了。希望這篇POST對您有幫助。
總結(jié)
以上是生活随笔為你收集整理的ajax 如何使 dropdownlist 无刷新,Jquery实现无刷新DropDownList联动实现代码的全部內(nèi)容，希望文章能夠幫你解決所遇到的問題。
- 上一篇： cnvd与cnnvd区别_漏洞都是怎么编
- 下一篇： chrome 内核CEF 编译和qt 封