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前后台交互过程中json格式如何解析以及如何生成_基础知识

WBOY
WBOYOriginal
2016-05-16 17:45:131209browse
前台
复制代码 代码如下:

$.ajax({
type: "POST",
url: "GetMenuRole.ashx",
data: "",
dataType: "json",
success: function (data) {
if (data.Status == false) {
alert(data.ErrorReason);
}
else {
//解析这个菜单列表
alert(data.MenuList);
}
},
error: function (data) {
alert(data);
}
});

后台:(需要使用双引号)
复制代码 代码如下:

string sReturnJson = " \"Status\":\"{0}\",\"MenuList\":\"{1}\",\"ErrorReason\":\"{2}\"";
sReturnJson = string.Format(sReturnJson, "True", sReturnMenuList,string.Empty).Replace("","{").Replace("","}");
context.Response.Write(sReturnJson);
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