Hierarchical Multitenancy is coming. Look busy.

Until we get CLI support for creating projects with parent relationships, we have to test via curl. This has given me a chance to clean up a few little techniques on using jq andd heredocs.

#!/usr/bin/bash -x
. ./keystonerc_admin

TOKEN=$( curl -si  -H "Content-type: application/json"  -d@- $OS_AUTH_URL/auth/tokens <<EOF | awk '/X-Subject-Token/ {print $2}'
{
    "auth": {
        "identity": {
            "methods": [
                "password"
            ],
            "password": {
                "user": {
                    "domain": {
                        "name": "$OS_USER_DOMAIN_NAME"
                    },
                    "name": "admin",
                    "password": "$OS_PASSWORD"
                }
            }
        },
        "scope": {
            "project": {
                "domain": {
                    "name": "$OS_PROJECT_DOMAIN_NAME"
                },
                "name": "$OS_PROJECT_NAME"
            }
        }
    }
}
EOF
)

PARENT_PROJECT=$( curl  -H "Content-type: application/json" -H"X-Auth-Token:$TOKEN"  -d@- $OS_AUTH_URL/projects <<EOF |  jq -r '.project  | {id}[]  '
{
    "project": {
        "description": "parent project",
        "domain_id": "default",
        "enabled": true,
        "name": "Parent"
    }
}
EOF
)

echo $PARENT_PROJECT


curl  -H "Content-type: application/json" -H"X-Auth-Token:$TOKEN"  -d@- $OS_AUTH_URL/projects <<EOF 
{
    "project": {
        "description": "demo-project",
        "parent_project_id": "$PARENT_PROJECT",
        "domain_id": "default",
        "enabled": true,
        "name": "child"
    }
}
EOF

Note that this uses V3 of the API. I have the following keystone_adminrc

export OS_USERNAME=admin
export OS_USER_DOMAIN_NAME=Default
export OS_PROJECT_DOMAIN_NAME=Default
export OS_PROJECT_NAME=admin
export OS_PASSWORD=cf8dcb8aae804722
export OS_AUTH_URL=http://192.168.1.80:5000/v3/

export OS_IDENTITY_API_VERSION=3

export OS_REGION_NAME=RegionOne
export PS1='[\u@\h \W(keystone_admin)]\$ '