Question

我有这个查询，其中MINUS运算符的两个操作数被计算两次。如何重写查询以便只创建一次？

（目的是检查两组是否相等）

SELECT DISTINCT t1.id
FROM t1
WHERE NOT EXISTS (SELECT t2.did FROM t1 AS t2
              WHERE t2.id = t1.id
              MINUS
              SELECT t3.did FROM t3
              WHERE t3.price > 500000)
AND NOT EXISTS (SELECT t3.did FROM t3
            WHERE t3.price > 500000
            MINUS
            SELECT t2.did FROM t1 AS t2
            WHERE t2.id = t1.id);

为了澄清事情，以下两个[sub]查询在查询中重复两次：

SELECT t2.did FROM t1 AS t2
WHERE t2.id = t1.id

和

SELECT t3.did FROM t3
WHERE t3.price > 500000

Answer 1

with
   ALL_DID as (
      select did from t3
      where price > 500000
   ),
   PAIRS as (
      select
         id, t3.did
      from t1
         left join ALL_DID t3
         on t1.did = t3.did
   )
select id from PAIRS
   group by id
   having count(did) = (
      select count(0) from ALL_DID
   )
minus
select id from PAIRS
   where did is null

fiddle

Answer 2

如果您只想重复使用同一个查询，请查看使用公用表表达式。或者您可以查看创建视图。这些方法可以快速“重用”相同的查询。

但是，我仍然不完全明白你想要的结果是什么。您是否想知道t1中的数据是否与t3中的数据相同，基于did / eid字段？如果是这种情况，似乎你可以使用LEFT JOIN来更轻松地解决这个问题。像这样：

SELECT DISTINCT t1.id
FROM t1 
  LEFT JOIN t3 ON t1.did = t3.did AND t3.price > 500000
WHERE t3.id IS NULL

这将告诉您t1中是否有任何不在t3中的记录（具有相同的do和价格> 500000）。

编辑 - 要查找销售价格超过500000的所有产品的所有人的ID，这应该有效：

SELECT t1.id, COUNT(DISTINCT t1.did) productcnt
FROM t1
  JOIN t3 ON t1.did = t3.did
WHERE t3.Price > 500000
GROUP BY t1.id
HAVING COUNT(DISTINCT t1.did) = (
  SELECT COUNT(DISTINCT did) cnt
  FROM t3
  WHERE Price > 500000 )

SQL Fiddle Demo

Answer 3

这不是正常的内连接查询吗？

SELECT t2.* 
from t1
INNER JOIN t3 as t3 on t1.did = t3.tid
INNER JOIN t1 as t2 on t2.did = t1.id
WHERE t3.price > 50000

在嵌套查询中创建和使用临时表

3 个答案: