问题描述

我有一个用户数据库.我想创建一个基于用户群增长的图表.我现在的查询是:

I have a database of users. I would like to create a graph based on userbase growth. The query I have now is:

SELECT DATE(datecreated), count(*) AS number FROM users 
WHERE DATE(datecreated) > '2009-06-21' AND DATE(datecreated) <= DATE(NOW())
GROUP BY DATE(datecreated) ORDER BY datecreated ASC

这返回几乎我想要的.如果我们一天有 0 个用户，那一天不会作为 0 值返回，它只是被跳过，并返回至少有一个用户的第二天.我怎样才能得到类似(伪响应):

This returns almost what I want. If we get 0 users one day, that day is not returned as a 0 value, it is just skipped and the next day that has at least one user is returned. How can I get something like (psuedo-response):

date1 5
date2 8
date3 0
date4 0
date5 9
etc...

带有零的日期与其余日期按顺序显示在哪里?

where the dates with zero show up in sequential order with the rest of the dates?

谢谢！

推荐答案

希望你能解决剩下的问题.

I hope you will figure out the rest.

select  * from (
select date_add('2003-01-01 00:00:00.000', INTERVAL n5.num*10000+n4.num*1000+n3.num*100+n2.num*10+n1.num DAY ) as date from
(select 0 as num
   union all select 1
   union all select 2
   union all select 3
   union all select 4
   union all select 5
   union all select 6
   union all select 7
   union all select 8
   union all select 9) n1,
(select 0 as num
   union all select 1
   union all select 2
   union all select 3
   union all select 4
   union all select 5
   union all select 6
   union all select 7
   union all select 8
   union all select 9) n2,
(select 0 as num
   union all select 1
   union all select 2
   union all select 3
   union all select 4
   union all select 5
   union all select 6
   union all select 7
   union all select 8
   union all select 9) n3,
(select 0 as num
   union all select 1
   union all select 2
   union all select 3
   union all select 4
   union all select 5
   union all select 6
   union all select 7
   union all select 8
   union all select 9) n4,
(select 0 as num
   union all select 1
   union all select 2
   union all select 3
   union all select 4
   union all select 5
   union all select 6
   union all select 7
   union all select 8
   union all select 9) n5
) a
where date >'2011-01-02 00:00:00.000' and date < NOW()
order by date

与

select n3.num*100+n2.num*10+n1.num as date

你会得到一列数字从 0 到 max(n3)*100+max(n2)*10+max(n1)

you will get a column with numbers from 0 to max(n3)*100+max(n2)*10+max(n1)

由于这里我们将最大 n3 设为 3，SELECT 将返回 399，加上 0 -> 400 条记录(日历中的日期).

Since here we have max n3 as 3, SELECT will return 399, plus 0 -> 400 records (dates in calendar).

您可以通过限制动态日历来调整它，例如，从 min(date) 到 now().

You can tune your dynamic calendar by limiting it, for example, from min(date) you have to now().

这篇关于MySQL:即使不存在记录，也选择范围内的所有日期的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！