MySQL：选择范围内的所有日期，即使没有记录存在答案

【问题标题】：MySQL: Select All Dates In a Range Even If No Records PresentMySQL：选择范围内的所有日期，即使没有记录存在
【发布时间】：2010-11-06 00:30:19
【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：

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

这几乎返回了我想要的。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（psuedo-response）的东西：

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

带零的日期与其余日期按顺序显示在哪里？

谢谢！

【问题讨论】：

标签： mysql date gaps-and-islands

【解决方案1】：

最好这样做：

-- 7 Days:
set @n:=date(now() + interval 1 day);
SELECT qb.day_series as days , COALESCE(col_byte, 0) as Bytes from tbl1 qa
    right join (
        select (select @n:= @n - interval 1 day) day_series from tbl1 limit 7 ) as qb 
    on date(qa.Timestamp) = qb.day_series and 
qa.Timestamp > DATE_SUB(curdate(), INTERVAL 7 day) order by qb.day_series asc

-- 30 Days:
set @n:=date(now() + interval 1 day);
SELECT qb.day_series as days , COALESCE(col_byte, 0) as Bytes from tbl1 qa
    right join (
        select (select @n:= @n - interval 1 day) day_series from tbl1 limit 30 ) as qb 
    on date(qa.Timestamp) = qb.day_series and 
qa.Timestamp > DATE_SUB(curdate(), INTERVAL 30 day) order by qb.day_series asc;

或者没有像这样的变量：

SELECT qb.day_series as days , COALESCE(col_byte, 0) as Bytes from tbl1 qa
right join (
    select curdate() - INTERVAL a.a day as day_series from(
        select 0 as a 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
    ) as a ) as qb
on date(qa.Timestamp) = qb.day_series and
qa.Timestamp > DATE_SUB(curdate(), INTERVAL 7 day) order by qb.day_series asc;

【讨论】：

【解决方案2】：

查询是：

SELECT qb.dy as yourday, COALESCE(count(yourcolumn), 0) as yourcount from yourtable qa 
right join (
    select curdate() as dy    union
    select DATE_SUB(curdate(), INTERVAL 1 day) as dy     union
    select DATE_SUB(curdate(), INTERVAL 2 day) as dy     union
    select DATE_SUB(curdate(), INTERVAL 3 day) as dy     union
    select DATE_SUB(curdate(), INTERVAL 4 day) as dy     union
    select DATE_SUB(curdate(), INTERVAL 5 day) as dy     union
    select DATE_SUB(curdate(), INTERVAL 6 day) as dy        
    ) as qb 
on qa.dates = qb.dy 
and qa.dates > DATE_SUB(curdate(), INTERVAL 7 day)
order by qb.dy asc;

结果是：

+------------+-----------+
| yourday    | yourcount |
+------------+-----------+
| 2015-06-24 | 274339    |
| 2015-06-25 |      0    |
| 2015-06-26 |      0    |
| 2015-06-27 |      0    |
| 2015-06-28 | 134703    |
| 2015-06-29 |  87613    |
| 2015-06-30 |      0    |
+------------+-----------+

【讨论】：

【解决方案3】：

我希望你能解决剩下的问题。

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)

因为这里我们将 max n3 设为 3，所以 SELECT 将返回 399，加上 0 -> 400 条记录（日历中的日期）。

您可以通过限制来调整您的动态日历，例如，从您必须的 min(date) 到 now()。

【讨论】：

我认为这是一个绝妙的技巧。得到我的投票。我假设，此查询可用于填充“日历”表。
这是我见过的最干净的解决方案。真棒

【解决方案4】：

进一步思考，这样的东西应该是你想要的：

CREATE TEMPORARY TABLE DateSummary1 ( datenew timestamp ) SELECT DISTINCT(DATE(datecreated)) as datenew FROM users;

CREATE TEMPORARY TABLE DateSummary2 ( datenew timestamp, number int ) SELECT DATE(datecreated) as datenew, count(*) AS number FROM users 
WHERE DATE(datecreated) > '2009-06-21' AND DATE(datecreated) <= DATE(NOW())
GROUP BY DATE(datecreated) ORDER BY datecreated ASC;

SELECT ds1.datenew,ds2.number FROM DateSummary1 ds1 LEFT JOIN DateSummary2 ds2 on ds1.datenew=ds2.datenew;

这将为您提供第一个表中的所有日期，以及第二个表中的 count 汇总数据。您可能需要将 ds2.number 替换为 IF(ISNULL(ds2.number),0,ds2.number) 或类似的东西。

【讨论】：

【解决方案5】：

对表进行右外部联接，将其命名为 tblCalendar，其中预先填充了您希望报告的日期。并加入日期字段。

保罗

【讨论】：

【解决方案6】：

This question 提出同样的问题。通常，公认的答案似乎是您要么在应用程序逻辑中执行此操作（将您拥有的内容读入数组，然后遍历数组并创建缺失的日期），要么使用填充了您希望的日期的临时表加入。

【讨论】：