Peewee 選擇具有多個連接和多個計數的查詢

SimonB 发表于 Dev

西蒙B

我一直在嘗試編寫一個 peewee 選擇查詢，它會生成一個包含 2 個計數的表（一個是與彩票相關的獎品數量，一個是與彩票相關的包裹數量），以及彩票模型。

我設法編寫了 1 個計數工作的選擇查詢（見下文），然後我不得不將 ModelSelects 轉換為列表並手動加入它們（我認為這很hacky）。

我確實設法編寫了一個連接結果的選擇查詢，但它會將包裹計數與獎品計數相乘（此後我丟失了該查詢）。

我也嘗試過使用 .switch(Lottery) 但我對此沒有任何運氣。

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']

雖然上面的代碼有效，但有沒有更簡潔的方法來編寫這個查詢？

科萊弗

最好的辦法是用子查詢來做到這一點。

# 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)