Querying a component based on resource identifier

[mark]

Hi,

I'd like to query all components published under a certain drive letter. Is there a way to do this?

My best guess at this point would be a query similar to:

'Component where component_locations.resource_identifier any (contains: "G:/")'

which I realize is not syntactically correct but hopefully you get the drift...

 

Thanks!

-Mark

[Lucas Correia]

Hi Mark,

How to do this depends a bit on the location used to publish the components. Three cases are:

1. Components are published to a location representing a storage G:/
2. Components are published to a location with a structure plugin that maps the component to different storages based on e.g. project.
3. Components are published to an location with an absolute path (e.g. ftrack.unmanaged in ftrack-python-api)

For case 3, your query is almost correct. The syntax is:

Component where component_locations any (resource_identifier like "G:%")

Here is a an example which lists all matching components:

# Query matching components, prefetching required attribtues
components = session.query(
    'select name, file_type, version.link from '
    'Component where component_locations any (resource_identifier like "G:%")'
)

# Format link atttributes
format_link = lambda entity: u' / '.join([item['name'] for item in entity['link']])

# Loop over results and print components
for component in components:
    print u'{}: {}{}'.format(
        format_link(component['version']),
        component['name'],
        component['file_type']
    )

Regards,
Lucas

[mark]

That did it. Thanks!