grafana/k6 · error · ErrInvalidTracesOutput

invalid traces output %q

Error message

invalid traces output %q

What it means

The -o/--traces-output value does not start with 'otel' (the only supported traces output). The config line is cut at the first '=' and the prefix checked; anything other than the literal 'otel' (or the bare 'otel' short form) wraps ErrInvalidTracesOutput with the offending prefix quoted in the message.

Source

Thrown at internal/lib/trace/otel.go:180

	params, err := tracerProviderParamsFromConfigLine(line)
	if err != nil {
		return nil, err
	}

	return NewTracerProvider(ctx, params)
}

func tracerProviderParamsFromConfigLine(line string) (tracerProviderParams, error) {
	params := defaultTracerProviderParams()

	if line == "otel" {
		// Use default params
		return params, nil
	}

	traceOutput, _, _ := strings.Cut(line, "=")
	if traceOutput != "otel" {
		return params, fmt.Errorf("%w %q", ErrInvalidTracesOutput, traceOutput)
	}

	tokens, err := strvals.Parse(line)
	if err != nil {
		return params, fmt.Errorf("error while parsing otel configuration %w", err)
	}

	for _, token := range tokens {
		key := token.Key
		value := token.Value

		var err error
		switch key {
		case "otel":
			err = params.parseURL(value)
			if err != nil {
				return params, fmt.Errorf("couldn't parse the otel URL: %w", err)
			}

View on GitHub (pinned to 01ffac6f24)

Solutions

1. Use -o otel or -o otel=key=value,... for OTLP trace export
2. Remove the unsupported output name shown in the error
3. Check for typos and leading whitespace in the output option

When it happens

Trigger: Thrown at internal/lib/trace/otel.go:180 when the library encounters an invalid state.

Common situations: See trigger scenarios.

AI-assisted analysis of grafana/k6@01ffac6f24 (2026-08-18). Data as JSON: /api/errors/15533af3f4b163bc. Report an issue: GitHub.