php - MySQL + PHP : Display items per week?

Question

这让我很困惑......

如果我有一个 MySQL 表:

UserId | Commission | Date Of Commission
   1   |  200.00    |   2014-02-12
   1   |  50.00     |   2014-04-01
   2   |  10.00     |   2014-04-05

我想显示特定用户从他/她的第一条记录开始每周的总佣金，如果没有记录，则显示该范围的 0。

我该怎么做？

示例输出

UserId |     Date Range      | Total Commission
   1   | 02/10/14 - 02/16/14 |     200.00
   1   | 02/17/14 - 02/23/14 |      0.00
  ...
   1   | 03/31/14 - 04/06/14 |     50.00

我不是一位经验丰富的编码员，因此我们将不胜感激任何帮助。

谢谢!

编辑:

我已经尝试过这个:

SELECT IFNULL(SUM(Commisssion),0) Total ,DATE_SUB(`DateOfCommission`,INTERVAL 7 DAY) 
  AS RangStart,DATE_SUB(`DateOfCommission`,INTERVAL 1 DAY) AS RangeEnd 
FROM `comms` WHERE `UserId` = '$UserID' GROUP BY DATE(`DateOfCommission`) DESC

但它从输入第一条记录的日期开始一周..

Answer 1

这很难实现。这是我通过一些小的修改设法做到的，它应该按需要的方式工作。我已经为 userid = 1 完成了此操作，也可以为其他用户完成此操作。

在查询中我有 2 行

 where a.Date BETWEEN (select min(date) from transactions where UserId = 1) AND NOW()

和

  WHERE date BETWEEN (select min(date) from transactions where UserId = 1) AND NOW()

查询将尝试使用用户截至今天的交易 min() 日期生成日期列表。这也可以用作用户的 max() 交易日期，而不是 now()。

select 
t1.date_range,
coalesce(SUM(t1.Commission+t2.Commission), 0) AS Commission
from
(
  select 
  a.Date as date,
  concat(
    DATE_ADD(a.Date, INTERVAL(1-DAYOFWEEK(a.Date)) +1 DAY),
    ' - ',
    DATE_ADD(a.Date, INTERVAL(7- DAYOFWEEK(a.Date)) +1 DAY)
  ) as date_range,
  '0' as  Commission
  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 (select min(date) from transactions where UserId = 1) AND NOW()
)t1
left join
(
  SELECT date ,
  coalesce(SUM(Commission), 0) AS Commission
  FROM transactions
  WHERE date BETWEEN (select min(date) from transactions where UserId = 1) AND NOW()
  AND UserId = 1
  GROUP BY date
)t2
on t2.date = t1.date
group by t1.date_range
order by t1.date_range asc

<强> DEMO