Expose ContainerName in Docker provider

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Thomas Quinot 2023-03-20 17:42:06 +01:00 committed by GitHub
parent 99d779a546
commit 4bc2305ed3
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2 changed files with 11 additions and 8 deletions

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@ -440,10 +440,11 @@ _Optional, Default=```Host(`{{ normalize .Name }}`)```_
The `defaultRule` option defines what routing rule to apply to a container if no rule is defined by a label. The `defaultRule` option defines what routing rule to apply to a container if no rule is defined by a label.
It must be a valid [Go template](https://pkg.go.dev/text/template/), and can use It must be a valid [Go template](https://pkg.go.dev/text/template/),
[sprig template functions](https://masterminds.github.io/sprig/). and can use [sprig template functions](https://masterminds.github.io/sprig/).
The container service name can be accessed with the `Name` identifier, The container name can be accessed with the `ContainerName` identifier.
and the template has access to all the labels defined on this container. The service name can be accessed with the `Name` identifier.
The template has access to all the labels defined on this container with the `Labels` identifier.
```yaml tab="File (YAML)" ```yaml tab="File (YAML)"
providers: providers:

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@ -75,9 +75,11 @@ func (p *Provider) buildConfiguration(ctx context.Context, containersInspected [
model := struct { model := struct {
Name string Name string
ContainerName string
Labels map[string]string Labels map[string]string
}{ }{
Name: serviceName, Name: serviceName,
ContainerName: strings.TrimPrefix(container.Name, "/"),
Labels: container.Labels, Labels: container.Labels,
} }