json result

2018-06-22 06:12:29来源：未知阅读 ()

public JsonResult JsonData()
        {
            HttpContext.Response.AppendHeader("Access-Control-Allow-Origin", "*");


            return Json(db.Weathers.ToList());
        }

json方法有一个重构：

protected internal JsonResult Json(object data); protected internal JsonResult Json(object data, JsonRequestBehavior behavior);

我们只需要使用第二种就行了，加上一个 json请求行为为Get方式就OK了

public JsonResult GetPersonInfo() { var person = new { Name = "张三", Age = 22, Sex = "男" }; return Json(person,JsonRequestBehavior.AllowGet); }

这样一来我们在前端就可以使用Get方式请求了：

view

$.ajax({ url: "/FriendLink/GetPersonInfo", type: "POST", dataType: "json", data: { }, success: function(data) { $("#friendContent").html(data.Name); } })

<!DOCTYPE html>
<html>
<head runat="server">
    <title>Index2</title>
    <script src="\Scripts\jquery-1.10.2.min.js" type="text/javascript"></script>
    <script type="text/javascript">
        var login = function () {
            $.ajax({ type: "post", url: "http://localhost:4968/Weathers/JsonData", data: null, success: function (res) {
                alert(JSON.stringify(res));
            }, dataType: "json"
            });
        }
    </script>
</head>
<body>
    <div id="nav">
        <a href="/Home/Index">ajax+Handler</a>&nbsp; <a>ajax+action</a>
    </div>
    <div>
        <h3>
            Login</h3>
        <button type="button" onclick="login()">Submit</button>
    </div>
</body>
</html>