我正在为团队合作游戏库构建API。这些游戏具有定义的属性,例如类型,大小和长度,我正在存储多对多关系。该模型如下所示:
class Game(db.Entity):
game_type = Set('Type')
game_length = Set('Length')
group_size = Set('GroupSize')
...
class Type(db.Entity): # similar for Length, GroupSize
game = Set(Game)
short = Required(str)
full = Required(str)
它们使用不同的类型/长度/大小值填充,然后分配给不同的游戏。效果很好。
我很难弄清楚如何让数据库用户查询例如其中两个与第三个未给出。例如,我希望所有游戏都使用type=race AND size=medium
而不是length=None
。
我之前在SQL中使用子查询和空字符串构建了该库。这是我第一次使用PonyORM:
def get_all_games(**kwargs):
game_type = kwargs.get('game_type', None)
group_size = kwargs.get('group_size', None)
game_length = kwargs.get('game_length', None)
query = select((g, gt, gs, gl) for g in Game
for gt in g.game_type
for gs in g.group_size
for gl in g.game_length)
if game_type:
query = query.filter(lambda g, gt, gs, gl: gt.short == game_type)
if group_size:
query = query.filter(lambda g, gt, gs, gl: gs.short == group_size)
if game_length:
query = query.filter(lambda g, gt, gs, gl: gl.short == game_length)
query = select(g for (g, gt, gs, gl) in query)
result = []
for g in query:
this_game = get_game(g)
result.append(this_game)
return result
对我来说似乎太复杂了。没有元组打包和拆包的方法吗?也许可以在查询中立即使用变量,而不使用if语句?
答案 0 :(得分:1)
您可以在exists
中使用in
或filter
。您也可以使用attribute lifting来简化复杂的联接:
query = Game.select()
if game_length:
# example with exists
query = query.filter(lambda g: exists(
gl for gl in g.game_length
if gl.min <= game_length and gl.max >= game_length))
if group_size:
# example with in
query = query.filter(lambda g: group_size in (
gs.value for gs in g.group_sizes))
if game_type:
# example with attribute lifting
query = query.filter(lambda g: game_type in g.game_types.short)