Question

有这样的几个问题，但我无法找到我特定问题的答案。

考虑下表：

|                DateOfSale |         food | customerid |
|---------------------------|--------------|------------|
| January, 03 2017 00:00:00 |        pizza |          1 |
| January, 03 2017 00:00:00 |        pizza |          2 |
| January, 03 2017 00:00:00 |        pizza |          3 |
| January, 03 2017 00:00:00 | fish & chips |          4 |
| January, 02 2017 00:00:00 |        pizza |          4 |
| January, 02 2017 00:00:00 |        pizza |          4 |
| January, 02 2017 00:00:00 |        pizza |          5 |
| January, 02 2017 00:00:00 | fish & chips |          6 |
| January, 01 2017 00:00:00 |        pizza |          7 |
| January, 01 2017 00:00:00 |        pizza |          8 |
| January, 01 2017 00:00:00 |        pizza |          9 |

我们出售比萨饼和鱼类＆amp;芯片。

您会注意到我们并没有出售任何鱼类和鱼类。 1月1日的筹码在下面的结果，但我希望看到0的销售数量，但我不是。此外，1月4日没有销售记录，我希望披萨和鱼类都能看到0。芯片。这就是我从日期参考表中获取日期的原因 - 即使销售人员没有这个日期也存在。

|                DateOfSale |         food | Sales |
|---------------------------|--------------|-------|
| January, 01 2017 00:00:00 |        pizza |     3 |
| January, 02 2017 00:00:00 | fish & chips |     1 |
| January, 02 2017 00:00:00 |        pizza |     3 |
| January, 03 2017 00:00:00 | fish & chips |     1 |
| January, 03 2017 00:00:00 |        pizza |     3 |

此结果的代码是：

SELECT DATE(d.thedate) as DateOfSale, f.food, count(s.customerid) as Sales
FROM ref_date d
LEFT JOIN sales s on d.thedate = s.saledate
INNER JOIN foods f on s.foodid = f.id
GROUP BY DATE(d.thedate), f.food

完整的数据小提琴在这里：http://sqlfiddle.com/#!9/ff9dc/2

Answer 1

你走了：

SELECT d.thedate AS DateOfSale, f.Food, COUNT(s.customerid) AS Sales
FROM ref_date d
JOIN foods f
LEFT JOIN sales s ON s.saledate = d.thedate AND s.foodid = f.id
GROUP BY DateOfSale, Food;

输出：

|                   thedate |         food | Sales |
|---------------------------|--------------|-------|
| January, 01 2017 00:00:00 | fish & chips |     0 |
| January, 01 2017 00:00:00 |        pizza |     3 |
| January, 02 2017 00:00:00 | fish & chips |     1 |
| January, 02 2017 00:00:00 |        pizza |     3 |
| January, 03 2017 00:00:00 | fish & chips |     1 |
| January, 03 2017 00:00:00 |        pizza |     3 |
| January, 04 2017 00:00:00 | fish & chips |     0 |
| January, 04 2017 00:00:00 |        pizza |     0 |

SQL Fiddle

Answer 2

您可以cross join包含所有可能日期的食物，然后离开加入销售表以获得0次的行。

SELECT DATE(d.thedate) as DateOfSale, f.food, count(s.customerid) as Sales
FROM ref_date d
cross join foods f
LEFT JOIN sales s on s.saledate=d.thedate and s.foodid=f.id   
GROUP BY DATE(d.thedate), f.food
order by 1,2

MySQL - 包括日期的零计数

2 个答案: