如何使用 Group By select 1 到 5 天的数据
How to select data from 1 to 5 days with Group By
我这里试过了,不行,只能显示总数。
我有一个下载 table 和一个程序 table。
每次下载一个程序我都会记录日期和时间,我需要对下载的程序进行分组,然后将日期分为 5 列,这是一个例子。
PROGRAMA | HOJE | ONTEM| 2 DIAS | 3 DIAS | 4 DIAS
Programa 1 11 110 55 66 12
Programa 2 25 140 60 90 12
Programa 3 10 20 20 10 10
TOTAL 46 270 135 166 32
下面是我的查询
select `k`.`app_id` AS `app_id`,`b`.`aplicativo` AS `aplicativo`,count(0) AS `HOJE`,
(select count(0) AS `count(*)` from (`registration` `a` join `aplicativos` `b`) where `k`.`app_id`= `b`.`id` and created_at > (cast(now() as date) - interval 1 day) and (`a`.`created_at` < cast(now() as date)- interval 0 day) ) as ONTEM ,
(select count(0) AS `count(*)` from (`registration` `a` join `aplicativos` `b`) where `k`.`app_id` = `b`.`id`
and created_at > (cast(now() as date) - interval 2 day) and (`a`.`created_at` < cast(now() as date)- interval 1 day) ) as 2_DIAS_ANTES ,
(select count(0) AS `count(*)` from (`registration` `a` join `aplicativos` `b`) where `k`.`app_id` = `b`.`id`
and created_at > (cast(now() as date) - interval 3 day) and (`a`.`created_at` < cast(now() as date)- interval 2 day) ) as 3_DIAS_ANTES ,
(select count(0) AS `count(*)` from (`registration` `a` join `aplicativos` `b`) where `k`.`app_id` = `b`.`id`
and created_at > (cast(now() as date) - interval 4 day) and (`a`.`created_at` < cast(now() as date)- interval 3 day) ) as 4_DIAS_ANTES ,
(select count(0) AS `count(*)` from (`registration` `a` join `aplicativos` `b`) where `k`.`app_id` = `b`.`id`
and created_at > (cast(now() as date) - interval 5 day) and (`a`.`created_at` < cast(now() as date)- interval 4 day) ) as 5_DIAS_ANTES
from (`registration` `k` join `aplicativos` `b`) where ((`k`.`app_id` = `b`.`id`) and (`k`.`created_at` > (cast(now() as date) - interval 0 day)))
group by `b`.`aplicativo`
Table结构
Table 应用程序
CREATE TABLE IF NOT EXISTS `aplicativos` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`id_usuario` int(11) NOT NULL,
`aplicativo` varchar(200) NOT NULL,
`link` varchar(400) NOT NULL,
`quantidade_notificacoes` int(11) NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=13 ;
Table 注册
CREATE TABLE IF NOT EXISTS `registration` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`gcm_regid` varchar(300) NOT NULL,
`app_id` int(11) NOT NULL,
`email` varchar(200) NOT NULL,
`created_at` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=73876 ;
这是否给了您预期的结果?
SELECT
a.aplicativo as PROGRAMA,
COUNT(IF(DATE(r.created_at) = CURDATE(), 1, 0)) AS HOJE,
COUNT(IF(DATE(r.created_at) = DATE_SUB(CURDATE(), INTERVAL 1 DAY), 1, 0)) AS ONTEM,
COUNT(IF(DATE(r.created_at) = DATE_SUB(CURDATE(), INTERVAL 2 DAY), 1, 0)) AS 2DIAS,
COUNT(IF(DATE(r.created_at) = DATE_SUB(CURDATE(), INTERVAL 3 DAY), 1, 0)) AS 3DIAS,
COUNT(IF(DATE(r.created_at) = DATE_SUB(CURDATE(), INTERVAL 4 DAY), 1, 0)) AS 4DIAS,
COUNT(IF(DATE(r.created_at) = DATE_SUB(CURDATE(), INTERVAL 5 DAY), 1, 0)) AS 5DIAS
FROM registration r
INNER JOIN aplicativos a
ON r.app_id = a.id
GROUP BY r.app_id, DATE(r.created_at) with ROLLUP;
这是 MySQL 的一种方法:
SELECT a.aplicativo as PROGRAMA,
sum(Date(r.created_at) = CURDATE()) AS HOJE,
sum(date(r.created_at) = DATE_SUB(CURDATE(), INTERVAL 1 DAY), 1, 0)) AS ONTEM,
...
FROM registration r INNER JOIN
aplicativos a
on r.app_id = a.id
GROUP BY r.app_id ;
我这里试过了,不行,只能显示总数。
我有一个下载 table 和一个程序 table。
每次下载一个程序我都会记录日期和时间,我需要对下载的程序进行分组,然后将日期分为 5 列,这是一个例子。
PROGRAMA | HOJE | ONTEM| 2 DIAS | 3 DIAS | 4 DIAS
Programa 1 11 110 55 66 12
Programa 2 25 140 60 90 12
Programa 3 10 20 20 10 10
TOTAL 46 270 135 166 32
下面是我的查询
select `k`.`app_id` AS `app_id`,`b`.`aplicativo` AS `aplicativo`,count(0) AS `HOJE`,
(select count(0) AS `count(*)` from (`registration` `a` join `aplicativos` `b`) where `k`.`app_id`= `b`.`id` and created_at > (cast(now() as date) - interval 1 day) and (`a`.`created_at` < cast(now() as date)- interval 0 day) ) as ONTEM ,
(select count(0) AS `count(*)` from (`registration` `a` join `aplicativos` `b`) where `k`.`app_id` = `b`.`id`
and created_at > (cast(now() as date) - interval 2 day) and (`a`.`created_at` < cast(now() as date)- interval 1 day) ) as 2_DIAS_ANTES ,
(select count(0) AS `count(*)` from (`registration` `a` join `aplicativos` `b`) where `k`.`app_id` = `b`.`id`
and created_at > (cast(now() as date) - interval 3 day) and (`a`.`created_at` < cast(now() as date)- interval 2 day) ) as 3_DIAS_ANTES ,
(select count(0) AS `count(*)` from (`registration` `a` join `aplicativos` `b`) where `k`.`app_id` = `b`.`id`
and created_at > (cast(now() as date) - interval 4 day) and (`a`.`created_at` < cast(now() as date)- interval 3 day) ) as 4_DIAS_ANTES ,
(select count(0) AS `count(*)` from (`registration` `a` join `aplicativos` `b`) where `k`.`app_id` = `b`.`id`
and created_at > (cast(now() as date) - interval 5 day) and (`a`.`created_at` < cast(now() as date)- interval 4 day) ) as 5_DIAS_ANTES
from (`registration` `k` join `aplicativos` `b`) where ((`k`.`app_id` = `b`.`id`) and (`k`.`created_at` > (cast(now() as date) - interval 0 day)))
group by `b`.`aplicativo`
Table结构
Table 应用程序
CREATE TABLE IF NOT EXISTS `aplicativos` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`id_usuario` int(11) NOT NULL,
`aplicativo` varchar(200) NOT NULL,
`link` varchar(400) NOT NULL,
`quantidade_notificacoes` int(11) NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=13 ;
Table 注册
CREATE TABLE IF NOT EXISTS `registration` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`gcm_regid` varchar(300) NOT NULL,
`app_id` int(11) NOT NULL,
`email` varchar(200) NOT NULL,
`created_at` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=73876 ;
这是否给了您预期的结果?
SELECT
a.aplicativo as PROGRAMA,
COUNT(IF(DATE(r.created_at) = CURDATE(), 1, 0)) AS HOJE,
COUNT(IF(DATE(r.created_at) = DATE_SUB(CURDATE(), INTERVAL 1 DAY), 1, 0)) AS ONTEM,
COUNT(IF(DATE(r.created_at) = DATE_SUB(CURDATE(), INTERVAL 2 DAY), 1, 0)) AS 2DIAS,
COUNT(IF(DATE(r.created_at) = DATE_SUB(CURDATE(), INTERVAL 3 DAY), 1, 0)) AS 3DIAS,
COUNT(IF(DATE(r.created_at) = DATE_SUB(CURDATE(), INTERVAL 4 DAY), 1, 0)) AS 4DIAS,
COUNT(IF(DATE(r.created_at) = DATE_SUB(CURDATE(), INTERVAL 5 DAY), 1, 0)) AS 5DIAS
FROM registration r
INNER JOIN aplicativos a
ON r.app_id = a.id
GROUP BY r.app_id, DATE(r.created_at) with ROLLUP;
这是 MySQL 的一种方法:
SELECT a.aplicativo as PROGRAMA,
sum(Date(r.created_at) = CURDATE()) AS HOJE,
sum(date(r.created_at) = DATE_SUB(CURDATE(), INTERVAL 1 DAY), 1, 0)) AS ONTEM,
...
FROM registration r INNER JOIN
aplicativos a
on r.app_id = a.id
GROUP BY r.app_id ;