0
嗨;
當我使用從jQuery ajax
,頁面凍結,直到請求結束。 這是我的JavaScript代碼:
function GetAboutContent(ID, from) {
var About = null;
if (from != true)
from = false;
$.ajax({
type: "POST",
url: "./ContentLoader.asmx/GetAboutContent",
contentType: "application/json; charset=utf-8",
dataType: "json",
data: JSON.stringify({ 'ID': ID, 'from': from }),
async: true,
success: function (msg) {
var Result = msg.d.Result;
if (Result == 'session') {
warning('Your session has expired, please login again!');
setTimeout(function() {
window.location.href = "Login.aspx";
}, 4000);
return;
}
if (Result == 'failed' || Result == false) {
About = false;
return;
}
About = JSON.parse(msg.d.About)[0];
}
});
return About;
}
,這是我WebService
[WebMethod(EnableSession = true)]
public object GetAboutContent(int ID, bool from = false)
{
try
{
if (HttpContext.Current.Session["MahdParent"] != null ||
HttpContext.Current.Session["MahdTeacher"] != null ||
from)
{
functions = new GlobalFunctions();
DataTable queryResult = new DataTable();
queryResult = functions.DoReaderTextCommand("SELECT Field FROM TT WHERE ID = " + ID);
if (queryResult.Rows.Count != 0)
return new { Result = true, About = JsonConvert.SerializeObject(queryResult.Rows[0].Table) };
else
return new { Result = false };
}
else
return new { Result = "session" };
}
catch (Exception ex)
{
return new { Result = "failed", Message = ex.Message };
}
}
我怎麼能解決這個問題? 請幫我
你***不能***返回一個**異步**函數的結果 – adeneo
可能重複[如何從AJAX調用返回響應?](http://stackoverflow.com/questions/ 14220321 /如何返回迴應從一個阿賈克斯調用) – adeneo
你的意思是我應該設置'async'爲'false' –