Warning: file_get_contents(/data/phpspider/zhask/data//catemap/4/json/14.json): failed to open stream: No such file or directory in /data/phpspider/zhask/libs/function.php on line 167

Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

Warning: array_chunk() expects parameter 1 to be array, null given in /data/phpspider/zhask/libs/function.php on line 181

Warning: file_get_contents(/data/phpspider/zhask/data//catemap/6/cplusplus/137.json): failed to open stream: No such file or directory in /data/phpspider/zhask/libs/function.php on line 167

Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

Warning: array_chunk() expects parameter 1 to be array, null given in /data/phpspider/zhask/libs/function.php on line 181
Asp.net mvc 如何将json字符串作为json对象发送到视图?_Asp.net Mvc_Json - Fatal编程技术网

Asp.net mvc 如何将json字符串作为json对象发送到视图?

Asp.net mvc 如何将json字符串作为json对象发送到视图?,asp.net-mvc,json,Asp.net Mvc,Json,我在MVC控制器的操作中有一个json字符串。我想将其作为JSON对象发送到视图。我怎样才能解决这个问题 public JsonResult Json() { ... some code here ... string jsonString = "{\"Success\":true, \"Msg\":null}"; // what should I do instead of assigning jsonString to Data. return new Js

我在MVC控制器的操作中有一个json字符串。我想将其作为JSON对象发送到视图。我怎样才能解决这个问题

public JsonResult Json()
{
    ... some code here ...
    string jsonString = "{\"Success\":true, \"Msg\":null}";
    // what should I do instead of assigning jsonString to Data. 
    return new JsonResult() { Data = jsonString, JsonRequestBehavior = JsonRequestBehavior.AllowGet };
}
但我建议您使用对象而不是字符串:

public ActionResult Json()
{
    ... some code here ...
    return Json(
        new 
        {
            Success = true,
            Msg = (string)null
        }, 
        JsonRequestBehavior.AllowGet
    );
}
但我建议您使用对象而不是字符串:

public ActionResult Json()
{
    ... some code here ...
    return Json(
        new 
        {
            Success = true,
            Msg = (string)null
        }, 
        JsonRequestBehavior.AllowGet
    );
}