Oracle SQL 中的高效四路连接

Question

我有两个要加入的表，Table1 和 Table2。每个表都有两个唯一的键，我们称它们为 Key1 和 Key2。我想要做的是 LEFT JOIN Table2 to Table1，其中任何键都匹配四种可能组合中的任何一种：

• Table1.Key1 = Table2.Key1
• Table1.Key1 = Table2.Key2
• Table1.Key2 = Table2.Key1
• Table1.Key2 = Table2.Key2

我的问题是：有没有有效的方法来做到这一点？现在我想出了类似的东西，但它需要很长时间才能运行。

CREATE TABLE NEW_TABLE AS
SELECT a.*,
       CASE WHEN a.Key1 = b.Key1 THEN 1 ELSE 0 END AS match1,
       CASE WHEN a.Key1 = c.Key2 THEN 1 ELSE 0 END AS match2,
       CASE WHEN a.Key2 = b.Key1 THEN 1 ELSE 0 END AS match3,
       CASE WHEN a.Key2 = c.Key2 THEN 1 ELSE 0 END AS match4
FROM Table1 a
LEFT JOIN (Select Key1 From Table2 Where Key1 is not null) b
    on a.Key1 = b.Key1 or a.Key2 = b.Key1
LEFT JOIN (Select Key2 From Table2 Where Key2 is not null) c
    on a.Key1 = c.Key2 or a.Key2 = c.Key2
;

没救了，我知道...

编辑：示例数据和所需结果如下：

表 1：

Key1     Key2     Sales    Revenue
qwer!@   dhfgfw   455      30005
asdf#$   dfg654   221      28711
edfr2#   gg%%^f   213      31667
gthy!2   awd^&5   133      13345
rf$#22   34ffgg   655      41237
bhjk%g   w3erff   122      10066
f&*yhj   dffghj   126      11004

表 2：

Key1     Key2 
qwer!@   {null}
{null}   dfg654
ffgww2   ppolkk
{null}   gthy!2
jjjj33   l00kjl
nmnmnm   34ffgg
awd^&5   {null}

期望的结果：

Key1     Key2     Sales    Revenue   match1    match2    match3    match4
qwer!@   dhfgfw   455      30005     1         0         0         0
asdf#$   dfg654   221      28711     0         0         0         1
edfr2#   gg%%^f   213      31667     0         0         0         0
gthy!2   awd^&5   133      13345     1         0         1         0
rf$#22   34ffgg   655      41237     0         0         0         1
bhjk%g   w3erff   122      10066     0         0         0         0
f&*yhj   dffghj   126      11004     0         0         0         0

Answer 1

连接条件中的or 可能是性能问题。我建议你改用exists：

SELECT a.*,
       (case when exists (select 1 from table2 b where a.Key1 = b.Key1) then 1 else 0 end) as match1,
       (case when exists (select 1 from table2 b where a.Key1 = b.Key2) then 1 else 0 end) as match2,
       (case when exists (select 1 from table2 b where a.Key2 = b.Key1) then 1 else 0 end) as match3,
       (case when exists (select 1 from table2 b where a.Key2 = b.Key2) then 1 else 0 end) as match4
FROM Table1 a;

过滤掉NULL 并不重要，因为NULL 无论如何都会使比较失败。

为了获得最佳性能，您需要在 table2(key1) 和 table2(key2) 上建立索引。

Answer 2

如果您只需要行的组合而不需要排列（即，如果table1.key 与table2.key1 和table2.key2 具有相同的值，则只返回一行），那么这应该有效：

SELECT a.*,
       CASE WHEN a.Key1 = b.Key1 THEN 1 ELSE 0 END AS match1,
       CASE WHEN a.Key1 = b.Key2 THEN 1 ELSE 0 END AS match2,
       CASE WHEN a.Key2 = b.Key1 THEN 1 ELSE 0 END AS match3,
       CASE WHEN a.Key2 = b.Key2 THEN 1 ELSE 0 END AS match4
FROM Table1 a
LEFT JOIN Table2 b
    on a.Key1 in (b.Key1, b.key2) or a.key2 in (b.key1, b.key2);

---

插入提供的数据，此解决方案确实有效，但需要对其进行汇总以提供您正在寻找的结果：

SELECT   a.key1,
         a.key2,
         a.sales,
         a.revenue,
         MAX (CASE WHEN a.key1 = b.key1 THEN 1 ELSE 0 END) AS match1,
         MAX (CASE WHEN a.key1 = b.key2 THEN 1 ELSE 0 END) AS match2,
         MAX (CASE WHEN a.key2 = b.key1 THEN 1 ELSE 0 END) AS match3,
         MAX (CASE WHEN a.key2 = b.key2 THEN 1 ELSE 0 END) AS match4
FROM     table1 a
         LEFT JOIN table2 b
            ON a.key1 IN (b.key1, b.key2) OR a.key2 IN (b.key1, b.key2)
GROUP BY a.key1,
         a.key2,
         a.sales,
         a.revenue;