Python Query.set_order_by_columns示例

示例#1

def test_query_order_by_and_results_order():
    query = Query()
    query.set_table("Movies")
    query.set_return_columns(["Name", "Studio"])
    query.set_order_by_columns(["Studio"])
    query.set_results_order(ResultsOrder.DESC)
    cmd = query.to_sql_query()

    assert cmd == "SELECT Name,Studio FROM boxofficementat.Movies ORDER BY Studio DESC LIMIT 0, 100"

示例#2

def test_inner_join_query():

    query = Query()
    query.set_table("Movies")
    query.set_return_columns(["Studios.Name"])
    query.add_inner_join('Studio', 'Studios', 'Id')
    query.add_aggregate_column(AggregateType.COUNT, 'Movies.Id', True)
    query.set_order_by_columns(['COUNT'])
    query.set_results_order(ResultsOrder.DESC)
    query.set_results_offset(80)
    query.set_max_results(20)
    cmd = query.to_sql_query()
    assert cmd == "SELECT Studios.Name,COUNT(DISTINCT Movies.Id) AS COUNT FROM boxofficementat.Movies " \
                  "INNER JOIN boxofficementat.Studios ON boxofficementat.Movies.Studio=boxofficementat.Studios.Id  " \
                  "GROUP BY Studios.Name ORDER BY COUNT DESC LIMIT 80, 20"