如何找到同一组连续行之间的差异？

Question

这个问题在 Facebook 采访中被问到：

(1) 有两张桌子

• 出席人数：日期 | student_id | 出勤率
• 学生：student_id | school_id | Grade_level | 出生日期 | 家乡

使用这些数据，您能否回答以下问题：

• 从昨天到今天，哪个年级的出勤率下降幅度最大？

到目前为止，我已经得到了这个，但我被卡住了。 你能帮助我吗？ 谢谢。

(SELECT grade_level, date, COUNT(attendance) AS att FROM attendances AS A
     LEFT JOIN students AS S
     ON A.student_id = S.student_id
     GROUP BY grade_level, date
     HAVING attendance == "yes"
     AND (date == "02-06-2020" OR date == "02-05-2020")
     ORDER BY grade_level, date DESC) AS t

Answer 1

您可以加入表，进行条件聚合，然后排序和限制：

select
    s.grade_level
    sum(a.attendance * case when a.date = date('now') then 1 else -1 end) attendance_drop
from attendances a
inner join students s on s.student_id = a.student_id
where a.date in (date('now'), date('now', '-1 day'))
group by s.grade_level
order by attendance_drop desc
limit 1

Answer 2

请尝试使用下面的脚本认为它工作正常。

SELECT TOP 1 s.grade_level, COUNT(a.attendance) AS dropCount 
FROM students  AS s
INNER JOIN attendances AS a ON  a.student_id = s.student_id
WHERE 
    a.attendance = 'no'
    AND (a.date <= DATEADD(DAY,-1, CAST(GETDATE() AS DATE)) AND a.date >= CAST(GETDATE() AS DATE))
GROUP BY s.grade_level
ORDER BY dropCount DESC