如果没有我向您展示表格,您是否有机会帮助优化此查询?
我的所有这些查询的原始表格都包含以下列,该表名为laterec-students
--------------------------------------------------------------
| studentid | name | class | latetime | waived |
--------------------------------------------------------------
| ID1111STU | Stu 1 | 1A |2012-01-09 08:09:00 |Waived |
SELECT A.class, NoStudentsLate, 1xLATE, 2xLATE FROM (
SELECT
class,
count(DISTINCT studentid) AS NoStudentsLate
FROM `laterec-students`
WHERE waived!="Waived"
GROUP BY class
) AS A
LEFT JOIN (
SELECT class, count(distinct studentid) AS 1xLATE from (
SELECT `laterec-students`.class, `laterec-students`.studentid
FROM `laterec-students`
WHERE waived!="Waived"
GROUP BY studentid
HAVING count(studentid)=1) as temp
GROUP BY class
) AS B ON A.class=B.class
LEFT JOIN (
SELECT class, count(distinct studentid) AS 2xLATE from (
SELECT `laterec-students`.class, `laterec-students`.studentid
FROM `laterec-students`
WHERE waived!="Waived"
GROUP BY studentid
HAVING count(studentid)=2) as temp
GROUP BY class
) AS C ON A.class=C.class
这就是我想要完成的事情
---------------------------------------------------------------------
| Class | Total # of students late | # late 1 times | # late 2 times |
---------------------------------------------------------------------
| 1A | 5 | 3 | 2 |
| 1B | 3 | 3 | 0 |
---------------------------------------------------------------------
所以这意味着,在1A级,总共有5名学生迟到,使用学生ID计算。在这5名学生中,有3名学生迟到了一次,2名学生迟到了两次。
同样在1B班,共有3名学生迟到,而且所有学生都只迟到一次。
答案 0 :(得分:3)
我希望我理解您的查询,但以下内容适用于我的SQL Fiddle example。
SELECT
class,
SUM(cnt > 0) AS NoStudentsLate,
SUM(cnt = 1) AS 1xLate,
SUM(cnt = 2) AS 2xLate
FROM
(
SELECT class, studentid, COUNT(*) AS cnt
FROM `laterec-students`
WHERE waived!='Waived'
GROUP BY class, studentid
) t
GROUP BY class;