Question

我有一个简单的表：user（id，date，task）

任务栏包含＆＃34; download＆＃34;或者＆＃34;上传＆＃34;

我想弄清楚每天执行每项操作的用户数量。

输出：日期，下载的用户数，上传的用户数

我首先遇到了在select的聚合计数函数中使用子查询的问题，所以我认为我应该在这里使用自联接来分解＆＃34;任务中的数据＆＃34;列。

我以为我可以为每个案例创建表格，然后结合这些和计数，但我无法完成这个：

SELECT id，date，task as task_download 来自用户任务=＆＃39;下载＆＃39;

SELECT id，date，task as task_upload 来自用户 WHERE task =＆＃39; upload＆＃39;

Answer 1

select  `date`, 
COUNT( distinct CASE WHEN task = 'download' then id end ) 'download', 
COUNT( distinct CASE WHEN task = 'upload' then id end ) 'upload'
from user
group by  `date`

Answer 2

我会说，既不是也不是。只需这样的查询即可完成任务：

select `date`, 
    count(distinct case when task = 'download' then id else null end) as downloads, 
    count(distinct case when task = 'upload' then id else null end) as uploads
from user
where  task in ('download', 'upload')
group by `date`

假设date是仅包含日期部分而不包含完整时间戳的列，id是用户ID。您可以在聚合函数中使用distinct关键字，这就是我在这里所做的。

要让此查询快速运行，我建议在task,date

上使用索引

但是，如果date包含完整的时间戳（即包括时间部分），则您希望以不同方式进行分组：

select `date`, 
    count(distinct case when task = 'download' then id else null end) as downloads, 
    count(distinct case when task = 'upload' then id else null end) as uploads
from user
where  task in ('download', 'upload')
group by date(`date`)

Answer 3

您可以使用子查询来执行此操作，例如：

SELECT `date` AS `day`,
(SELECT COUNT(*) FROM activity WHERE date = day AND activity = 'upload') AS upload_count,
(SELECT COUNT(*) FROM activity WHERE date = day AND activity = 'download') AS download_count
FROM activity
GROUP BY date;

这里是SQL Fiddle。

Answer 4

首先按日期和任务统计不同的用户，然后根据日期的每个任务对用户求和。

select date,
       sum(case when task = 'upload' then num_users else 0 end) as "upload",
       sum(case when task = 'download' then num_users else 0 end) as "download"
from  (       
       select   date, task, count(distinct id) num_users
       from     usert
       group by date, task
      ) x
group by date
;

在此处查看：http://rextester.com/ZACFB64945

Answer 5

如果您需要不同的用户，那么建议使用count(distinct)：

SELECT date, 
       COUNT(DISTINCT CASE WHEN task = 'upload' THEN userid END) as uploads,
       COUNT(DISTINCT CASE WHEN task = 'download' THEN userid END) as downloads
FROM user
GROUP BY date
ORDER BY date;

如果您需要不同的操作，则可以执行以下操作：

SELECT date, 
       SUM( (task = 'upload')::int ) as uploads,
       SUM( (task = 'download')::int) as downloads
FROM user
GROUP BY date
ORDER BY date;

这使用方便的Postgres简写来计算布尔表达式。

Answer 6

我使用条件聚合。

要计算在给定日期至少执行一次上传的用户数量的计数（但仅在该日期为该用户增加一次计数，即使该用户执行了更多操作我们可以使用COUNT(DISTINCT user)表达式。

要计算上传总数，我们可以使用COUNT或SUM。

SELECT DATE(t.date) AS `date`
     , COUNT(DISTINCT IF(t.task='upload'  ,t.user,NULL)) AS cnt_users_who_uploaded
     , COUNT(DISTINCT IF(t.task='download',t.user,NULL)) AS cnt_users_who_downloaded
     , SUM(IF(t.task='upload'  ,1,0))                    AS cnt_uploads
     , SUM(IF(t.task='download',1,0))                    AS cnt_downloads
  FROM user t
 GROUP BY DATE(t.date)
 ORDER BY DATE(t.date)

注意：对于没有行date未显示在表中的日期，这不会返回零计数。

SQL：联盟或自我加入

6 个答案: