Question

我需要查询两个表来打印列表 - PURCHASE HISTORY和PEOPLE

PURCHASE HISTORY
----------------
purchase_txnid...purchase_date...purchase_userid...purchase_productid

PEOPLE
------
people_householdid...people_userid...people_street...people_city...people_state...(etc)

我需要让所有人householdid让某人购买productid = "X"（基本上是SOMEONE已购买产品的所有人的清单，然后显示他们的每次购买历史。

PURCHASE HISTORY有七种不同的产品和320,000条记录。

现在，我正在查询所有人，RIGHT JOIN购买历史记录以选择购买"X"的人，然后遍历结果，并在每个结果中查询{{1}这个家庭中的每个人以及他们的每个购买历史。由于有SELECT人和45,000次购买，因此速度非常慢。

关于如何将其合并为单个查询或优化的任何想法？

更新

以下是查询：

320,000

Answer 1

SELECT DISTINCT peopleB.userid, peopleB.* FROM purchase
JOIN people AS peopleA ON peopleA.people_userid = purchase_userid
JOIN people AS peopleB ON peopleB.people_householdid = peopleA.householdid
WHERE purchase.purchase_productid = "X"

您可以通过向表中添加这些索引来加速此查询：

CREATE INDEX productid ON purchase (purchase_productid)
CREATE INDEX householdid ON people (people_householdid)

我认为people_userid已经是people中的主要密钥，否则您应该将其设为唯一：

CREATE UNIQUE INDEX userid ON people (people_userid)

编辑 - 您要求提供完整的购买历史记录：

SELECT peopleB.*, phB.* FROM purchase_history AS phA
JOIN people AS peopleA ON peopleA.people_userid = phA.purchase_userid
JOIN people AS peopleB ON peopleB.people_householdid = peopleA.householdid
LEFT JOIN purchase_history AS phB ON phB.purchase_userid = peopleB.userID
GROUP BY peopleB.id, phB.purchase_txnid
WHERE purchase.purchase_productid = "X"

请注意，每次购买时，这都会为人们提供一次数据。

此外，我不知道该查询是否真的按照我想要的方式运行，或者它的速度有多快/慢。您也可以尝试倒转GROUP BY行中的列，看看是否更快。

Answer 2

您可以使用子查询和join：

执行此操作

select p.*
from people p join
     (select distinct p.people_householdid
      from PurchaseHistory ph join
           People p
           on ph.purchase_userid = p.people_userid
      where ph.purchase_productid = 'X'
     ) ph
     on ph.people_userid = p.people_userid;

编辑：

如果您想标记进行购买的成员：

select p.*,
       max(p.people_userid = ph.people_userid) as IsPurchaser
from people p join
     (select p.people_householdid, p.people_userid
      from PurchaseHistory ph join
           People p
           on ph.purchase_userid = p.people_userid
      where ph.purchase_productid = 'X'
     ) ph
     on ph.people_userid = p.people_userid
group by p.people_userid

MySQL选择所有匹配X的人，然后选择每个人家中的所有人

更新

2 个答案: