46

我有一个看起来像这样的报告:

CompanyA      Workflow27     June5
CompanyA      Workflow27     June8
CompanyA      Workflow27     June12
CompanyB      Workflow13     Apr4
CompanyB      Workflow13     Apr9
CompanyB      Workflow20     Dec11
CompanyB      Wofkflow20     Dec17

这是使用 SQL(特别是 T-SQL 版本 Server 2005)完成的:

SELECT company
   , workflow
   , date
FROM workflowTable

我希望报告仅显示每个工作流程的最早日期:

CompanyA      Workflow27     June5
CompanyB      Workflow13     Apr4
CompanyB      Workflow20     Dec11

有任何想法吗?我想不通。我尝试使用返回最早托盘日期的嵌套选择,然后在 WHERE 子句中设置它。如果只有一家公司,这很有效:

SELECT company
   , workflow
   , date
FROM workflowTable
WHERE date = (SELECT TOP 1 date
              FROM workflowTable
              ORDER BY date)

但如果该表中有多个公司,这显然行不通。任何帮助表示赞赏!

4

3 回答 3

65

只需使用min()

SELECT company, workflow, MIN(date) 
FROM workflowTable 
GROUP BY company, workflow
于 2011-04-20T21:27:38.607 回答
28

在这种情况下,一个相对简单的GROUP BY方法可以工作,但一般来说,当有其他列无法排序但您希望它们来自与它们关联的特定行时,您可以使用所有部分关键或用途OVER()

可运行示例(原始数据中的 Wofkflow20 错误已更正)

;WITH partitioned AS (
    SELECT company
        ,workflow
        ,date
        ,other_columns
        ,ROW_NUMBER() OVER(PARTITION BY company, workflow
                            ORDER BY date) AS seq
    FROM workflowTable
)
SELECT *
FROM partitioned WHERE seq = 1
于 2011-04-20T21:42:20.500 回答
8
SELECT company
   , workflow
   , MIN(date)
FROM workflowTable
GROUP BY company
       , workflow
于 2011-04-20T21:28:12.927 回答