Question

我们有一个存储每月员工价值的数据库（例如兼职百分比）：

+-----+------+-------+----------+
| emp | year | month | parttime |
+-----+------+-------+----------+
|   1 | 2015 |     1 |      100 |
|   1 | 2015 |     2 |      100 |
|   1 | 2015 |     3 |      100 |
|   1 | 2015 |     4 |      100 |
|   2 | 2015 |     1 |       80 |
|   2 | 2015 |     2 |      100 |
|   2 | 2015 |     3 |      100 |
|   2 | 2015 |     4 |       80 |
|   3 | 2015 |     1 |       60 |
|   3 | 2015 |     2 |       60 |
|   3 | 2015 |     3 |       80 |
|   3 | 2015 |     4 |      100 |
+-----+------+-------+----------+

出于报告目的，我需要以from / until形式显示值：

+-----+---------+---------+----------+
| emp |  from   |   to    | parttime |
+-----+---------+---------+----------+
|   1 | 2015.01 | 2015.04 |      100 |
|   2 | 2015.01 | 2015.01 |       80 |
|   2 | 2015.02 | 2015.03 |      100 |
|   2 | 2015.04 | 2015.04 |       80 |
|   3 | 2015.01 | 2015.02 |       60 |
|   3 | 2015.03 | 2015.03 |       80 |
|   3 | 2015.04 | 2015.04 |      100 |
+-----+---------+---------+----------+

我的第一次尝试是用简单的最小/最大方法解决它。但是员工nr。 2有点棘手，循环值为80。

任何想法/例子？数据库基于db / 2或microsoft。

感谢

菲利普

Answer 1

这被称为差距和群岛问题。一个快速的解决方案：

DECLARE @Employee TABLE
(emp int, year int, month int, parttime int)

INSERT INTO @Employee
VALUES
(1, 2015, 1, 100),
(1, 2015, 2, 100),
(1, 2015, 3, 100),
(1, 2015, 4, 100),
(2, 2015, 1,  80),
(2, 2015, 2, 100),
(2, 2015, 3, 100),
(2, 2015, 4,  80),
(3, 2015, 1,  60),
(3, 2015, 2,  60),
(3, 2015, 3,  80),
(3, 2015, 4, 100)


;WITH cte
AS 
(
    SELECT *
        ,e.[month] - ROW_NUMBER() OVER (ORDER BY e.emp, e.[parttime]) AS Grp
    FROM @Employee e
)
SELECT 
    emp, 
    CAST([year] AS varchar(50)) + '.' + CAST(MIN([month])AS varchar(50)) AS [from],
    CAST([year] AS varchar(50)) + '.' + CAST(MAX([month])AS varchar(50)) AS [to],
    parttime 
FROM cte
GROUP BY emp, parttime, year, Grp
ORDER BY emp, [from]

Answer 2

如果您的数据库存储完整日期而不仅仅是年/月（或至少是等效的组合类型），这将更容易。或者，如果您可以操作原始基础数据：

SELECT emp, partTime, MIN(monthStart) AS monthStart, MAX(monthNext) AS monthEnd
FROM (SELECT emp, partTime,
             DATEADD(month, month - 1, DATEADD(year, year - 1, CAST('00010101' AS DATE))) AS monthStart,
             DATEADD(month, month, DATEADD(year, year - 1, CAST('00010101' AS DATE))) AS monthNext,
             ROW_NUMBER() OVER(PARTITION BY emp ORDER BY year, month)  -
             ROW_NUMBER() OVER(PARTITION BY emp, partTime ORDER BY year, month) AS groupId
      FROM Monthly_Hours) AS Grouping
GROUP BY emp, partTime, groupId
ORDER BY emp, monthStart

SQL Fiddle Example

请注意，我特别使用了该范围的独占上限。日期/时间/时间戳类型，如所有正，连续范围类型（除了显式整数之外的任何类型）应始终以这种方式处理（它使得推理和查询更加容易）。

这个答案略有不足，因为缺少的月份没有直接报告（不显示为0） - 如果有必要，有办法纠正这个问题，尽管需要更多的工作

Answer 3

我已根据您的示例数据在Postgres上测试了此解决方案，但我几乎可以肯定这将适用于DB2。它可能需要一些细微的改变，但不确定。

要逐步了解它是如何工作的，您可以从执行最内部的块开始。

SELECT 
    emp, 
    (year||'.'||CASE WHEN length(min_month::text) = 1 THEN '0'||min_month::text ELSE min_month::text END) AS from, 
    (year||'.'||CASE WHEN length(max_month::text) = 1 THEN '0'||max_month::text ELSE max_month::text END) AS to, 
    parttime 
FROM(
    SELECT 
        emp,
        year,
        parttime,
        first_different,
        min(month) AS min_month,
        max(month) AS max_month 
    FROM( 
        SELECT 
            a.*,
            b.* 
        FROM(
            SELECT *
            FROM tablename 
            ) a,
            LATERAL 
            (
            SELECT 
                min(CASE WHEN a.parttime IS DISTINCT FROM b.parttime THEN b.month END) AS first_different
            FROM 
                tablename b 
            WHERE 
                a.emp = b.emp 
                AND a.year = b.year 
                AND a.month < b.month 
            ) b 
        ) foo 
    GROUP BY 1,2,3,4
    ORDER BY 1 
    ) goo 
ORDER BY 1,2;

结果：

 emp |  from   |   to    | parttime 
-----+---------+---------+----------
   1 | 2015.01 | 2015.04 |      100
   2 | 2015.01 | 2015.01 |       80
   2 | 2015.02 | 2015.03 |      100
   2 | 2015.04 | 2015.04 |       80
   3 | 2015.01 | 2015.02 |       60
   3 | 2015.03 | 2015.03 |       80
   3 | 2015.04 | 2015.04 |      100

Answer 4

第一步：检测用户或兼职变化发生的位置（1 =变化，0 =与最后一行相同的值）。您可以使用分析函数LAG执行此操作。

第二步：使用分析函数SUM基于更改标记构建组。

第三步：在每组中显示一条记录，并在组中找到最小和最大年/月。

+-----+------+-------+----------+-------+-------+
| emp | year | month | parttime | step1 | step2 |
|     |      |       |          |  chg  |  grp  |
+-----+------+-------+----------+-------+-------+
|   1 | 2015 |     1 |      100 |     1 |     1 |
|   1 | 2015 |     2 |      100 |     0 |     1 |
|   1 | 2015 |     3 |      100 |     0 |     1 |
|   1 | 2015 |     4 |      100 |     0 |     1 |
|   2 | 2015 |     1 |       80 |     1 |     2 |
|   2 | 2015 |     2 |      100 |     1 |     3 |
|   2 | 2015 |     3 |      100 |     0 |     3 |
|   2 | 2015 |     4 |       80 |     1 |     4 |
|   3 | 2015 |     1 |       60 |     1 |     5 |
|   3 | 2015 |     2 |       60 |     0 |     5 |
|   3 | 2015 |     3 |       80 |     1 |     6 |
|   3 | 2015 |     4 |      100 |     1 |     7 |
+-----+------+-------+----------+-------+-------+

select
  emp,
  min(format(year, '0000') + '.' + format(month, '00')) as from_month,
  max(format(year, '0000') + '.' + format(month, '00')) as to_month,
  parttime
from
(
  select
    emp, year, month, parttime,
    sum(chg) over (order by emp, year, month) as grp
  from
  (
    select 
      emp, year, month, parttime, 
      case when lag(emp) over (order by emp, year, month) = emp 
           and lag(parttime) lag(emp) over (order by emp, year, month) = parttime 
        then 0
        else 1
      end as chg
    from mytable
  ) changes
) groups
group by grp, emp, parttime
order by grp;

sql转换自＆＃34;每月记录＆＃34;到＆＃34;记录/直到＆＃34;

4 个答案: