Question

我正在尝试获取用户第一个视图的日期和IP地址列表。这个2天的清单（实际数据中会有更多的ip）：

"day"           "ip"
"2014-07-01"    "173.8.118.153"
"2014-07-01"    "173.8.118.153"
"2014-07-01"    "173.8.118.153"
"2014-07-02"    "173.8.118.153"
"2014-07-02"    "173.8.118.153"
"2014-07-02"    "173.8.118.153"
"2014-07-02"    "173.8.118.153"
"2014-07-02"    "173.8.118.153"

我想在这个用户的第一天返回2014-07-01。如何在第一天获取所有用户的列表？

我在这里的查询有效，但我知道这不正确，因为如果我将order by从asc更改为desc，反之亦然，我仍然会在同一天。基本上这告诉我它正在为这个ip找到它的第一天，并且谁知道数据可能在某些用户的数据库中倒退。我该怎么做才能保证我第一次来访？

select date(viewed) as day, inet_ntoa(ip) as user_ip
from ad_views
where ad_id = 3058440
and viewed > date_sub(now(), interval 30 day)
group by user_ip order by day;

Answer 1

Select min(date_col) from table_name

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Answer 2

获取日期MIN()：

select MIN(date(viewed)) as day, inet_ntoa(ip) as user_ip
from ad_views
where ad_id = 3058440
group by user_ip;

Answer 3

由于您已经进行了分组，请使用min

的viewed

select date(min(viewed)) as day, inet_ntoa(ip) as user_ip
from ad_views
where ad_id = 3058440
and viewed > date_sub(now(), interval 30 day)
group by user_ip order by day;

Answer 4

SELECT day, ip
FROM MyTable t1 LEFT OUTER JOIN MyTable t2
    ON (t1.ip = t2.ip AND t1.day <= t2.day)

让所有用户首次访问日期

4 个答案: