asp.net-mvc – 将JSON对象作为参数传递给MVC控制器

public string GetData()
    {
        // InputStream contains the JSON object you've sent
        String jsonString = new StreamReader(this.Request.InputStream).ReadToEnd();

        // Deserialize it to a dictionary
        var dic = 
          Newtonsoft.Json.JsonConvert.DeserializeObject<Dictionary<String,dynamic>>(jsonString);

        string result = "";

        result += dic["firstname"] + dic["lastname"];

        // You can even cast your object to their original type because of 'dynamic' keyword
        result += ",Age: " + (int)dic["age"];

        if ((bool)dic["married"])
            result += ",Married";


        return result;
    }

此解决方案的真正好处是您不需要为每个参数组合定义新类,除此之外,您可以轻松地将对象转换为其原始类型.

更新

现在,你甚至可以合并你的GET和POST动作方法,因为你的post方法不再有任何参数,就像这样：

public ActionResult GetData()
 {
    // GET method
    if (Request.HttpMethod.ToString().Equals("GET"))
        return View();

    // POST method 
    .
    .
    .

    var dic = GetDic(Request);
    .
    .
    String result = dic["fname"];

    return Content(result);
 }

并且您可以使用这样的帮助方法来促进您的工作

public static Dictionary<string,dynamic> GetDic(HttpRequestBase request)
{
    String jsonString = new StreamReader(request.InputStream).ReadToEnd();
    return Newtonsoft.Json.JsonConvert.DeserializeObject<Dictionary<string,dynamic>>(jsonString);
}