0
我真的面对奇怪的行为,当我尝试运行在Linux操作系统我泊坞窗图像 - 图像中包含的.Net核心Web API项目运行码头工人形象造成隼异常
这里是我的Program.cs
var host = new WebHostBuilder()
.UseKestrel().UseStartup<Startup>().Build();
host.Run();
我的码头工人,compose.yml文件
version: '2'
services:
server:
image: repo.testCompany.com:1443/testCompany:6
ports:
- "60000:60000"
hostname: ucp.${HOST_HOSTNAME}
restart: unless-stopped
的形象是建立和推动successfully.The问题就来当Web API的尝试运行容器内的项目,并且运行时异常为t hrown -
Unhandled Exception: System.AggregateException: One or more errors occurred.
(Error -99 EADDRNOTAVAIL address not available) --->
Microsoft.AspNetCore.Server.Kestrel.Internal.Networking.UvException: Error
-99 EADDRNOTAVAIL address not available
at Microsoft.AspNetCore.Server.Kestrel.Internal.Networking.Libuv.Check(Int32
statusCode)
at Microsoft.AspNetCore.Server.Kestrel.Internal.Networking.Libuv.tcp_bind(UvTcpHandle handle, SockAddr& addr, Int32 flags)
at Microsoft.AspNetCore.Server.Kestrel.Internal.Networking.UvTcpHandle.Bind(ServerAddress address)
at Microsoft.AspNetCore.Server.Kestrel.Internal.Http.TcpListener.CreateListenSocket()
at Microsoft.AspNetCore.Server.Kestrel.Internal.Http.Listener.<>c.<StartAsync>b__6_0(Object state)
--- End of inner exception stack trace ---
at System.Threading.Tasks.Task.ThrowIfExceptional(Boolean includeTaskCanceledExceptions)
at System.Threading.Tasks.Task.Wait(Int32 millisecondsTimeout, CancellationToken cancellationToken)
at System.Threading.Tasks.Task.Wait()
at Microsoft.AspNetCore.Server.Kestrel.Internal.KestrelEngine.CreateServer(ServerAddress address)
at Microsoft.AspNetCore.Server.Kestrel.KestrelServer.Start[TContext](IHttpApplication`1 application)
at Microsoft.AspNetCore.Hosting.Internal.WebHost.Start()
at Microsoft.AspNetCore.Hosting.WebHostExtensions.Run(IWebHost host, CancellationToken token, String shutdownMessage)
at Microsoft.AspNetCore.Hosting.WebHostExtensions.Run(IWebHost host)
at Exactor.XeroConnector.API.Program.Main(String[] args)
的主要问题是,错误不会重现我的本地机器上,但只有当图像是建立并通过詹金斯上传。 你能请别人提出建议吗。
何你指定的主机和端口,在您的应用程序正在运行? –
我试图添加 .UseUrls(“http:// *:60000”) 但错误仍然相同。 – NDym