Question

我有以下查询（在此论坛上找到here的示例），该查询生成序列号以标记位置中的每个更改。

WITH t(ID, col1  ,Location) AS (
select 1, 1 , 1 union all  
select 1, 2 , 1 union all  
select 1, 3 , 2 union all  
select 1, 4 , 2 union all  
select 1, 5 , 1 union all  
select 1, 6 , 2 union all  
select 1, 7 , 2 union all  
select 1, 8 , 3 union all 
select 2, 1 , 1 union all  
select 2, 2 , 2 union all  
select 2, 3 , 2 union all  
select 2, 4 , 2 union all  
select 2, 5 , 1 union all  
select 2, 6 , 1 union all  
select 2, 7 , 2 union all  
select 2, 8 , 3
)
SELECT t.ID, t.col1, t.Location,
    sum(x) OVER (partition by ID order by col1) sequence
FROM (
    SELECT t.*,
        CASE WHEN Location = lag(Location) OVER (order by ID, col1) THEN 0
            ELSE 1
        END x
    FROM t
    ) t
ORDER BY ID, col1
;

现在我想只保留那些指示每个ID通过不同位置的顺序路径的行。如何相应地过滤数据，以便生成以下结果：

ID  Location
1   1
1   2
1   1
1   2
1   3
2   1
2   2
2   1
2   2
2   3

有没有办法实现他的？

Answer 1

您似乎想删除相邻的重复项：

SELECT t.ID, t.col1, t.Location
FROM (SELECT t.*,
             (CASE WHEN Location = lag(Location) OVER (order by ID, col1) THEN 0
                   ELSE 1
              END) x
      FROM t
     ) t
WHERE x = 1
ORDER BY ID, col1;

我使用了你的查询结构。我实际上会把它写成：

SELECT t.ID, t.col1, t.Location
FROM (SELECT t.*,
            lag(Location) OVER (order by ID, col1) as prev_location
      FROM t
     ) t
WHERE prev_location is NULL or prev_location <> location
ORDER BY ID, col1;

过滤顺序数据

1 个答案: