SELECT student_id, section, count( * ) as total
FROM raw_data r
WHERE response = 1
GROUP BY student_id, section
测试共有4个部分,每个部分都有不同的问题。我想知道,对于每个学生和每个部分,他们正确回答了多少问题(响应= 1)。
但是,使用此查询,如果学生在给定部分中没有得到任何问题,那么我的结果集中将完全丢失该行。我怎样才能确保每个学生总共返回4行,即使一行的“总数”为0?
以下是我的结果集:
student_id section total
1 DAP--29 3
1 MEA--16 2
1 NNR--13 1 --> missing the 4th section for student #1
2 DAP--29 1
2 MEA--16 4
2 NNR--13 2 --> missing the 4th section for student #2
3 DAP--29 2
3 MEA--16 3
3 NNR--13 3 --> missing the 4th section for student #3
4 DAP--29 5
4 DAP--30 1
4 MEA--16 1
4 NNR--13 2 --> here, all 4 sections show up because student 4 got at least one question right in each section
感谢您的任何见解!
更新:我试过
SELECT student_id, section, if(count( * ) is null, 0, count( * )) as total
并没有改变结果。其他想法?
更新2:由于以下回复,我得到了它的工作:
SELECT student_id, section, SUM(CASE WHEN response = '1' THEN 1 ELSE 0 END ) AS total
FROM raw_data r
WHERE response = 1
GROUP BY student_id, section
答案 0 :(得分:9)
SELECT student_id, section, sum(case when response=1 then 1 else 0 end) as total
FROM raw_data_r GROUP BY student_id, section
请注意,没有WHERE
条件。
答案 1 :(得分:1)
SELECT r.student_id,
r.subject,
sum( r.response ) as total
FROM raw_data r
GROUP BY student_id, subject
答案 2 :(得分:0)
如果您有一个包含学生信息的单独表格,您可以从该表格中选择学生并将结果连接到data_raw
表格:
SELECT si.student_name, rd.student_id, rd.section, rd.count(*) AS total
FROM student_info AS si LEFT JOIN raw_data AS rd USING rd.student_id = si.student_id
这样,它首先选择所有学生,然后执行计数命令。