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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]

Updated about a year ago

Finding project by name


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