我有3张桌子:
users(id, account_balance)
grocery(user_id, date, amount_paid)
fishmarket(user_id, date, amount_paid)
fishmarket
和 grocery
表可能会多次出现同一 user_id,且具有不同的日期和支付金额,或者对于任何给定用户根本没有任何内容。
当我尝试以下查询时:
SELECT
t1."id" AS "User ID",
t1.account_balance AS "Account Balance",
count(t2.user_id) AS "# of grocery visits",
count(t3.user_id) AS "# of fishmarket visits"
FROM users t1
LEFT OUTER JOIN grocery t2 ON (t2.user_id=t1."id")
LEFT OUTER JOIN fishmarket t3 ON (t3.user_id=t1."id")
GROUP BY t1.account_balance,t1.id
ORDER BY t1.id
它会产生不正确的结果:
"1", "12", "12"
。LEFT JOIN
时,它会为 grocery
或 fishmarket
访问产生正确的结果,即 "1", "3", "4"
。
我在这里做错了什么?
我正在使用 PostgreSQL 9.1。
连接从左到右处理(除非括号另有规定)。如果您向一名用户
LEFT JOIN
(或只是 JOIN
,类似的效果)购买 3 种杂货,您将获得 3 行 (1 x 3)。如果您随后为同一用户加入 4 个鱼市,您将获得 12 (3 x 4) 行,乘以结果中的先前计数,而不是像您希望的那样添加。你可以让它像这样工作:
SELECT u.id
, u.account_balance
, g.grocery_visits
, f.fishmarket_visits
FROM users u
LEFT JOIN (
SELECT user_id, count(*) AS grocery_visits
FROM grocery
GROUP BY user_id
) g ON g.user_id = u.id
LEFT JOIN (
SELECT user_id, count(*) AS fishmarket_visits
FROM fishmarket
GROUP BY user_id
) f ON f.user_id = u.id
ORDER BY u.id;
要获取一个或几个用户的聚合值,相关子查询像@Vince提供的就可以了。对于整个表或其中的主要部分,聚合 n 个表并连接到结果once(更)有效。这样,我们在外部查询中也不需要另一个
GROUP BY
。
grocery_visits
和 fishmarket_visits
是 NULL
,适用于相应表中没有任何相关条目的用户。如果您需要 0
(或任何任意数字),请在外部 COALESCE
中使用 SELECT
:
SELECT u.id
, u.account_balance
, COALESCE(g.grocery_visits , 0) AS grocery_visits
, COALESCE(f.fishmarket_visits, 0) AS fishmarket_visits
FROM ...
对于您的原始查询,如果您取消分组依据以查看预先分组的结果,您将了解为什么创建您收到的计数。
也许以下使用子查询的查询可以达到您的预期结果:
SELECT
t1."id" AS "User ID",
t1.account_balance AS "Account Balance",
(SELECT count(*) FROM grocery t2 ON (t2.user_id=t1."id")) AS "# of grocery visits",
(SELECT count(*) FROM fishmarket t3 ON (t3.user_id=t1."id")) AS "# of fishmarket visits"
FROM users t1
ORDER BY t1.id
因为用户表连接到杂货表时,匹配到了3条记录。然后这 3 条记录中的每一条都与 Fishmarket 中的 4 条记录相匹配,产生 12 条记录。您需要子查询来获取您要查找的内容。
SELECT t1."id" AS "User ID", t1.account_balance AS "Account Balance", Sum(Case When t2.user_id is null then 0 else 1 end) AS "# of grocery visits", Sum(Case When t3.user_id is null then 0 else 1 end) AS "# of fishmarket visits" FROM users t1 LEFT OUTER JOIN grocery t2 ON (t2.user_id=t1."id") LEFT OUTER JOIN fishmarket t3 ON (t3.user_id=t1."id") GROUP BY t1.account_balance,t1.id ORDER BY t1.id
上述内容还允许您根据需要添加其他标准