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Expose ContainerName in Docker provider
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@ -440,10 +440,11 @@ _Optional, Default=```Host(`{{ normalize .Name }}`)```_
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The `defaultRule` option defines what routing rule to apply to a container if no rule is defined by a label.
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It must be a valid [Go template](https://pkg.go.dev/text/template/), and can use
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[sprig template functions](https://masterminds.github.io/sprig/).
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The container service name can be accessed with the `Name` identifier,
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and the template has access to all the labels defined on this container.
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It must be a valid [Go template](https://pkg.go.dev/text/template/),
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and can use [sprig template functions](https://masterminds.github.io/sprig/).
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The container name can be accessed with the `ContainerName` identifier.
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The service name can be accessed with the `Name` identifier.
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The template has access to all the labels defined on this container with the `Labels` identifier.
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```yaml tab="File (YAML)"
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providers:
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@ -74,11 +74,13 @@ func (p *Provider) buildConfiguration(ctx context.Context, containersInspected [
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serviceName := getServiceName(container)
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model := struct {
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Name string
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Labels map[string]string
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Name string
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ContainerName string
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Labels map[string]string
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}{
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Name: serviceName,
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Labels: container.Labels,
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Name: serviceName,
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ContainerName: strings.TrimPrefix(container.Name, "/"),
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Labels: container.Labels,
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}
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provider.BuildRouterConfiguration(ctx, confFromLabel.HTTP, serviceName, p.defaultRuleTpl, model)
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