如何通过SQL进行连接和一匹配一

Question

我有两张桌子，如下所示。 当 table_a.name=table_b.name 且 table_a.value=table_b.value 并且 table_b.id 未设置为其他 table_a.fid 时，我想将 table_a.fid 更新为 table_b.id

表_a（a）：

id(int) name(varchar) value(varchar) fid(int)
1       name_a        value_1        0
2       name_a        value_1        0
3       name_a        value_1        0
4       name_b        value_2        0
5       name_c        value_3        0

表_b (b):

id(int) name(varchar) value(varchar)
1       name_a        value_1
2       name_a        value_1
3       name_b        value_2
4       name_b        value_2

table_a预期结果：

id name    value   fid
1  name_a  value_1  1 // b.id(1,2) matched, so set the first b.id(1) to a.fid(1)
2  name_a  value_1  2 // b.id(1,2) matched and b.id(1) already set to a.fid(1), so set the second b.id(2) to a.fid(2)
3  name_a  value_1  0 // b.id(1,2) matched and b.id(1,2) already set to a.fid(1,2), so is 0
4  name_b  value_2  3 // b.id(3,4) matched. so set the first b.id(3) to a.fid
5  name_c  value_3  0 // b not matched

Answer 1

如果您使用 mysql 8.0 或更高版本，那么您可以使用窗口函数

ROW_NUMBER()

为相关记录集中的行分配订单号，然后也使用此唯一标识符进行连接:

with cte_a as (
  select *, row_number() over (partition by name, val order by id) as rn
  from table_a
),
cte_b as (
  select *, row_number() over (partition by name, val order by id) as rn
  from table_b
)
select cte_a.id, cte_a.name, cte_a.val, coalesce(cte_b.id, 0) as fid
from cte_a
left join cte_b using(name, val, rn);

结果：

id  name    val      fid
1   name_a  value_1  1
2   name_a  value_1  2
3   name_a  value_1  0
4   name_b  value_2  3
5   name_c  value_3  0

演示在这里