Question

我每个月有4张桌子; oct，dec，feb，apr。它们具有相同的架构。

我需要使用相同的查询从所有表中提取数据，并将其分组为一个属性。

我知道如何为一个表执行此操作，但我想将结果合并到一个查询中，然后将输出转储到文件中。

以下是我的示例查询：

select TIMESTAMPDIFF(hour,startTime,endTime) as duration, count(*) from feb 
where type='gen' group by duration 
into outfile "/feb";

由于我正在为持续时间属性对此进行分组，因此我最终得到4个不同的输出文件，然后明确地合并和排序数据是我想要避免的。

任何帮助将不胜感激！

欧麦

Answer 1

您可以使用UNION：

select TIMESTAMPDIFF(hour,startTime,endTime) as duration, count(*) from feb 
where type='gen' group by duration 
UNION
select TIMESTAMPDIFF(hour,startTime,endTime) as duration, count(*) from apr
where type='gen' group by duration 
UNION
select TIMESTAMPDIFF(hour,startTime,endTime) as duration, count(*) from oct 
where type='gen' group by duration 
UNION
select TIMESTAMPDIFF(hour,startTime,endTime) as duration, count(*) from dec 
where type='gen' group by duration 
into outfile "/all";

请注意，您只在最后使用INFO OUTFILE子句而不是每个SELECT语句。

Answer 2

使用UNION ALL。它提供了比UNION更好的性能，无论如何在这种情况下删除重复都是错误的。您还可以在将所有表放在一起后执行GROUP BY，以便可以跨表对行进行分组。这将为您提供所有4个表中相同持续时间的行计数。

select everything.duration, count(*)
from
((select TIMESTAMPDIFF(hour,startTime,endTime) as duration
  from feb where type='gen')
 union all
 (select TIMESTAMPDIFF(hour,startTime,endTime) as duration
  from apr where type='gen')
 union all
 (select TIMESTAMPDIFF(hour,startTime,endTime) as duration
  from oct where type='gen')
 union all
 (select TIMESTAMPDIFF(hour,startTime,endTime) as duration
  from dec where type='gen')) everything
group by everything.duration
into outfile "/all";

从多个表中分组数据

2 个答案: