LeetCode（数据库）- 好友申请l：总体通过率

题目链接：​​点击打开链接​​

题目大意：略。

解题思路：1/0, 0/0 结果都为 NULL。

AC 代码

-- 解决方案(1)
WITH t1 AS(SELECT 1 id, COUNT(DISTINCT requester_id, accepter_id) up FROM RequestAccepted),
t2 AS(SELECT 1 id, COUNT(DISTINCT sender_id, send_to_id) down FROM FriendRequest)

SELECT IF(down = 0, 0.00, ROUND(up/down, 2)) accept_rate
FROM t1, t2
WHERE t1.id = t2.id

-- 解决方案(2)
select
round(
    ifnull(
    (select count(*) from (select distinct requester_id, accepter_id from request_accepted) as A)
    /
    (select count(*) from (select distinct sender_id, send_to_id from friend_request) as B),
    0)
, 2) as accept_rate;