sql—合并的最终表等价于什么?

neekobn8  于 2021-07-26  发布在  Java
关注(0)|答案(1)|浏览(278)

我正在尝试使用“合并到”仅插入新记录。我想收集插入的新记录的ID和被忽略的重复记录的ID。
下面是表的create语句:

drop table SSZ_ME_MIS.test_update_table;
create table ssz_me_mis.test_update_table (
    ID_col int not null generated always as identity, -- Primary Key
    val_col_1 int not null,
    val_col_2 varchar(255) not null,
    constraint pk_test_update_table primary key (ID_col),
    constraint uq_test_update_table unique (val_col_1, val_col_2)
);

然后,填充一些初始值:

insert into ssz_me_mis.test_update_table (val_col_1, val_col_2)
select *
from (values 
    (231, 'Value 1'),
    (481, 'Value 2'),
    (813, 'Value 3')
);

所以,最后,我想试着做这样的插入:

select ID_col from final table (
    merge into ssz_me_mis.test_update_table t using (
        select *
        from (values 
            (231, 'Value 1'),
            (481, 'Value 2'),
            (513, 'Value 4')
        )
    ) as s (val_col_1, val_col_2)
    on
        t.val_col_1 = s.val_col_1
        and t.val_col_2 = s.val_col_2
    when not matched then 
        insert (val_col_1, val_col_2)
        values (s.val_col_1, s.val_col_2)
    else
        ignore
);

有什么方法可以做到这一点吗?

wlsrxk51

wlsrxk511#

类似的内容将在db2luw上运行(假设您正在使用 ORGANIZE BY ROW 表格)。

with s (val_col_1, val_col_2) AS  (values 
            (231, 'Value 1'),
            (481, 'Value 2'),
            (513, 'Value 4')
        )
, i as (select * from final table(
    INSERT INTO ssz_me_mis.test_update_table ( val_col_1 , val_col_2) 
     select * from s where not exists (select 1 from ssz_me_mis.test_update_table t
        where 
        t.val_col_1 = s.val_col_1
        and t.val_col_2 = s.val_col_2
        )
))
, u as (select count(*) as dummy from new table(
    update ssz_me_mis.test_update_table t
    set val_col_1 = (select val_col_1 from s where t.val_col_1 = s.val_col_1 and t.val_col_2 = s.val_col_2)
    ,   val_col_2 = (select val_col_2 from s where t.val_col_1 = s.val_col_1 and t.val_col_2 = s.val_col_2)
    where exists    (select val_col_2 from s where t.val_col_1 = s.val_col_1 and t.val_col_2 = s.val_col_2)
))
select ID_col from i, u

我包含了一个用于更新的分支,但从逻辑上讲,您需要一些非键列才能理解。你的例子只是实践中的一个插页,所以我有点搞不懂你为什么要用 MERGE 完全。

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