mysql SELECT COUNT(*)... GROUP BY ...不返回计数为零的行

时间:2009-09-25 16:47:17

标签: sql mysql

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

3 个答案:

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。

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