选择存在的数据[关闭]

Question

我有

LEFT JOIN

ed 的三张桌子。在这种情况下，我需要使用

COALESCE(ID, ID1)

用 ID1 替换丢失的 ID。

目标是仅获取 ID 及其所有相关数据条目，给定 OPS 列中的值“w”。

我尝试使用

EXISTS

子句，但它只返回列 OPS 的值为 'w' 的行，而不返回与同一 ID 相关的其他行：

SELECT t1.id, t1.age, t2.operation, t3.ops
FROM table1 AS t1
LEFT JOIN table2 AS t2 ON COALESCE(t2.id, t2.id1) = t1.id 
LEFT JOIN table3 AS t3 ON COALESCE(t3.id, t3.id1) = t1.id
WHERE EXISTS (SELECT 1 
              FROM table3 
              WHERE id = t1.id
                AND (t3.ops = 'w'))

现在我试着弄清楚为什么我的

EXISTS

子句不能正常工作。

资料：
db<>小提琴

这是我想用代码产生的结果：

id	年龄	操作	操作
1	23	A	q
1	23	A	w
1	23	A	e
1	23	乙	q
1	23	乙	w
1	23	乙	e
1	23	C	q
1	23	C	w
1	23	C	e
2	25	A	q
2	25	A	w
2	25	乙	q
2	25	乙	w
4	43	A	q
4	43	A	w
4	43	乙	q
4	43	乙	w

Answer 1

如果我理解正确的话，我会推荐窗口函数而不是

exist

来检查每个组是否有操作

w

：

SELECT *
FROM (
    SELECT t1.id AS t1id, 
        t1.age AS t1age, 
        t2.id AS t2id, 
        t2.id1 AS t2id1, 
        t2.operation AS t2operation, 
        t3.id AS t3id, 
        t3.id1 AS t3id1, 
        t3.ops AS t3ops,
        MAX(CASE WHEN t3.ops = 'w' then 1 else 0 end) OVER(PARTITION BY t1.id) has_ops_w
    FROM table1 AS t1
    LEFT JOIN table2 AS t2 ON COALESCE(t2.id, t2.id1) = t1.id 
    LEFT JOIN table3 AS t3 ON COALESCE(t3.id, t3.id1) = t1.id
) t
WHERE has_ops_w = 1

在你的 DB Fiddle:

t1id	t1age	t2id	t2id1	t2操作	t3id	t3id1	t3ops	has_ops_w
1	23	1	1	A	1	1	q	1
1	23	1	1	A	空	1	w	1
1	23	1	1	A	空	1	e	1
1	23	空	1	乙	1	1	q	1
1	23	空	1	乙	空	1	w	1
1	23	空	1	乙	空	1	e	1
1	23	空	1	C	1	1	q	1
1	23	空	1	C	空	1	w	1
1	23	空	1	C	空	1	e	1
2	25	2	2	A	2	2	q	1
2	25	2	2	A	空	2	w	1
2	25	2	2	乙	2	2	q	1
2	25	2	2	乙	空	2	w	1
4	43	4	4	A	4	4	q	1
4	43	4	4	A	空	4	w	1
4	43	空	4	乙	4	4	q	1
4	43	空	4	乙	空	4	w	1

Answer 2

GMB 指定的窗口函数可以工作，但是，我认为使用 EXISTS 子句有些混乱。

SELECT t1.id, t1.age, t2.operation, t3.ops
FROM table1 AS t1
LEFT JOIN table2 AS t2
ON COALESCE(t2.id, t2.id1) = t1.id 
LEFT JOIN table3 AS t3 ON COALESCE(t3.id, t3.id1) = t1.id
WHERE EXISTS (
  SELECT 1 
  FROM table3 inner_t3
  WHERE COALESCE(inner_t3.id, inner_t3.id1) = t1.id -- your join above wasn't 
                                                    -- on ID alone. This one 
                                                    -- shouldn't be either
  AND inner_t3.ops = 'w'  -- Make sure you have proper reference to inner 
                          -- table using alias.
)

Answer 3

行得通。那么是不是因为从 ID1 替换 ID 的 CLAUSE 需要在 exists 子句中再次重复？

SELECT t1.id, t1.age, t2.operation, t3.ops
FROM table1 AS t1
LEFT JOIN table2 AS t2 ON t2.id = t1.id OR (t2.id IS NULL AND t2.id1 = t1.id)
LEFT JOIN table3 AS t3 ON t3.id = t1.id OR (t3.id IS NULL AND t3.id1 = t1.id)
WHERE EXISTS (
  SELECT 1 
  FROM table2 AS t2 
  LEFT JOIN table3 AS t3 ON t3.id = t1.id OR (t3.id IS NULL AND t3.id1 = t1.id)
  WHERE t1.id = t1.id
  AND (t3.ops = 'w')
);