我使用ajax时仍然不稳定因此我的实现中存在一些漏洞。我试图发布一个控制器动作,它将调用存储过程来更新我的视图模型,然后重新加载将显示信息的div。
Ajax Post:
$("#order-summary-panel").click(function(){
$.ajax({
url: '@Url.Action("GetOrderSummary", "Home")',
type: 'POST',
success: function() {
alert("It Worked!")
}
});
});
控制器操作:
[HttpPost]
public ActionResult GetOrderSummary
{
using (var entity = new OrderEntities())
{
var user = User.Identity.Name;
var orderSummary = entity.uspGetOrderSummary(user).ToList();
var viewModel = new OrderViewModel
{
OrderSummary = orderSummary
};
return View("Index", viewModel)
}
}
Div重新加载:
<div id="order-summary-panel">
@if (Model != null && Model.OrderSummary != null)
{
foreach (var order in Model.OrderSummary)
{
// Display Orders
}
}
</div>
我相信我不应该返回完整视图,但我不确定如何在不这样做的情况下更新视图模型。任何帮助,将不胜感激。
答案 0 :(得分:8)
您可以返回PartialView并将结果放入div中,用于示例:
[HttpPost]
public ActionResult GetOrderSummary
{
using (var entity = new OrderEntities())
{
var user = User.Identity.Name;
var orderSummary = entity.uspGetOrderSummary(user).ToList();
var viewModel = new OrderViewModel
{
OrderSummary = orderSummary
};
return PartialView("Index", viewModel);
}
}
并在您的javascript中:
$("#order-summary-panel").click(function(){
$.ajax({
url: '@Url.Action("GetOrderSummary", "Home")',
type: 'POST',
success: function(data) {
if (data) { // check if data is defined
$("#order-summary-panel").html(data);
}
}
});
});
您还可以尝试返回json
操纵html(可以提高性能),样本:
[HttpPost]
public ActionResult GetOrderSummary
{
using (var entity = new OrderEntities())
{
var user = User.Identity.Name;
var orderSummary = entity.uspGetOrderSummary(user).ToList();
var viewModel = new OrderViewModel
{
OrderSummary = orderSummary
};
return Json(new { success = true, order = viewModel }, JsonRequestBehavior.DenyGet);
}
}
在您的Javascript中,您可以尝试阅读这些属性:
$("#order-summary-panel").click(function(){
$.ajax({
url: '@Url.Action("GetOrderSummary", "Home")',
type: 'POST',
success: function(data) {
if (data) { // check if data is defined
if (data.success) { // if success is true
alert("It Worked!");
// you could read data.order
}
}
}
});
});