从C#中的JsonResult中删除一个元素

Question

我有一个JsonResult从MVC方法返回的对象,但我需要在发送之前从中删除一个元素.

更新:
我试图在没有映射的情况下完成它,因为对象庞大而且非常复杂.

我怎样才能做到这一点？

例如:

public class MyClass {

   public string PropertyToExpose {get; set;}
   public string PropertyToNOTExpose {get; set;}
   public string Otherthings {get; set;}

}

和

JsonResult result = new JsonResult();
result = Json(myObject, JsonRequestBehavior.AllowGet);

然后从结果中删除 PropertyToNOTExpose.

真实代码的更新:

public JsonResult GetTransaction(string id)
{    
    //FILL UP transaction Object

    JsonResult resultado = new JsonResult();

    if (CONDITION USER HAS NOT ROLE) {
        var jObject = JObject.FromObject(transaction);
        jObject.Remove("ValidatorTransactionId");
        jObject.Remove("Validator");
        jObject.Remove("WebSvcMethod");
        resultado = Json(jObject, JsonRequestBehavior.AllowGet);
    } else {
        //etc.
    }
    return resultado;
}

Answer 1

您可以创建一个新对象,不包括您不希望在结果中发送的属性...

var anonymousObj = new {
   myObject.PropertyToExpose,
   myObject.Otherthings
};
JsonResult result = Json(anonymousObj, JsonRequestBehavior.AllowGet);

另一个选项可能是将对象转换为Newtonsoft.Json.Linq.JObject并使用JObject删除该属性.Remove方法(String)

var jObject = JObject.FromObject(myObject);
jObject.Remove("PropertyToNOTExpose");
var json = jObject.ToString(); // Returns the indented JSON for this token.
var result = Content(json,"application/json");