问题描述
所以如果我们有下表:
runner ran
Carol 2011-02-01
Alice 2011-02-01
Bob 2011-02-01
Carol 2011-02-02
Bob 2011-02-02
Bob 2011-02-03
Bob 2011-02-04
Carol 2011-02-07
Alice 2011-02-08
我如何编写查询(没有任何子查询)来查找每个跑步者在两次运行之间必须等待的平均天数(即 Carol 等了 1 天，然后 5 天，所以平均值是 3；Bob 每天跑步；Alice 等了7 天)?
How can I write a query (without any subquery) to find the average number of days each runner has to wait between runs (i.e, Carol waited 1 day, then 5, so average is 3; Bob runs everyday; Alice waited 7 days)?
我正在考虑对表本身进行连接，然后找到每个跑步者的最大值和最小值，减去它们并除以运行次数 - 1.但是我如何在没有任何子查询的情况下组合所有这些?
I was thinking about a join on the table itself, then finding the max and min for each runner, subtracting them and dividing by the number of runs - 1. But how do I combine all these without any subquery?
推荐答案
Sorin，公平地说，你已经有了答案 - (max-min)/(count-1) 确实是正确的无需详细说明运行相距多远.
Sorin, to be fair, you already have the answer - (max-min)/(count-1) is indeed correct without going into the specifics of how far apart the runs are.
select runner, datediff(max(ran),min(ran)) / (count(ran)-1)
from running
group by runner;
注意:MySQL 会将 X/0(对于一个跑步者只有一条记录的情况)变成 NULL，因为它不能被 0 整除.
Note: MySQL will turn X / 0 (for where there is only one record for a runner) into NULL because it is indivisable by 0.
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