MySQL显示无记录的日期计数为0

Question

我正在尝试通过当前一周（过去7天）的列表获取所有用户尝试的COUNT次

此查询有效，但如果日期不存在，则不会返回0：

SELECT COUNT(*) AS attempt_count, 
    DATE_FORMAT(attempt_date,'%Y/%m/%d') AS attempt_date 
FROM users_attempts 
WHERE DATE_SUB(attempt_date, INTERVAL 1 DAY) > DATE_SUB(DATE(NOW()), INTERVAL 1 WEEK) 
GROUP BY DAY(attempt_date) DESC;

此查询返回每天上一个当前周的所有尝试次数COUNT，我知道了这一点（我只有1条记录）：

attempt_count | attempt_date
1               2014/06/19

我想要这个结果：

attempt_count | attempt_date
1               2014/06/19
0               2014/06/18
0               2014/06/17
0               2014/06/16
0               2014/06/15
0               2014/06/14
0               2014/06/13

非常感谢

演示 ：http : **//sqlfiddle.com/#!2** /
b58bb/1/0

Answer 1

好的，我从线程MySql Single
Table的
上一个答案中选择了过去7天，并包含空行

在这里，您可以做些什么来使日期选择动态化

select 
t1.attempt_date,
coalesce(SUM(t1.attempt_count+t2.attempt_count), 0) AS attempt_count
from
(
  select DATE_FORMAT(a.Date,'%Y/%m/%d') as attempt_date,
  '0' as  attempt_count
  from (
    select curdate() - INTERVAL (a.a + (10 * b.a) + (100 * c.a)) DAY as Date
    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 union all select 8 union all select 9) as a
    cross join (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 union all select 8 union all select 9) as b
    cross join (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 union all select 8 union all select 9) as c
  ) a
  where a.Date BETWEEN NOW() - INTERVAL 7 DAY AND NOW()
)t1
left join
(
  SELECT DATE_FORMAT(attempt_date,'%Y/%m/%d') AS attempt_date, 
  COUNT(*) AS attempt_count
  FROM users_attempts
  WHERE DATE_SUB(attempt_date, INTERVAL 1 DAY) > DATE_SUB(DATE(NOW()), INTERVAL 1 WEEK) 
  GROUP BY DAY(attempt_date) DESC
)t2
on t2.attempt_date = t1.attempt_date
group by DAY(t1.attempt_date)
order by t1.attempt_date desc;

演示