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我有 3 张桌子：

shops, 主键cid,zbid
shop_items, 主键id
shop_inventory, 主键id

shops a与shop_items b以下a.cid=b.cid AND a.zbid=b.szbid shops相关：与shop_inventory

shop_items b与shop_inventory c以下有关：b.cid=c.cid AND b.id=c.iid

现在，我想运行一个返回a.*（来自的所有列shops）的查询。那将是：

SELECT a.* FROM shops a WHERE a.cid=1 AND a.zbid!=0

请注意，该WHERE子句是必要的。

接下来，我要返回每个商店的商品数量：

SELECT
  a.*,
  COUNT(b.id) items
FROM shops a
LEFT JOIN shop_items b ON b.cid=a.cid AND b.szbid=a.zbid
WHERE a.cid=1
GROUP BY b.szbid,b.cid

如您所见，我添加了一个GROUP BY子句以使其起作用。

接下来，我想返回商店中每件商品的平均价格。这并不难：

SELECT
  a.*,
  COUNT(b.id) items,
  AVG(COALESCE(b.price,0)) average_price
FROM shops a
LEFT JOIN shop_items b ON b.cid=a.cid AND b.szbid=a.zbid
WHERE a.cid=1
GROUP BY b.szbid,b.cid

我的下一个标准是它变得复杂的地方。我还想返回每个商店的唯一买家。这可以通过查询shop_inventory c、获取COUNT(DISTINCT c.zbid). 现在记住这些表是如何相关的；这只应针对与相应商店拥有c的商品相关的行进行。ba

我尝试执行以下操作：

SELECT
  a.*,
  COUNT(b.id) items,
  AVG(COALESCE(b.price,0)) average_price,
  COUNT(DISTINCT c.zbid)
FROM shops a
LEFT JOIN shop_items b     ON b.cid=a.cid AND b.szbid=a.zbid
LEFT JOIN shop_inventory c ON c.cid=b.cid AND c.iid=b.id
WHERE a.cid=1
GROUP BY b.szbid,b.cid

然而，这并没有奏效，因为它弄乱了items价值。实现此结果的正确方法是什么？

我还希望能够返回每家商店的购买总数。这将通过查看shop_inventory c并累加c.quantity每家商店的价值来完成。我该如何添加呢？

4

2 回答 2

2

试试这个解决方案：

SELECT    a.*,
          COALESCE(b.item_cnt,  0) AS item_cnt,
          COALESCE(b.avg_price, 0) AS avg_price,
          COALESCE(b.buyer_cnt, 0) AS buyer_cnt
FROM      shops a
LEFT JOIN (
          SELECT    a.cid, 
                    a.szbid, 
                    COUNT(*)     AS item_cnt, 
                    AVG(a.price) AS avg_price,
                    b.buyer_cnt
          FROM      shop_items a
          LEFT JOIN (
                    SELECT   cid,
                             iid,
                             COUNT(DISTINCT zbid) AS buyer_cnt
                    FROM     shop_inventory
                    WHERE    cid = 1
                    GROUP BY cid, 
                             iid
                    ) b ON a.cid = b.cid AND a.id = b.iid
          WHERE     a.cid = 1 AND
                    a.szbid <> 0
          GROUP BY  a.cid,
                    a.szbid
          ) b ON a.cid = b.cid AND a.zbid = b.szbid
WHERE     a.cid = 1 AND 
          a.zbid <> 0
于 2012-08-29T21:49:31.593 回答
1

而不是COUNT(DISTINCT c.zbid)+LEFT JOIN shop_inventory你可以写一个子选择：

SELECT
  a.*,
  COUNT(b.id) items,
  AVG(COALESCE(b.price,0)) average_price,

  ( SELECT COUNT(DISTINCT c.zbid)
    FROM shop_inventory c
    WHERE c.cid=b.cid AND c.iid=b.id
  )

FROM shops a
LEFT JOIN shop_items b     ON b.cid=a.cid AND b.szbid=a.zbid
WHERE a.cid=1
GROUP BY b.szbid,b.cid
于 2012-08-29T21:49:32.717 回答