从非异步方法运行异步代码
我想从Azure AD获取访问令牌,并且异步调用永远不会结束。我不确定自己做错了什么,也许有人可以指引我走向正确的方向。这是代码:从非异步方法运行异步代码
private static void GetAccessTokenNonAsync()
{
Func<System.Threading.Tasks.Task> task = async() => { await GetAccessToken().ConfigureAwait(false); };
task().Wait();
}
private static async Task<AuthenticationResult> GetAccessToken()
{
string aadInstance = ConfigurationManager.AppSettings["ida:AADInstance"];
string clientId = ConfigurationManager.AppSettings["ida:ClientId"];
string clientSecret = ConfigurationManager.AppSettings["ida:ClientSecret"];
string source = ConfigurationManager.AppSettings["ExchangeOnlineId"];
var authContext = new AuthenticationContext(aadInstance, false);
var credentials = new ClientCredential(clientId, clientSecret);
var appRedirectUrl = HttpContext.Current.GetOwinContext().Request.Scheme + "://" + HttpContext.Current.GetOwinContext().Request.Host + HttpContext.Current.GetOwinContext().Request.PathBase + "/";
var res =
await
authContext.AcquireTokenByAuthorizationCodeAsync(
((ClaimsIdentity)HttpContext.Current.User.Identity).FindFirst("AuthenticationCode").Value,
new Uri(appRedirectUrl), credentials, source).ConfigureAwait(false);
Current.AccessToken = res.AccessToken;
return res;
}
private static string AccessToken
{
get
{
return HttpContext.Current.Session["AccessToken"].ToString();
}
set { HttpContext.Current.Session["AccessToken"] = value; }
}
而在我的主要方法我称之为GetAccessTokenNonAsync()
您使用async关键字不当。像这样的东西应该工作:
public static void GetAccessTokenNonAsync()
{
// Call the async method and get the resulting task.
Task<AuthenticationResult> accessTokenTask = GetAccessToken();
// Wait for the task to finish.
accessTokenTask.Wait();
}
private async static Task<AuthenticationResult> GetAccessToken()
{
return await Task.Factory.StartNew<AuthenticationResult>
(
() =>
{
// Insert code here.
return new AuthenticationResult();
}
);
}
由于您的异步代码返回的东西(使用任务签名),尝试,如果你的目标最新的.NET Framework这样做:
private static void GetAccessTokenNonAsync()
{
var result = GetAccessToken().ConfigureAwait(false).Result;
}
您正在运行到deadlock scenario because you're blocking on async code 。我在我的博客上描述了细节。
理想情况下,您不应该在任何新代码中使用HttpContext.Current
。它被认为是很差的练习,并且已经在ASP.NET Core中完全删除(并且不会回来)。
它看起来像你想要做的是将令牌存储在会话状态,并(如果有必要,(异步))创建它。适当的方式做,这是使访问异步:
public static async Task<string> GetAccessTokenAsync()
{
var result = HttpContext.Current.Session["AccessToken"] as string;
if (result != null)
return result;
...
var res = await authContext.AcquireTokenByAuthorizationCodeAsync(...);
result = res.AccessToken.ToString();
HttpContext.Current.Session["AccessToken"] = result;
return result;
}
我不认为这解决了我的问题。我需要在用户需要使用它时重新生成访问令牌,所以我需要它同步。我还在所有的异步调用上使用ConfigureAwait(false),但它仍然锁定用户界面... –
@OanaMarina:这不是问题;你只是异步而非同步地重新生成它。 'ConfigureAwait(false)'只在整个调用堆栈中使用时才有效; 'AcquireTokenByAuthorizationCodeAsync'可能没有使用它。 –
如果我异步重新生成它,我不能确定它会在我使用它的下一行代码中可用。我需要等待它生成,然后使用它... –
终于工作的代码是这样的:
private static void GetAccessTokenNonAsync()
{
var userId = ((ClaimsIdentity)HttpContext.Current.User.Identity).FindFirst("UserId").Value;
var task = System.Threading.Tasks.Task.Run(async() =>
{
return await GetAccessToken(userId);
});
task.Wait();
Current.AccessToken = task.Result.AccessToken;
Current.AccessTokenExpiresOn = task.Result.ExpiresOn.ToString();
}
private static async Task<AuthenticationResult> GetAccessToken(string userId)
{
return await System.Threading.Tasks.Task.Factory.StartNew<AuthenticationResult>
(
() =>
{
string aadInstance = ConfigurationManager.AppSettings["ida:AADInstance"];
string clientId = ConfigurationManager.AppSettings["ida:ClientId"];
string clientSecret = ConfigurationManager.AppSettings["ida:ClientSecret"];
string source = ConfigurationManager.AppSettings["ExchangeOnlineId"];
var authContext = new AuthenticationContext(aadInstance, false);
var credentials = new ClientCredential(clientId, clientSecret);
var res =
authContext.AcquireTokenSilentAsync(source, credentials,
new UserIdentifier(userId, UserIdentifierType.UniqueId)).Result;
return res;
}
);
}
的可能的复制[我将如何运行的异步任务方法同步?] (http://stackoverflow.com/questions/5095183/how-would-i-run-an-async-taskt-method-synchronously) –
Igor
@Igor:在这个问题上接受的答案有严重的重入问题。请不要推荐它。 –