Mysql 为应具有值的列连接返回零

我相信我的查询格式不正确，但我的谷歌技能还没有找到一个有效的解决方案。这是我的疑问：

use movement;
select t1.StoreNumber, t1.saleDate, t1.departmentNumber, sum(t1.dollarsSold), sum(ifnull(t2.loss,0))
from movement.movement t1
left join knownLoss.producekl t2 ON t1.StoreNumber = t2.Store
where t1.StoreNumber = 3
AND (t1.departmentNumber = 10 OR t1.departmentNumber = 20 OR t1.departmentNumber = 27 OR t1.departmentNumber = 30 OR t1.departmentNumber = 40 OR t1.departmentNumber = 50 OR t1.departmentNumber = 51 OR t1.departmentNumber = 60 OR t1.departmentNumber = 70 OR t1.departmentNumber = 80 OR t1.departmentNumber = 81 OR t1.departmentNumber = 82 OR t1.departmentNumber = 90 OR t1.departmentNumber = 95 OR t1.departmentNumber = 96 OR t1.departmentNumber = 97)
AND t1.saleDate > date_sub(curdate(), interval 7 day)
group by t1.saleDate, t1.StoreNumber, t1.departmentNumber;

我的预期输出是在最后两列中提供一个按销售日期、商店和部门分组的表格，汇总销售总额和销售损失

我的预期输出应该如下所示：

-------------------------------------------------------------
| Store Number | Sale Date | Dept Num | Dollars Sold | Loss |
-------------------------------------------------------------
|      1       | 2017-3-9  |    10    |    7435.26   | 0.00 |
-------------------------------------------------------------
|      1       | 2017-3-9  |    20    |    30.50     | 3.00 |
-------------------------------------------------------------
|      1       | 2017-3-10 |    10    |    2503.22   | 4.00 |
-------------------------------------------------------------
|      2       | 2017-3-11 |    10    |    5685.49   |17.50 |
-------------------------------------------------------------
|      3       | 2017-3-9  |    10    |    75.26     | 4.55 |
-------------------------------------------------------------
|      4       | 2017-3-9  |    10    |    7435.26   | 0.00 |
-------------------------------------------------------------
|      4       | 2017-3-13 |    30    |      20.45   | 1.99 |
-------------------------------------------------------------

损失列不是我的预期输出，而是全部为零

-------------------------------------------------------------
| Store Number | Sale Date | Dept Num | Dollars Sold | Loss |
-------------------------------------------------------------
|      1       | 2017-3-9  |    10    |    7435.26   | 0.00 |
-------------------------------------------------------------
|      1       | 2017-3-9  |    20    |    30.50     | 0.00 |
-------------------------------------------------------------
|      1       | 2017-3-10 |    10    |    2503.22   | 0.00 |
-------------------------------------------------------------
|      2       | 2017-3-11 |    10    |    5685.49   | 0.00 |
-------------------------------------------------------------
|      3       | 2017-3-9  |    10    |    75.26     | 0.00 |
-------------------------------------------------------------
|      4       | 2017-3-9  |    10    |    7435.26   | 0.00 |
-------------------------------------------------------------
|      4       | 2017-3-13 |    30    |      20.45   | 0.00 |
-------------------------------------------------------------

即使当我单独查询t2时，它也会返回结果

use knownLoss;
select t2.Store, t2.Department, t2.klDate, sum(t2.loss)
from producekl t2
group by klDate, Store, Department;

---

所以我的加入一定是一团糟，但我不知道怎么办。有什么想法吗？

FWIW，我觉得这更容易阅读

我不熟悉curdatee函数，所以我用它代替了其他函数

SELECT t1.StoreNumber
     , t1.saleDate
     , t1.departmentNumber
     , SUM(t1.dollarsSold)
     , SUM(IFNULL(t2.loss,0))
  FROM movement.movement t1
  LEFT
  JOIN knownLoss.producekl t2 
    ON t1.StoreNumber = t2.Store
 WHERE t1.StoreNumber = 3
   AND t1.departmentNumber IN(10,20,27,30,40,50,51,60,70,80,81,82,90,95,96,97)
   AND t1.saleDate > DATE_SUB(CURDATE(), INTERVAL 7 DAY)
 GROUP BY t1.saleDate
     , t1.StoreNumber
     , t1.departmentNumber;

---

请注意，组BY中缺少saledate。糟糕。

我终于解决了这个问题。我加入的表都没有唯一的值，因此我必须使用子查询来生成唯一的值。这是最终的查询，它产生了我所期望的输出

SELECT t2.StoreNumber
 , t2.departmentNumber
 , sum(t2.Sales) AS netSales
 , ifnull(sum(t3.loss),0) AS knownLoss
 , ifnull(round(avg(t3.loss / t2.Sales),4),0) AS percentLoss
from ( SELECT t1.StoreNumber
   , t1.departmentNumber
   , SUM(t1.dollarsSold) AS Sales
  FROM movement.movement t1
  WHERE t1.StoreNumber = 1
  AND t1.departmentNumber = 80
  AND t1.saleDate >= date_sub(SUBDATE(curdate(), WEEKDAY(curdate())), interval 7 day)
  AND t1.saleDate < SUBDATE(curdate(), WEEKDAY(curdate()))
  GROUP BY t1.StoreNumber
   , t1.departmentNumber) t2
   left join
   (select Store, Department, sum(loss) AS loss
    from producekl
    where Department = 80
    AND Store = 1
    AND klDate >= date_sub(SUBDATE(curdate(), WEEKDAY(curdate())), interval 7 day)
    AND klDate < SUBDATE(curdate(), WEEKDAY(curdate()))
    group by Store, Department) t3 ON t2.StoreNumber = t3.Store
GROUP BY t2.StoreNumber, t2.departmentNumber

---

感谢草莓鼓励良好的格式，它帮助我更快地解决了问题。

我怀疑你的t1.StoreNumber和t2.Store是不相关的。您如何查询producekl退货？请参阅中的。一根没有棒棒糖的棍子一点也不好玩。请注意，格式化更容易阅读，但它返回相同的结果。我不希望按日期分组，这只是我用来返回结果的标准，这应该是在日期范围内出现的值的总和，如果我包括日期，它将把我要查找的总数分成我不想要的部分。虽然格式很好，但我需要记住，格式对易读性很重要。