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