MVC 4中的Ajax异步调用 [英] Ajax Async call in MVC 4
本文介绍了MVC 4中的Ajax异步调用的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我有一个JS函数可以进行AJAX异步调用.
I have a JS function that makes an AJAX Async call.
function PostCall() {
$('#txtRes').val('Please Wait......');
var reqval = $('#txtReq').val();
$.ajax({
url: "@urlsMenu",
type: 'post',
data: "{'name':'" + reqval + "'}",
async: true,
contentType: "application/json",
success: function (result) {
debugger;
$('#txtRes').val(result);
}
});
}
我在Controller类中有一个函数可以进行Service调用.
I have a function in Controller class that makes a Service call.
public JsonResult PostAsync(string name)
{
string val = "Not Found ";
MySr.Service1Client scc = new MySr.Service1Client();
scc.GetDataCompleted += (result, e) =>
{
val = e.Result;
string str = val;
};
scc.GetDataAsync(name);
return Json(val, JsonRequestBehavior.AllowGet);
}
但是,我的问题是,服务器端函数已经返回了值(如未找到"),而不是在webservice调用之后收到的值.
我希望函数在Web服务返回值后返回值.
But, my problem is, the Server-Side function already returns the value (as "Not Found") instead of the value received after the webservice call.
I want the function to return the value after the webservice returns the value.
推荐答案
(' #txtRes').val(' 请稍候.... ..'); var reqval =
('#txtRes').val('Please Wait......'); var reqval =
('
('#txtReq').val();
.ajax({ url:" , 类型:' post', 数据:" + reqval + " , 异步: true , contentType:" , 成功:功能(结果){ 调试器;
.ajax({ url: "@urlsMenu", type: 'post', data: "{'name':'" + reqval + "'}", async: true, contentType: "application/json", success: function (result) { debugger;
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