具有多个连接和多个计数的 Peewee 选择查询

Question

我一直在尝试编写一个 peewee 选择查询，它会生成一个包含 2 个计数的表（一个是与彩票相关的奖品数量，一个是与彩票相关的包裹数量），以及Lottery 模型中的字段。

我设法编写了 1 个计数有效的选择查询（见下文），然后我不得不将 ModelSelects 转换为列表并手动加入它们（我认为这很 hacky）。

我确实设法编写了一个连接结果的选择查询，但它会将包裹数乘以奖品数（我已经丢失了该查询）。

我也尝试过使用 .switch(Lottery) 但我没有任何运气。

query1 = (Lottery.select(Lottery,fn.count(Package.id).alias('packages'))
          .join(LotteryPackage)
          .join(Package)
          .order_by(Lottery.id)
          .group_by(Lottery)
          .dicts())

query2 = (Lottery.select(Lottery.id.alias('lotteryID'), fn.count(Prize.id).alias('prizes'))
          .join(LotteryPrize)
          .join(Prize)
          .group_by(Lottery)
          .order_by(Lottery.id)
          .dicts())

lottery = list(query1)
query3 = list(query2)

for x in range(len(lottery)):
    lottery[x]['prizes'] = query3[x]['prizes']

虽然上面的代码有效，但有没有更简洁的方法来编写这个查询？

Answer 1

最好的办法是使用子查询。

# Create query which gets lottery id and count of packages.
L1 = Lottery.alias()
subq1 = (L1
         .select(L1.id, fn.COUNT(LotteryPackage.package).alias('packages'))
         .join(LotteryPackage, JOIN.LEFT_OUTER)
         .group_by(L1.id))

# Create query which gets lottery id and count of prizes.
L2 = Lottery.alias()
subq2 = (L2
         .select(L2.id, fn.COUNT(LotteryPrize.prize).alias('prizes'))
         .join(LotteryPrize, JOIN.LEFT_OUTER)
         .group_by(L2.id))

# Select from lottery, joining on each subquery and returning
# the counts.
query = (Lottery
         .select(Lottery, subq1.c.packages, subq2.c.prizes)
         .join(subq1, on=(Lottery.id == subq1.c.id))
         .join(subq2, on=(Lottery.id == subq2.c.id))
         .order_by(Lottery.name))

for row in query.objects():
    print(row.name, row.packages, row.prizes)