MySql单表，选择最近7天，包括空行

我在stackoverflow上搜索过类似的问题，但我不明白如何使其工作，我正在尝试做什么

所以，我想从数据库中获取最近7天的交易记录，并获取总销售额，如果某一天并没有数据，还包括空行

到目前为止，我所拥有的：

这将产生：

| PURCHASE_DATE | AMOUNT |
|---------------|--------|
|    2014-04-25 |     19 |
|    2014-04-24 |     38 |
|    2014-04-22 |     19 |
|    2014-04-19 |     19 |

我想要的是：

| PURCHASE_DATE | AMOUNT |
|---------------|--------|
|    2014-04-25 |     19 |
|    2014-04-24 |     38 |
|    2014-04-23 |      0 |
|    2014-04-22 |     19 |
|    2014-04-21 |      0 |
|    2014-04-20 |      0 |
|    2014-04-19 |     19 |

---

感谢您的帮助：

只需将一个子查询与您想要的日期放在一起，然后使用左外联接：

---

这并不容易。我从这个线程获取帮助，并将其与您的查询相结合

因此，我们的想法是获取过去7天的日期列表，然后左键将静态数量为0的这些日期加入到您的查询中，最后对它们求和。这可以用于任何日期范围，只需在两个查询中更改它们

select 
t1.purchase_date,
coalesce(SUM(t1.amount+t2.amount), 0) AS amount
from
(
  select DATE_FORMAT(a.Date,'%Y-%m-%d') as purchase_date,
  '0' as  amount
  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(purchase_date, '%Y-%m-%d') as purchase_date,
  coalesce(SUM(amount), 0) AS amount
  FROM transactions
  WHERE purchase_date BETWEEN NOW() - INTERVAL 7 DAY AND NOW()
  AND vendor_id = 0
  GROUP BY purchase_date
)t2
on t2.purchase_date = t1.purchase_date
group by t1.purchase_date
order by t1.purchase_date desc

---

这是最后7天

select d.thedate, coalesce(SUM(amount), 0) AS amount
from (select DATE(NOW()) as thedate union all
      select DATE(DATE_SUB( NOW(), INTERVAL 1 DAY)) union all
      select DATE(DATE_SUB( NOW(), INTERVAL 2 DAY)) union all
      select DATE(DATE_SUB( NOW(), INTERVAL 3 DAY)) union all
      select DATE(DATE_SUB( NOW(), INTERVAL 4 DAY)) union all
      select DATE(DATE_SUB( NOW(), INTERVAL 5 DAY)) union all
      select DATE(DATE_SUB( NOW(), INTERVAL 6 DAY))) d left outer join
     transactions t
     on t.purchase_date = d.thedate and vendor_id = 0
GROUP BY d.thedate
ORDER BY d.thedate DESC;

---

可能重复感谢，但它给出了一些错误异常->string282 SQLSTATE[42000]：语法错误或访问冲突：1064您的SQL语法有错误；查看与MySQL服务器版本对应的手册，了解第3行“union all select date”2014-04-23“union all select date”附近使用的正确语法。这应该是公认的答案。。。很不错的。它也是动态的，所以你不必硬编码日期。虽然这可能会回答这个问题，但你应该在你的答案中包含一个简短的解释，说明这个脚本是如何回答这个问题的。这有助于提供上下文，并使答案对以后可能遇到相同问题的人更有用。

select d.thedate, coalesce(SUM(amount), 0) AS amount
from (select DATE(NOW()) as thedate union all
      select DATE(DATE_SUB( NOW(), INTERVAL 1 DAY)) union all
      select DATE(DATE_SUB( NOW(), INTERVAL 2 DAY)) union all
      select DATE(DATE_SUB( NOW(), INTERVAL 3 DAY)) union all
      select DATE(DATE_SUB( NOW(), INTERVAL 4 DAY)) union all
      select DATE(DATE_SUB( NOW(), INTERVAL 5 DAY)) union all
      select DATE(DATE_SUB( NOW(), INTERVAL 6 DAY))) d left outer join
     transactions t
     on t.purchase_date = d.thedate and vendor_id = 0
GROUP BY d.thedate
ORDER BY d.thedate DESC;

with recursive all_dates(dt) as (
    select '2014-04-19' as dt
    union all 
    select dt + interval 1 day
    from all_dates
    where dt + interval 1 day <= '2014-04-25'
)
select d.dt as purchase_date, coalesce(m.amount, 0) as purchased
from all_dates as d
left join mytable m
on d.dt = m.purchase_date
order by purchase_date desc;