Question

我的MariaDb中有一个表，该表如下表所示，需要创建“组”列（请参见输出）。如果KEY_A和KEY_B的组合按LOAD_DATE的顺序更改，则组会更改（请参见输出）。我找不到合适的SQL来生成“组”列。

也许有人有一个想法如何在MariaDB中解决这个问题？

TestData：

CREATE OR REPLACE TABLE test AS 
SELECT 'AAA' AS KEY_A,'A' AS KEY_B,DATE('2010-01-01') AS LOAD_DATE
UNION
SELECT 'AAA' AS KEY_A,'B' AS KEY_B,DATE('2010-01-02') 
UNION
SELECT 'AAA' AS KEY_A,'A' AS KEY_B,DATE('2010-01-03') 
UNION
SELECT 'AAA' AS KEY_A,'A' AS KEY_B,DATE('2010-01-04') 
UNION
SELECT 'BBB' AS KEY_A,'C' AS KEY_B,DATE('2010-01-04') 
UNION
SELECT 'CCC' AS KEY_A,'D' AS KEY_B,DATE('2010-01-04') ;

输出：

KEY_A; KEY_B; LOAD_DATE;  group

AAA;   A;     2010-01-01;  1
AAA;   B;     2010-01-02;  2
AAA;   A;     2010-01-03;  3
AAA;   A;     2010-01-04;  3
BBB;   C;     2010-01-04;  1
CCC;   D;     2010-01-04;  1

Answer 1

您的测试数据实际上写错了。

测试数据：

CREATE TABLE TestDB.test AS 
SELECT 'AAA' AS KEY_A,'A' AS KEY_B, DATE('2010-01-01') AS LOAD_DATE FROM DUAL
UNION
SELECT 'AAA' AS KEY_A,'B' AS KEY_B, DATE('2010-01-02') AS LOAD_DATE FROM DUAL
UNION
SELECT 'AAA' AS KEY_A,'A' AS KEY_B, DATE('2010-01-03') AS LOAD_DATE FROM DUAL
UNION
SELECT 'AAA' AS KEY_A,'A' AS KEY_B, DATE('2010-01-04') AS LOAD_DATE FROM DUAL
UNION
SELECT 'BBB' AS KEY_A,'C' AS KEY_B, DATE('2010-01-04') AS LOAD_DATE FROM DUAL
UNION
SELECT 'CCC' AS KEY_A,'D' AS KEY_B, DATE('2010-01-04') AS LOAD_DATE FROM DUAL
;

Answer 2

我认为这是一个孤岛问题。

您可以使用窗口函数解决此问题：

select t.*,
       sum( case when key_b = prev_key_b then 0 else 1 end ) over (partition by key_a order by load_date) as grp
from (select t.*,
             lag(key_b) over (partition by key_a order by load_data) as prev_key_b
      from t
     ) t;

如何在MariaDB中创建更改组标识符

2 个答案: