SQL查询在日期范围内匹配日期的记录数？

Question

我有一个表格,其中包含如下记录:

CREATE TABLE sample (
  ix int unsigned auto_increment primary key,
  start_active datetime,
  last_active datetime
);

我需要知道过去30天每天有多少记录活跃.日期也应该按递增顺序进行排序,以便将它们最旧的返回到最新.

我正在使用MySQL,查询将从PHP运行,但我不需要PHP代码,只需要查询.

这是我的开始:

SELECT COUNT(1) cnt, DATE(?each of last 30 days?) adate
FROM sample
WHERE adate BETWEEN start_active AND last_active
GROUP BY adate;

Answer 1

做外连接.

没有桌子？做一张桌子.我总是为此保留一张假桌子.

create table artificial_range( 
  id int not null primary key auto_increment, 
  name varchar( 20 ) null ) ;

-- or whatever your database requires for an auto increment column

insert into artificial_range( name ) values ( null )
-- create one row.

insert into artificial_range( name ) select name from artificial_range;
-- you now have two rows

insert into artificial_range( name ) select name from artificial_range;
-- you now have four rows

insert into artificial_range( name ) select name from artificial_range;
-- you now have eight rows

--etc.

insert into artificial_range( name ) select name from artificial_range;
-- you now have 1024 rows, with ids 1-1024

现在使用方便,并将其限制为30天,并具有以下视图:

编辑:JR Lawhorne指出:

您需要将"date_add"更改为"date_sub"以在创建的视图中获取前30天.

谢谢JR!

create view each_of_the_last_30_days as
select date_sub( now(), interval (id - 1) day ) as adate
from artificial_range where id < 32;

现在在你的查询中使用它(我实际上没有测试过你的查询,我只是假设它正常工作):

编辑:我应该加入另一种方式:

SELECT COUNT(*) cnt, b.adate
FROM  each_of_the_last_30_days b
left outer join sample a 
 on ( b.adate BETWEEN a.start_active AND a.last_active)
GROUP BY b.adate;