Finding project by name

Not optimized for rate limit

Iterate over all projects and compare names.

project = [ p for p in api.projects.query().all() if p.name == project_name ][0]

Optimized for rate limit

Use 'name' query parameter in search to restrict results. Query parameter performs partial match, so name comparison is still required to ensure the exact match.

project = [ p for p in api.projects.query(name=project_name).all() if p.name == project_name ][0]