题目链接：点击打开链接
题目大意：略。
解题思路：略。
AC 代码
-- 解决方案(1)
WITH t1 AS(SELECt '[0-5>' bin
UNIOn ALL
SELECT '[5-10>' bin
UNIOn ALL
SELECT '[10-15>' bin
UNIOn ALL
SELECT '15 or more' bin),
t2 AS(SELECT CASE
WHEN duration BETWEEN 0 AND 299 THEN '[0-5>'
WHEN duration BETWEEN 300 AND 599 THEN '[5-10>'
WHEN duration BETWEEN 600 AND 899 THEN '[10-15>'
ELSE '15 or more'
END bin
FROM Sessions),
t3 AS(SELECt bin, COUNT(*) total
FROM t2
GROUP BY bin)
SELECt t1.bin, IFNULL(total, 0) total
FROM t1 LEFT JOIN t3 ON t3.bin = t1.bin
-- 解决方案(2)
select '[0-5>' as bin, count(*) as total from Sessions where duration/60>=0 and duration/60<5
union
select '[5-10>' as bin, count(*) as total from Sessions where duration/60>=5 and duration/60<10
union
select '[10-15>' as bin, count(*) as total from Sessions where duration/60>=10 and duration/60<15
union
select '15 or more'as bin, count(*) as total from Sessions where duration/60>=15
