Question

在我的表中，我记录了各种信息，但是这个问题的关键字段是subject_id和lesson_time，我想按主题分组，给出每个主题的每一天的总行数和总行数对于每个主题（每周总数），所以从这个样本数据：

id    subject_id   lesson_time
1        4         2015-04-28
2        4         2015-04-28
3        3         2015-04-28
4        1         2015-04-28
5        4         2015-04-27

我想计算星期一到星期五和星期总数的每个主题的总数，因此上述数据示例的输出将为：

subject_id    monday_total   tuesday_total   wednesday_total  ... week_total
  1               0                1               0                  1
  3               0                1               0                  1
  4               1                2               0                  3

我可以通过主题轻松获得总数，因为它只是在分组后计数（*），我正在努力的是每个日期的计数，我当前（非工作）查询

SELECT
    subject_id,
    COUNT( DAYOFWEEK(lesson_time)=2) AS monday_total,
    COUNT( DAYOFWEEK(lesson_time)=3) AS tuesday_total,
    COUNT( DAYOFWEEK(lesson_time)=4) AS wednesday_total,
    COUNT( DAYOFWEEK(lesson_time)=5) AS thursday_total,
    COUNT( DAYOFWEEK(lesson_time)=6) AS friday_total,
    COUNT(*) AS week_total
FROM
    tbl_lessons
GROUP BY
    subject_id

非常感谢任何帮助。

Answer 1

试试这个

SELECT
    subject_id,
    SUM(DAYOFWEEK(lesson_time)=2) AS monday_total,
    SUM(DAYOFWEEK(lesson_time)=3) AS tuesday_total,
    SUM(DAYOFWEEK(lesson_time)=4) AS wednesday_total,
    SUM(DAYOFWEEK(lesson_time)=5) AS thursday_total,
    SUM(DAYOFWEEK(lesson_time)=6) AS friday_total,
    COUNT(*) AS week_total
FROM
    tbl_lessons
GROUP BY
    subject_id

在mysql中COUNT(n)计算每一行（n不是NULL的每一行都为+1）。 SUM(n)将所有n值相加（每行+ n）。

当布尔表达式返回1或0时，SUM(DAYOFWEEK(lesson_time)=2)将返回DAYOFWEEK(lesson_time)=2的行数（它就像汇总布尔值1 + 0 + 0 + 1 + 1 + 1 + 0 + 0 1 + ...）

Answer 2

SELECT date(lesson_time) AS lesson_time, count( * ) AS count FROM tbl_lessons GROUP BY date(lesson_time)

按星期几和每个科目的总周数计算总计

2 个答案: