技術標籤：mysqlmysql資料庫

Mysql練習

-- 12.22 求每個使用者相鄰兩次瀏覽時間之差小於三分鐘的次數

T1221
+---------+---------------------+
| user_id | times               |
+---------+---------------------+
|       1 | 2020-12-07 21:13:07 |
|       1 | 2020-12-07 21:15:26 |
|       1 | 2020-12-07 21:17:44 |
|       2 | 2020-12-13 21:14:06 |
 

|       2 | 2020-12-13 21:18:19 |
|       2 | 2020-12-13 21:20:36 |
|       3 | 2020-12-21 21:16:51 |
|       4 | 2020-12-16 22:22:08 |
|       4 | 2020-12-02 21:17:22 |
|       4 | 2020-12-30 15:15:44 |
|       4 | 2020-12-30 15:17:57 |
+---------+---------------------+
結果：
+---------+------+
| user_id | cnt  |
+---------+------+
 

|       1 |    2 |
|       2 |    1 |
|       3 |    0 |
|       4 |    1 |
+---------+------+

方法一: 
1. 首先兩張相同的表 按照相同的id 做左連線 找到符合條件的相鄰兩次時間小於三分鐘的
2. 對每個使用者進行分組 計算滿足條件的次數 
3. 為了讓user_id = 3的也顯示出來 再做一個左連線 如果是null 則返回0

select t4.user_id,IFNULL(t3.c,0) as cnt
from T1222 t4 left join (
  select t1.user_id,
 
count(t1.times)as c
	from t1222 t1 left join t1222 t2 
	on t1.user_id = t2.user_id
	where t1.times < t2.times and DATE_ADD(t1.times,INTERVAL 3 MINUTE) > t2.times
  group by t1.user_id
)t3
on t4.user_id = t3.user_id
group by t4.user_id;
------------------------------------------------------
select t1.user_id,sum(case when DATE_ADD(t1.times,INTERVAL 3 MINUTE) > t2.times 
					  and t1.times < t2.times
                      then 1 else 0 end) as cnt
from t1222 t1 left join t1222 t2 
on t1.user_id = t2.user_id
group by t1.user_id

方法二:
1. 先建立一個檢視 運用視窗函式 按照user分組 按照時間排序 產生排序數字rn
2. 對這兩個檢視做左連線 要求比較的時間是連續的 避免了方法1中全部連線
3. 計算兩個相鄰時間的差值 為cn
4. 統計差值小於3的個數

create view tt as
(select row_number() over(partition by user_id order by times) rn,user_id,times
from T1222 group by user_id,times)

-- mysql 的TIMESTAMPDIFF（MINUTE,starttime,endtime）函式獲得時間差，
-- 得到的可以是DAY/天，HOUR/小時，MINUTE/分鐘，SECOND/秒。
-- 其中starttime為時間小的那個時間，endtime為時間大的時間。

select c.user_id,sum(case when c.cn < 3 then 1 else 0 end) cnt
from (
	select a.user_id,
  ABS(TIMESTAMPDIFF(minute,IFNULL(a.times,'1970-01-01 00:00:00'),IFNULL(b.times,'1970-01-01 00:00:00'))) cn
  from tt as a left join tt as b 
  on a.rn = b.rn+1 and a.user_id = b.user_id
)c
group by c.user_id;