make span names more generic by using Route().Path instead of Path
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031177f30c
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7dd3dc70d5
@ -51,6 +51,7 @@ func main() {
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})
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main.Get("/", Handler)
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main.Get("/just/some/more/complex/path/:with/params", Handler)
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// handle interrupts (shutdown)
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c := make(chan os.Signal, 1)
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@ -1,2 +1,5 @@
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GET http://127.0.0.1:3344/
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HTTP 200
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HTTP 500
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GET http://127.0.0.1:3344/just/some/more/complex/path/:with/params
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HTTP 500
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@ -48,7 +48,7 @@ func Start(ctx context.Context, spanName string, opts ...trace.SpanStartOption)
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func FStart(c *fiber.Ctx, opts ...trace.SpanStartOption) (context.Context, trace.Span) {
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span := trace.SpanFromContext(fiberOpentelemetry.FromCtx(c))
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c.SetUserContext(trace.ContextWithSpan(c.UserContext(), span))
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return fiberOpentelemetry.Tracer.Start(c.UserContext(), c.Method()+" "+c.Path(), opts...)
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return fiberOpentelemetry.Tracer.Start(c.UserContext(), c.Method()+" "+c.Route().Path, opts...)
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}
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// Retrieve span using [fiber.Ctx]
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@ -81,7 +81,7 @@ func NewMiddleware(config Config) func(*fiber.Ctx) error {
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return fiberOpentelemetry.New(
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fiberOpentelemetry.Config{
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Tracer: tracer,
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SpanName: "{{ .Method }} {{ .Path }}",
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SpanName: "{{ .Method }} {{ .Route.Path }}",
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TracerStartAttributes: []trace.SpanStartOption{
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trace.WithSpanKind(trace.SpanKindServer),
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trace.WithNewRoot(),
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