Question

我在异步方法中返回变量时遇到麻烦。我可以执行该代码，但是无法返回该电子邮件地址。

    public async Task<string> GetSignInName (string id)
    {

        RestClient client = new RestClient("https://graph.windows.net/{tenant}/users");
        RestRequest request = new RestRequest($"{id}");
        request.AddParameter("api-version", "1.6");
        request.AddHeader("Authorization", $"Bearer {token}");
        //string emailAddress = await client.ExecuteAsync<rootUser>(request, callback);

        var asyncHandler = client.ExecuteAsync<rootUser>(request, response =>
        {
            CallBack(response.Data.SignInNames);
        });

        return "test"; //should be a variable
    }

Answer 1

RestSharp内置了用于执行基于任务的异步模式（TAP）的方法。这是通过public void PPLChoiceOption(int selection) { buttons[selection].GetComponent<Button>().interactable = false; PPLTextBox.GetComponent<Text>().text = hints[numberOfWrongAnswers - 1]; showRandomHintImage(); PPLChoiceMade = selection; }方法调用的。这将给您一个响应，并且RestClient.ExecuteTaskAsync<T>属性将具有通用参数的反序列化版本（在您的情况下为rootUser）。

response.Data

请注意，public async Task<string> GetSignInName (string id) { RestClient client = new RestClient("https://graph.windows.net/{tenant}/users"); RestRequest request = new RestRequest($"{id}"); request.AddParameter("api-version", "1.6"); request.AddHeader("Authorization", $"Bearer {token}"); var response = await client.ExecuteTaskAsync<rootUser>(request); if (response.ErrorException != null) { const string message = "Error retrieving response from Windows Graph API. Check inner details for more info."; var exception = new Exception(message, response.ErrorException); throw exception; } return response.Data.Username; }对于C＃中的类不是一个好名字。我们的常规约定是PascalCase类名，因此它应该是RootUser。

如何在RestSharp中使用ExecuteAsync返回变量

1 个答案: