解决:JavaScript 在函数中使用Ajax获取的值作为函数的返回值,结果无法获取到返回值 原因:ajax默认使用异步方式,要将异步改为同步方式 案例:通过区域ID,获取该区域下所有的学校 var School = new Object(); //保存区下面所有学校 School.GetAllInArea = function (areaId) { var returnData; $.ajax({ url: "/Area/Home/Schools/" + areaId, type:
先看一个场景 var arr = ["a","b","c"]; for (var i in arr) { $.get("h.html", function (data) {//1 console.log(data); console.log(arr[i]); console.log("----"); })
Function GetHttpPage(HttpUrl,endoce) If endoce = "" Then endoce = "GB2312" If IsNull(HttpUrl)=True Or Len(HttpUrl)<18 Or HttpUrl="$False$" Then GetHttpPage="$False$" Exit Function End If Dim Http Set Http=server.