DML 中的自增逻辑

Question

我有两个表 test2 和 test_hist。我想从 test2 将数据加载到 test_hist，但由于唯一约束而失败。

CREATE TABLE TEST2 (ID NUMBER , TEXT VARCHAR2(10));
create table test_hist (id number , text varchar2(10) , constraint t_pk PRIMARY key (id , text));

INSERT INTO TEST2 VALUES(100 , '20180909-I');
INSERT INTO TEST2 VALUES(101 , '20180909-I');
INSERT INTO TEST2 VALUES(102 , '20180809-I');
INSERT INTO TEST2 VALUES(100 , '20180909-I');
COMMIT;

INSERT INTO test_hist SELECT ID , TEXT FROM TEST2;

我想在 TEXT 字段中附加自动递增编号，每当出现如下重复项时。

expected OUTPUT 

ID         TEXT
100      20180909-I
101      20180909-I
102      20180809-I
100      20180909-I-1
100      20180909-I-2
102      20180809-I-1

任何人都可以帮助我实现这一目标。提前致谢

如果我多次执行插入语句，它应该插入到带有自动增量文本的 test_hist 中。例如

insert into test_hist
select id,
  text || case when row_number() over (partition by id, text order by null) > 1
          then (1 - row_number() over (partition by id, text order by null)) end
from test2;

9 rows inserted.

select *
from test_hist
order by id, text;

        ID TEXT        
---------- ------------
       100 20180909-I  
       100 20180909-I-1
       100 20180909-I-2
       100 20180909-I-3
       101 20180909-I  
       101 20180909-I-1
       102 20180809-I  
       102 20180809-I-1
       102 20180809-I-2

Answer 1

您可以尝试以下方法：

select ID, text||decode(rn-1,0,null,'-'||(rn-1)) as text
from
(
with test2(rnk,ID,text) as
(
    select 1, 100 , '20180909-I' from dual union all
    select 2, 101 , '20180909-I' from dual union all
    select 3, 102 , '20180809-I' from dual union all
    select 4, 100 , '20180909-I' from dual union all
    select 5, 100 , '20180909-I' from dual union all
    select 6, 102 , '20180909-I' from dual
    )
    select t.ID, t.rnk, 
           t.text, row_number() over (partition by ID order by Text,ID) as rn
      from test2 t
)
 order by rn, rnk

Rextester Demo

Answer 2

与@Barbaros 的基本思想相同，但排列略有不同，第二个 table 的列大小增加，因此它可以容纳修改后的值：

create table test2 (
  id number, text varchar2(10)
);

create table test_hist (id number,
  text varchar2(12), -- increased size to 12; may need to be larger
  constraint t_pk primary key (id , text)
);

insert into test2 values(100 , '20180909-I');
insert into test2 values(101 , '20180909-I');
insert into test2 values(102 , '20180809-I');
insert into test2 values(100 , '20180909-I');
insert into test2 values(100 , '20180909-I');
insert into test2 values(102 , '20180809-I');

然后是同一个解析函数，如果你不介意重复它，在一个层次上，并在partition-by子句中包括所有的PK列：

insert into test_hist
select id,
  text || case when row_number() over (partition by id, text order by null) > 1
          then (1 - row_number() over (partition by id, text order by null)) end
from test2;

6 rows inserted.

select *
from test_hist
order by id, text;

        ID TEXT        
---------- ------------
       100 20180909-I  
       100 20180909-I-1
       100 20180909-I-2
       101 20180909-I  
       102 20180809-I  
       102 20180809-I-1

如果您在真实场景中实际上有更多列，并且原始 table 中还有另一个非 PK 列，您想要影响历史记录行的递增顺序，您可以在函数的 order by 而不是 null，这实际上只是一个虚拟占位符。

如果它需要在多次插入时继续做同样的事情（这表明数据模型问题甚至比原始要求更严重）那么您也需要计算历史记录中的现有匹配项 table .

从与之前相同的原始数据和空历史记录开始table:

insert into test_hist
select id,
  text || case when appearance > 1 then (1 - appearance) end
from (
  select t.id,
    t.text,
    row_number() over (partition by t.id, t.text order by null) + (
      select count(*) from test_hist th
      where th.id = t.id
      and th.text like t.text || '%'
    ) as appearance
  from test2 t
);

6 rows inserted.

select *
from test_hist
order by id, text;

        ID TEXT        
---------- ------------
       100 20180909-I  
       100 20180909-I-1
       100 20180909-I-2
       101 20180909-I  
       102 20180809-I  
       102 20180809-I-1

6 rows selected.

和运行第二次相同的语句：

insert into test_hist
select id,
  text || case when appearance > 1 then (1 - appearance) end
from (
  select t.id,
    t.text,
    row_number() over (partition by t.id, t.text order by null) + (
      select count(*) from test_hist th
      where th.id = t.id
      and th.text like t.text || '%'
    ) as appearance
  from test2 t
);

6 rows inserted.

select *
from test_hist
order by id, text;

        ID TEXT        
---------- ------------
       100 20180909-I  
       100 20180909-I-1
       100 20180909-I-2
       100 20180909-I-3
       100 20180909-I-4
       100 20180909-I-5
       101 20180909-I  
       101 20180909-I-1
       102 20180809-I  
       102 20180809-I-1
       102 20180809-I-2
       102 20180809-I-3

12 rows selected.

可能有一些方法可以对其进行优化，这样您就不必经常访问历史记录 table，但这可能会给您一个工作起点。

like 的使用实际上意味着这仅在文本始终具有相同长度的情况下才有效，或者至少您不能拥有其他值的扩展值；否则你会得到比你想要的更多的比赛。至少从您的示例数据来看，这看起来不是问题。如果需要，您可以通过切换到 regexp_like 来解决这个问题，具体取决于您的实际数据。

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