環境Windows 10 專業版.NET Core 控制臺應用程序代碼我有一個看起來像這樣的抽象消息接收器。在此代碼中，entity是Subscription（例如user）的名稱。public class AzureMessageReceiver : ITdlMessageReceiver{ private readonly ServiceBusConnection serviceBusConnection; private readonly ILogger<AzureMessageReceiver> logger; public AzureMessageReceiver(ServiceBusConnection serviceBusConnection, ILogger<AzureMessageReceiver> logger) { this.serviceBusConnection = serviceBusConnection; this.logger = logger; } public async Task<TdlMessage<T>> ReceiveAsync<T>(string topic, string entity) where T : class { try { var subscriptionPath = EntityNameHelper.FormatSubscriptionPath(topic, entity); var messageReceiver = new MessageReceiver(serviceBusConnection, subscriptionPath, ReceiveMode.ReceiveAndDelete); var message = await messageReceiver.ReceiveAsync(); if (message == null) { return null; } var messageString = Encoding.UTF8.GetString(message.Body); return JsonConvert.DeserializeObject<TdlMessage<T>>(messageString); } catch (Exception ex) { logger.LogError(ex, "Error receiving Azure message."); return null; } }}注入ServiceBusConnection的構造是這樣的。注：此相同的連接的初始化工作，以寫郵件到同Topic和Subscription。services.AddSingleton(serviceProvider => new ServiceBusConnection(configuration[$"{DurableCommunicationKey}:AzureConnectionString"]));問題每次我執行這一行時，var message = await messageReceiver.ReceiveAsync();它都會使控制臺應用程序崩潰。沒有Exception，也沒有Event Viewer。我試過的使用Secondary Connection String來自 ASB提供超時，如 messageReceiver.ReceiveAsync(TimeSpan.FromMinutes(1));將注入的內容topic從主題名稱更改為主題的整個 URL（例如https://{...}.servicebus.windows.net/{topicName}）更改ReceiveMode為PeekLock后跟蹤ConfigureAwait(false)到ReceiveAsync的呼叫。將超時更改為TimeSpan.Zero. 注意：這并沒有崩潰的應用程序，但實際上拋出Exception該得到記錄。
1 回答
侃侃爾雅
TA貢獻1801條經驗 獲得超16個贊
async void應該轉換為 anasync Task以及您應該等待Task.Delay而不是調用Thread.Sleep. 如果要異步，則需要一直異步
static async Task Receive(ITdlMessageReceiver receiver, ILogger logger) {
while (true) {
var message = await receiver.ReceiveAsync<TdlMessage<object>>(topic, entity);
if (message != null) {
logger.LogDebug($"Message received. Topic: {topic}. Action: {Enum.GetName(typeof(TopicActions), message.Action)}. Message: {JsonConvert.SerializeObject(message)}.");
}
await Task.Delay(sleepTime);
}
}
嘗試使代碼一直異步，是的，但是作為控制臺應用程序（單線程），您將被允許調用Wait()該Receive方法，Main因為它不會混合調用會導致異步流出現問題。
public static void Main(string[] args) {
//...
//...
//...
Receive(receiver, logger).Wait();
}
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