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我的任务是生成一些关于我们的请求跟踪器使用情况的报告。Request Tracker是我工作的几个部门使用的票务系统。为此，我每晚将有关当天更改的门票的详细信息拍摄到另一个数据库中。这种方法将我的报告与 RT 使用的内部数据库模式分离。

在报告的许多其他问题中，我需要报告每个部门每个月解决了多少票。在 RT 中，部门存储为 CustomField，我的建模遵循这种趋势，正如您在下面的查询中看到的那样。但是，由于我每天晚上抓取快照的方式，一张票有多行，并且部门字段可能会在一个月内发生变化。我只对最近的部门领域感兴趣。我不知道如何在查询中得到它。

我知道我可以使用“GROUP BY”将查询结果减少到每张票一个，但是当我这样做时，我不知道如何获取最后一个部门设置。由于部门都是字符串，因此 MAX() 不会得到最后一个。MySQL 不要求您对正在选择的字段使用聚合函数，但结果是不确定的（从我的测试看来，它可能会抢占我的 MySQL 版本中的第一个）。

为了说明，这里是一个查询的结果，它显示了两张票，以及所有它的部门字段设置：

"ticket_num","date","QueueName","CF","CFValue","closed"
35750,"2009-09-22","IT_help","Department","",""
35750,"2009-09-23","IT_help","Department","",""
35750,"2009-09-24","IT_help","Department","",""
35750,"2009-09-25","IT_help","Department","",""
35750,"2009-09-26","IT_help","Department","",""
35750,"2009-10-02","IT_help","Department","",""
35750,"2009-10-03","IT_help","Department","",""
35750,"2009-10-12","IT_help","Department","",""
35750,"2009-10-13","IT_help","Department","",""
35750,"2009-10-26","IT_help","Department","Conference/Visitors","2009-10-26 10:10:32"
35750,"2009-10-27","IT_help","Department","Conference/Visitors","2009-10-26 10:10:32"
36354,"2009-10-20","IT_help","Department","",""
36354,"2009-10-21","IT_help","Department","",""
36354,"2009-10-22","IT_help","Department","FS Students",""
36354,"2009-10-23","IT_help","Department","FS Students",""
36354,"2009-10-26","IT_help","Department","FS Students","2009-10-26 12:23:00"
36354,"2009-10-27","IT_help","Department","FS Students","2009-10-26 12:23:00"

如我们所见，两张票都在 26 日关闭，并且两张票在首次出现时的几天内都有空的 Department 字段。我在下面包含了我的查询，您可以看到我人为地限制了 where 语句后半部分返回的列数：

SELECT d.ticket_num, d.date, q.name as QueueName, cf.name as CF, cfv.value as CFValue, d.closed
FROM daysCF dcf
INNER JOIN daily_snapshots d on dcf.day_id = d.id
INNER JOIN Queues q on d.queue_id = q.id
INNER JOIN CustomFieldValues cfv on dcf.cfv_id = cfv.id
INNER JOIN CustomFields cf on cf.id = cfv.field_id
WHERE cf.name = 'Department' and (d.ticket_num = 35750 or d.ticket_num = 36354)
ORDER by d.ticket_num, d.date

如何修改该查询，以便我得到一个结果集，告诉我在 10 月份有一张为“FS 学生”关闭的票和一张为“会议/访客”关闭的票？

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2 回答 2

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Mysql 没有 LAST 运算符，因此您确实需要使用临时表来执行此操作。

CREATE TEMPORARY TABLE last_dates SELECT ticket_num, MAX(date) AS date
  FROM daily_snapshots GROUP BY ticket_num

这会为您提供一张包含每张票的最后日期的表格。然后在您的主查询中，使用 ticket_num 和 date 字段加入这个表。这将过滤掉相应票号日期不是最新的所有行。

您可能需要该临时表上的索引，我将把它留给您。

于 2009-10-28T06:23:28.163 回答
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这是 Stack Overflow 上经常出现的“greatest-n-per-group”问题。

在您的情况下，我将如何解决它：

SELECT d1.ticket_num, d1.date, q.name as QueueName, 
  cf.name as CF, cfv.value as CFValue, d1.closed
FROM daysCF dcf
INNER JOIN daily_snapshots d1 ON (dcf.day_id = d1.id)
INNER JOIN Queues q ON (d1.queue_id = q.id)
INNER JOIN CustomFieldValues cfv ON (dcf.cfv_id = cfv.id)
INNER JOIN CustomFields cf ON (cf.id = cfv.field_id)
LEFT OUTER JOIN daily_snapshots d2 ON (d1.ticket_num = d2.ticket_num AND d1.date < d2.date)
WHERE d2.id IS NULL AND cf.name = 'Department'
ORDER by d1.ticket_num, d1.date;
于 2009-10-28T06:29:00.603 回答