C# 如何从c中的异步操作方法中获取值#

我有一个

async

方法，比如

[HttpPost]
public async Task<JsonResult> AddTwoIntegers(int param1, int param2)
{         
   var result = await (param1 + param2);
   return Json(new {finalValue: result}, JsonRequestBehavior.AllowGet)            
}

   Type data = null; //Note that I put Type because I don't know your type for result.Data 
  jsonData.ContinueWith(task =>
  {
       JsonResult result = task.Result;
       if (result.Data.ToString() == "") {
             data = result.Data;
       }            
  });

  // I want to retrieve the value returned and use that value in some operation.
  //Now you can use your variable data here!!
  return Json("Success", JsonRequestBehavior.AllowGet);

---

如何从

操作结果

获取返回值

应该是这样的：

JsonResult = await AddTwoIntegers(5,10);

   Type data = null; //Note that I put Type because I don't know your type for result.Data 
  jsonData.ContinueWith(task =>
  {
       JsonResult result = task.Result;
       if (result.Data.ToString() == "") {
             data = result.Data;
       }            
  });

  // I want to retrieve the value returned and use that value in some operation.
  //Now you can use your variable data here!!
  return Json("Success", JsonRequestBehavior.AllowGet);

---

检查它应该是这样的：

JsonResult = await AddTwoIntegers(5,10);

   Type data = null; //Note that I put Type because I don't know your type for result.Data 
  jsonData.ContinueWith(task =>
  {
       JsonResult result = task.Result;
       if (result.Data.ToString() == "") {
             data = result.Data;
       }            
  });

  // I want to retrieve the value returned and use that value in some operation.
  //Now you can use your variable data here!!
  return Json("Success", JsonRequestBehavior.AllowGet);

---

选中

则需要声明在操作外部存储结果的变量。然后使用变量

   Type data = null; //Note that I put Type because I don't know your type for result.Data 
  jsonData.ContinueWith(task =>
  {
       JsonResult result = task.Result;
       if (result.Data.ToString() == "") {
             data = result.Data;
       }            
  });

  // I want to retrieve the value returned and use that value in some operation.
  //Now you can use your variable data here!!
  return Json("Success", JsonRequestBehavior.AllowGet);

Task jsonData=this.addTwoInteger（5,10）

   Type data = null; //Note that I put Type because I don't know your type for result.Data 
  jsonData.ContinueWith(task =>
  {
       JsonResult result = task.Result;
       if (result.Data.ToString() == "") {
             data = result.Data;
       }            
  });

  // I want to retrieve the value returned and use that value in some operation.
  //Now you can use your variable data here!!
  return Json("Success", JsonRequestBehavior.AllowGet);

---

}

然后需要声明在操作外部存储结果的变量。然后使用变量

   Type data = null; //Note that I put Type because I don't know your type for result.Data 
  jsonData.ContinueWith(task =>
  {
       JsonResult result = task.Result;
       if (result.Data.ToString() == "") {
             data = result.Data;
       }            
  });

  // I want to retrieve the value returned and use that value in some operation.
  //Now you can use your variable data here!!
  return Json("Success", JsonRequestBehavior.AllowGet);

Task jsonData=this.addTwoInteger（5,10）

   Type data = null; //Note that I put Type because I don't know your type for result.Data 
  jsonData.ContinueWith(task =>
  {
       JsonResult result = task.Result;
       if (result.Data.ToString() == "") {
             data = result.Data;
       }            
  });

  // I want to retrieve the value returned and use that value in some operation.
  //Now you can use your variable data here!!
  return Json("Success", JsonRequestBehavior.AllowGet);

---

}

您需要更改返回

任务的方法，并将其标记为
async
，然后等待
AddTwoIntegers（）
的结果

   Type data = null; //Note that I put Type because I don't know your type for result.Data 
  jsonData.ContinueWith(task =>
  {
       JsonResult result = task.Result;
       if (result.Data.ToString() == "") {
             data = result.Data;
       }            
  });

  // I want to retrieve the value returned and use that value in some operation.
  //Now you can use your variable data here!!
  return Json("Success", JsonRequestBehavior.AllowGet);

public async Task<ActionResult> SomeFunction(string userSettingsViewModel)
{          

    JsonResult result = await this.AddTwoIntegers(5, 10);  
    var jsonData = result.data;

    // ...

    return Json("Success", JsonRequestBehavior.AllowGet);
}

public async Task SomeFunction（字符串usersetingsviewmodel）
{          
JsonResult=wait this.addTwoInteger（5,10）；
var jsonData=result.data；
// ...
返回Json（“Success”，JsonRequestBehavior.AllowGet）；
}

为了完整性，示例方法应该如下所示

   Type data = null; //Note that I put Type because I don't know your type for result.Data 
  jsonData.ContinueWith(task =>
  {
       JsonResult result = task.Result;
       if (result.Data.ToString() == "") {
             data = result.Data;
       }            
  });

  // I want to retrieve the value returned and use that value in some operation.
  //Now you can use your variable data here!!
  return Json("Success", JsonRequestBehavior.AllowGet);

[HttpPost]
public Task<JsonResult> AddTwoIntegers(int param1, int param2)
{         
   var result = param1 + param2;
   return Task.FromResult(Json(new {finalValue: result}, 
                               JsonRequestBehavior.AllowGet));            
}

[HttpPost]
公共任务addTwoInteger（int-param1，int-param2）
{         
var结果=参数1+参数2；
返回Task.FromResult（Json）（新的{finalValue:result}，
JsonRequestBehavior.AllowGet））；
}
您需要更改返回
任务的方法，并将其标记为
async
，然后等待
AddTwoIntegers（）
的结果

   Type data = null; //Note that I put Type because I don't know your type for result.Data 
  jsonData.ContinueWith(task =>
  {
       JsonResult result = task.Result;
       if (result.Data.ToString() == "") {
             data = result.Data;
       }            
  });

  // I want to retrieve the value returned and use that value in some operation.
  //Now you can use your variable data here!!
  return Json("Success", JsonRequestBehavior.AllowGet);

public async Task<ActionResult> SomeFunction(string userSettingsViewModel)
{          

    JsonResult result = await this.AddTwoIntegers(5, 10);  
    var jsonData = result.data;

    // ...

    return Json("Success", JsonRequestBehavior.AllowGet);
}

public async Task SomeFunction（字符串usersetingsviewmodel）
{          
JsonResult=wait this.addTwoInteger（5,10）；
var jsonData=result.data；
// ...
返回Json（“Success”，JsonRequestBehavior.AllowGet）；
}

为了完整性，示例方法应该如下所示

   Type data = null; //Note that I put Type because I don't know your type for result.Data 
  jsonData.ContinueWith(task =>
  {
       JsonResult result = task.Result;
       if (result.Data.ToString() == "") {
             data = result.Data;
       }            
  });

  // I want to retrieve the value returned and use that value in some operation.
  //Now you can use your variable data here!!
  return Json("Success", JsonRequestBehavior.AllowGet);

[HttpPost]
public Task<JsonResult> AddTwoIntegers(int param1, int param2)
{         
   var result = param1 + param2;
   return Task.FromResult(Json(new {finalValue: result}, 
                               JsonRequestBehavior.AllowGet));            
}

[HttpPost]
公共任务addTwoInteger（int-param1，int-param2）
{         
var结果=参数1+参数2；
返回Task.FromResult（Json）（新的{finalValue:result}，
JsonRequestBehavior.AllowGet））；
}
等待（param1+param2）酷！这甚至可以编译吗？您不能等待
int
<代码>等待（param1+param2）
我的用例与我给出的示例代码不同。我实际需要的是，我希望在另一个操作方法中访问该值。但它只是从那里跳出来的。我没有得到应有的价值，最好展示一下你所做的。因为使用
async
和
await
编写的代码是可以避免的。你不能那样使用它。@lnanikian-yah这一切都是关于更改密码之类的。我不能在这里写它。
wait（param1+param2）酷！这甚至可以编译吗？您不能等待
int
<代码>等待（param1+param2）
我的用例与我给出的示例代码不同。我实际需要的是，我希望在另一个操作方法中访问该值。但它只是从那里跳出来的。我没有得到应有的价值，最好展示一下你所做的。因为使用
async
和
await
编写的代码是可以避免的。你不能那样使用它。@lnanikian-yah这一切都是关于更改密码之类的。我不能写在这里，我的问题是我能把这两个值相加吗。当我返回json Result时。我的问题是，我能把这两个值相加吗。当我返回json Result时。如何访问返回整数值的函数