Question

我需要它来返回每个城市每次更改的最大日期，所以基本上是他们每次访问城市的最后一天

我已尝试SELECT userid, city, Max(date) from table group by userid, city，但它将单独的住宿与同一个城市合并为一个。

样本表

userid  city       date
------  ---------  --------------
Francis Cambridge  06/15/2017
Francis Cambridge  07/01/2017
Francis London     07/02/2017
Francis Cambridge  07/03/2017
Francis NewYork    07/05/2017
Francis NewYork    07/20/2017
Francis NewYork    07/25/2017
Francis London     07/31/2017

所需的查询结果

userid  city       date
------  ---------  --------------
Francis Cambridge  07/01/2017
Francis London     07/02/2017
Francis Cambridge  07/03/2017
Francis New York   07/25/2017
Francis London     07/31/2017

Answer 1

使用Array ( [0] => Array ( [product_id] => 1 ) [1] => Array ( [product_id] => 3 ) [2] => Array ( [product_id] => 4 ) [3] => Array ( [product_id] => 5 ) )将上一行的用户ID和城市与当前行进行比较，并使用运行总和根据指定的条件分配组。然后你需要一个lag来获取最新的日期。

group by

Answer 2

所以，你想要留下没有后来留在给定城市的住宿

调用表stay_log：

select userid, city, date from stay_log last_date LEFT JOIN stay_log next_date ON last_date.userid = next_date.userid and last_date.city = next_date.city and DateAdd(last_date, "1", 1) = next_date.date WHERE next_date.userid is null

每次出现字段时返回最大日期

2 个答案: