[英]Query giving undesirable results
我有以下两个表:
lab_assistant
la_id - la_firstname - la_lastname
1 - Dennis - Rodman
2 - Michael - Jordan
3 - Horace - Grant
hours_worked
hw_semester - hw_date - hw_hours - la_id
Fall 2012 - 2012-11-01 - 4 - 2
Fall 2012 - 2012-11-04 - 5 - 3
Spring 2012 - 2012-02-12 - 4 - 1
Spring 2012 -2012-03-10 - 4 - 1
我的查询结果应如下所示:
我正在尝试运行以下查询,以列出Lab Assistant及其每个学期的工作时间
SELECT DISTINCT(CONCAT(la.la_firstname, ' ', la.la_lastname)) AS 'Lab Assistant',
hw.hw_semester AS 'Semester Worked', hw.hw_hours AS 'Hours Worked'
FROM lab_assistant la
JOIN hours_worked hw
ON la.la_id = hw.la_id;
我的查询结果应如下所示:
Michael Jordan - Fall 2012 - 4
Horace Grant - Spring 2012 - 5
Dennis Rodman - Fall 2012 - 8
但是我得到的结果如下:
Michael Jordan - Fall 2012 - 4
Horace Grant - Fall 2012 - 5
Dennis Rodman - Spring 2012 - 4
所以从根本上讲,这只是在计算Dennis Rodman的正常工作时间。 应该是8,而不是4。
我可能不应该在这里使用DISTINCT,因为同一个人可能在不同的学期工作。
DISTINCT
需要在SELECT
的开头进行:
SELECT DISTINCT CONCAT(la.la_firstname, ' ', la.la_lastname) AS 'Lab Assistant',
hw.hw_semester, COUNT(w.hw_hours)
FROM lab_assistant la
JOIN hours_worked hw
ON la.la_id = hw.la_id;
但是,由于您正在使用聚合函数,因此您将需要使用GROUP BY
:
SELECT CONCAT(la.la_firstname, ' ', la.la_lastname) AS 'Lab Assistant',
hw.hw_semester, COUNT(w.hw_hours)
FROM lab_assistant la
JOIN hours_worked hw
ON la.la_id = hw.la_id
GROUP BY hw.hw_semester;
根据您的编辑,您需要以下内容:
SELECT CONCAT(la.la_firstname, ' ', la.la_lastname) AS 'Lab Assistant',
hw.hw_semester AS 'Semester Worked',
sum(hw.hw_hours) AS 'Hours Worked'
FROM lab_assistant la
JOIN hours_worked hw
ON la.la_id = hw.la_id
GROUP BY la.la_id, hw.hw_semester;
结果:
| LAB ASSISTANT | SEMESTER WORKED | HOURS WORKED |
---------------------------------------------------
| Dennis Rodman | Spring 2012 | 8 |
| Michael Jordan | Fall 2012 | 4 |
| Horace Grant | Fall 2012 | 5 |
这是因为DISTINCT
是SELECT
的修饰符,不能那样使用。 在您的情况下,最好的方法是在hw.hw_semester
上应用GROUP BY
进行修复。
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