SQL：如何选择最早的行 [英] SQL: How To Select Earliest Row

问题描述

我有一个类似这样的报告：

I have a report that looks something like this:

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（具体来说，T-SQL版本Server 2005）完成的：

This is done with SQL (specifically, T-SQL version Server 2005):

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：

I would like the report to show just the earliest dates for each workflow:

CompanyA      Workflow27     June5
CompanyB      Workflow13     Apr4
CompanyB      Workflow20     Dec11

任何想法？我无法理解这一点。我尝试使用一个嵌套选择返回最早的托盘日期，然后在WHERE子句中进行设置。如果只有一家公司，这样做很棒：

Any ideas? I can't figure this out. I've tried using a nested select that returns the earliest tray date, and then setting that in the WHERE clause. This works great if there were only one company:

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

但是如果该表中有多个公司，这显然不会奏效。任何帮助不胜感激！

but this obviously won't work if there is more than one company in that table. Any help is appreciated!

推荐答案

只需使用 min（） p>

Simply use min()

SELECT company, workflow, MIN(date) 
FROM workflowTable 
GROUP BY company, workflow

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