我有以下查询。JOIN不显示所有行
SELECT WEEK(Review.created) AS Week,
CONCAT(Employee.firstname, ' ', Employee.lastname) AS Name,
AVG(Rating.scale) AS Average
FROM employees Employee
LEFT JOIN reviews Review
ON Employee.id = Review.reviewee_id
LEFT JOIN ratings AS Rating
ON Rating.id = Review.rating_id
WHERE Employee.id IN (71, 72)
GROUP BY WEEK(Review.created),
Employee.id
ORDER BY WEEK(Review.created),
Employee.id
结果是这样的:
36, Employee1, 2.9091
37, Employee2, 3.5000
37, Employee1, 3.7143
38, Employee2, 4.2000
38, Employee1, 4.0000
39, Employee2, 2.0000
40, Employee2, 2.8333
40, Employee1, 3.8571
41, Employee1, 2.6667
43, Employee2, 2.5000
43, Employee1, 1.5714
44, Employee2, 3.8333
44, Employee1, 4.4000
45, Employee2, 3.2500
45, Employee1, 4.8571
46, Employee2, 2.1667
46, Employee1, 2.2000
48, Employee2, 2.6667
49, Employee2, 1.4000
49, Employee1, 3.5000
50, Employee2, 2.0000
50, Employee1, 1.5000
51, Employee2, 2.7143
51, Employee1, 2.7500
我需要的是返回的员工,即使他们没有为这一周的评级。
所以前两行会返回
36, Employee1, 2.9091
36, Employee2, NULL
任何帮助,在此将不胜感激。
您需要外连接到包含所有星期的源。 – 2011-12-27 17:34:01
那么这意味着我需要创建一个虚拟日历表? – Jeff 2011-12-27 18:39:12
这将是最有效的方法。你也可以做'DISTINCT WEEK(创建)FROM Review',但我不会建议。 – 2011-12-27 19:03:37