Question

我有一张桌子订单：

id    total_amount       date_order 
1        12.000          2018-09-01
2        10.000          2018-09-01
3         5.000          2018-09-03
4         2.000          2018-09-05

我有按日期查询SUM数据组：

select SUM(total_amount) as 'total', DATE(date_order) as 'date'
from Order
where date_order >= '2018-09-01' and date_order <= '2018-09-06'
group by (date_order)

它显示：

   total           date       
  22.000        2018-09-01      
   5.000        2018-09-03   
   2.000        2018-09-05

因为 2018-09-02 和 2018-09-04 中的数据均无数据，因此未在结果中显示。但我想查询显示具有预期结果的表：

   total           date       
  22.000        2018-09-01    
       0        2018-09-02      
   5.000        2018-09-03
       0        2018-09-04   
   2.000        2018-09-05

每个人都可以帮助我编写查询以显示预期结果吗？

Answer 1

创建另一个要联接的表。

它可能是日期表，例如2000年1月1日至2099年12月31日之间的所有日期...

SELECT
  dt.`date`,
  SUM(total_amount) as `total`
FROM
  yourDatesTable     AS dt
LEFT JOIN
  Order              AS o
    ON  o.`date_order` = dt.`date`
WHERE
      dt.`date` >= '2018-09-01'
  AND dt.`date` <  '2018-09-08'
GROUP BY
  dt.`date`
ORDER BY
  dt.`date`

或者它可能是一个数值表，其值介于-1023到+1024 （或其他有用范围） ...

SELECT
  n.id,
  '2018-09-01' + INTERVAL n.id DAYS,
  SUM(total_amount) as `total`
FROM
  yourNumbersTable   AS n
LEFT JOIN
  Order              AS o
    ON  o.`date_order` = '2018-09-01' + INTERVAL n.id DAYS
WHERE
      n.id >= 0
  AND n.id <  8
GROUP BY
  n.id
ORDER BY
  n.id

Answer 2

生成日期并与订单表联接

    select SUM(O.total_amount) as total, DATE(gen_date) as oreder_date
        from    
        (
         select * from 
            (select adddate('1970-01-01',t4*10000 + t3*1000 + t2*100 + t1*10 + t0) gen_date from
             (select 0 t0 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t0,
             (select 0 t1 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t1,
             (select 0 t2 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t2,
             (select 0 t3 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t3,
             (select 0 t4 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t4) v
            where gen_date between '2018-09-01' and '2018-09-30'
        ) as t1 left join `Order` O on t1.gen_date= DATE(O.date_order)

where gen_date >= '2018-09-01' and gen_date <= '2018-09-06'
         group by oreder_date

关于在MySQL中按日期显示数据筛选器的问题

2 个答案: