以下将在一个查询中为你提供所需的对象：
q = (session.query(Group, Member, Item, Version) .join(Member) .join(Item) .join(Version) .filter(Version.name == my_version) .order_by(Group.number) .order_by(Member.number) ).all()print_tree(q)
但是，你得到的结果将是元组列表
(Group, Member, Item, Version)。现在由你决定以树形形式显示它。下面的代码可能被证明是有用的：
def print_tree(rows): def get_level_diff(row1, row2): """ Returns tuple: (from, to) of different item positions. """ if row1 is None: # first row handling return (0, len(row2)) assert len(row1) == len(row2) for col in range(len(row1)): if row1[col] != row2[col]: return (col, len(row2)) assert False, "should not have duplicates" prev_row = None for row in rows: level = get_level_diff(prev_row, row) for l in range(*level): print 2 * l * " ", row[l] prev_row = row
Update-1：如果你愿意放弃
lazy = 'dynamic'前两个关系，则可以执行查询以
object network使用以下代码加载整个对象（与上述元组相对）：
q = (session.query(Group) .join(Member) .join(Item) .join(Version) # @note: here we are tricking sqlalchemy to think that we loaded all these relationships, # even though we filter them out by version. Please use this only to get data and display, # but not to continue working with it as if it were a regular UnitOfWork .options( contains_eager(Group.member). contains_eager(Member.items). contains_eager(Item.version) ) .filter(Version.name == my_version) .order_by(Group.number) .order_by(Member.number) ).all()# print tree: easy navigation of relationshipsfor g in q: print "", g for m in g.member: print 2 * " ", m for i in m.items: print 4 * " ", i
