Add the Azure subscription-id to the sample delete command given by inletsctl create

[eltariel]

When you run inletsctl create, it prints a command to run in order to delete the server again. The command generated for an Azure deployment is incomplete and will not run without additional parameters.

Expected Behaviour

• Run the inletsctl delete command generated by inletsctl create
• The inlets server is deleted.

Current Behaviour

• The provided command fails:

$ inletsctl delete --provider azure --id "inlets-flamboyant-buck2|deployment-c1f95ea9-6957-4aca-be24-26274269e52d"
Using provider: azure
give a value for --access-token-file, --access-token or set the environment variable "INLETS_ACCESS_TOKEN"

• Adding --access-token-file with the same file used for create also fails:

$ inletsctl delete --provider azure --id "inlets-flamboyant-buck2|deployment-c1f95ea9-6957-4aca-be24-26274269e52d" --access-token-file az-sp.json
Using provider: azure
Deleting host: inlets-flamboyant-buck2|deployment-c1f95ea9-6957-4aca-be24-26274269e52d from azure
resources.GroupsClient#Delete: Failure sending request: StatusCode=400 -- Original Error: Code="InvalidSubscriptionId" Message="The provided subscription identifier 'resourcegroups' is malformed or invalid."

• Adding the subscription id succeeds:

$ inletsctl delete --provider azure --id "inlets-flamboyant-buck2|deployment-c1f95ea9-6957-4aca-be24-26274269e52d" --access-token-file az-sp.json --subscription-id $SUBSCRIPTION_ID
Using provider: azure
Deleting host: inlets-flamboyant-buck2|deployment-c1f95ea9-6957-4aca-be24-26274269e52d from azure

Possible Solution

Maybe including at least the --subscription-id in the generated command would be a good idea. Alternatively, the subscription ID is already in the access token file so if it's not specified from the command line it could just be read from there too.

[alexellis]

Hi,

This seems like a reasonable improvement in user-experience. We'd be open to a pull request / contribution to inletsctl (which is an open source component).

Alex