MySQL按第一周值分组，包括第二周值

我正在努力实现以下目标：

我有一个包含如下数据的表：

    INSERT INTO `control_leads` (`LeadID`, `Type`, `Business`, `Date`, `Multiplier`, `UserID`)
VALUES
    (1026, 'Bags', 'AB', '2017-01-18', 1, 11),
    (1030, 'Ice Scraper', 'AB', '2017-01-18', 1, 8),
    (1029, 'Drawstring ', 'AB', '2017-01-18', 1, 8),
    (1028, 'Umbrellas', 'AB', '2017-01-18', 1, 11),
    (1027, 'Lanyards', 'LP', '2017-01-18', 2, 11),
    (1025, 'Bags', 'AB', '2017-01-18', 2, 8),
    (1031, 'Lanyards', 'LP', '2017-01-18', 2, 8),
    (1018, 'Lanyards', 'LP', '2017-01-17', 1, 8),
    (1017, 'Bags', 'AB', '2017-01-17', 1, 8),
    (1023, 'Hand Warmer', 'AB', '2017-01-17', 1, 8),
    (1020, 'Lanyards', 'LP', '2017-01-17', 2, 11),
    (1021, 'Pens', 'AB', '2017-01-17', 1, 11),
    (1022, 'Bags', 'AB', '2017-01-17', 1, 8),
    (1024, 'Headphones', 'AB', '2017-01-17', 1, 11),
    (1007, 'Balls', 'AB', '2017-01-16', 1, 8),
    (1008, 'Lanyards', 'LP', '2017-01-16', 2, 8),
    (1009, 'Mugs', 'AB', '2017-01-16', 1, 11),
    (1010, 'Bags', 'AB', '2017-01-16', 1, 8),
    (1011, 'Bags', 'AB', '2017-01-16', 1, 11),
    (1013, 'Phone/Table', 'AB', '2017-01-16', 1, 8),
    (1015, 'Pens', 'AB', '2017-01-16', 1, 11),
    (1016, 'Selfie Stic', 'AB', '2017-01-16', 1, 8);

以下是我的sql查询：

SELECT IFNULL( SUM(Multiplier), 0) AS val, control_leads.Date AS date
FROM control_leads
WHERE Date > '2017-01-04'
GROUP BY control_leads.Date
ORDER BY date ASC 

结果：

10  2017-01-05
6   2017-01-06
8   2017-01-09
14  2017-01-10
9   2017-01-11
14  2017-01-12
10  2017-01-13
10  2017-01-16
8   2017-01-17
10  2017-01-18

但我试图得到本周的值，但也包括上周的比较

像这样：

val1  week1        val2  week2
10    2017-01-05   14    2017-01-12
6     2017-01-06   10    2017-01-1
8     2017-01-09   10    2017-01-16
14    2017-01-10   8     2017-01-17
9     2017-01-11   10    2017-01-18

编辑：

我没有很好地解释，我想按天分组，并将其与当天上周的值进行比较

      val1  week1        val2  week2
   day1 10    2017-01-05   14    2017-01-12
   day2 6     2017-01-06   10    2017-01-1
   day3 8     2017-01-09   10    2017-01-16
   day4 14    2017-01-10   8     2017-01-17
   day5 9     2017-01-11   10    2017-01-18

---

这应该适用于与上周进行比较，按天计算

SELECT
cl.date
, sum(cl.multiplier) val
, cl2.date last_week
,  ifnull(cl2.last_week_val,0) last_week_val
, sum(cl.multiplier) - ifnull(cl2.last_week_val,0) diff 
FROM control_leads cl 
  LEFT JOIN (select t.date, sum(t.multiplier) last_week_val from control_leads t group by t.date) cl2 
  ON (cl2.date = cl.date-7)
GROUP BY date
ORDER BY date;

---

差不多了。您插入的数据行中没有一行是在2017年，您是如何获得您用

WHERE Date>'2017-01-04'

描述的结果的？@CharlesBretana它被用作示例。实际的表太大了。很抱歉，我想我没有解释清楚，我正在尝试按天分组。@jrobilliard这更简单、更干净@我更新了我的答案。您还可以单击该链接以查看它如何处理数据集