小编Ole*_*hev的帖子

大家好我怎样才能得到这个输出

{
    "@t": "2020-11-03T16:40:21.6133998Z",
    "@m": "HTTP \"POST\" \"/api/Email/SendEmail\" responded 200 in 1.0358 ms",
    "@i": "62d0885c",
    "CorrelationId": "",
    "Host": "localhost:32768",
    "Protocol": "HTTP/1.1",
    "Scheme": "http",
    "ContentType": null,
    "EndpointName": "Email.Controllers.EmailController.SendEmail (Email)",
    "RequestMethod": "POST",
    "RequestPath": "/api/Email/SendEmail",
    "StatusCode": 200,
    "Elapsed": 1.0358,
    "SourceContext": "Serilog.AspNetCore.RequestLoggingMiddleware",
    "RequestId": "",
    "SpanId": "|55f14a36-4918a3efd4f265b9.",
    "TraceId": "55f14a36-4918a3efd4f265b9",
    "ParentId": "",
    "ConnectionId": "0HM402S7EC249"
}

但是当我调用我的控制器时，我看到并运行应用程序，我看到

这就是我设置 Serilog 的方式，我做错了什么

 public class LogConfig
    {
        public static void Configure()
        {
            Log.Logger = new LoggerConfiguration()
               .WriteTo.Console(new RenderedCompactJsonFormatter())
              .CreateLogger();
        }

        public static void EnrichFromRequest(IDiagnosticContext diagnosticContext, HttpContext httpContext)
        {
            diagnosticContext.Set("RequestScheme", httpContext.Request.Host.Value); …