獲取所有用戶的首次訪問日期

Question

我正在嘗試獲取用戶首次查看的日期和ip地址列表。取2天這個名單（會有更多的IP的實際數據）：

"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" 

我想返回2014-07-01該用戶的第一天。我能做些什麼來獲得所有用戶第一天的清單？

查詢我這裏工作，但我知道這是不正確的，因爲如果我改變order by從asc到desc，反之亦然，我仍然得到同樣的一天。基本上，這告訴我它正在獲得該ip的第一天，並且誰知道某些用戶的數據可能在數據庫中倒退。我能做些什麼來保證我獲得他們的第一次訪問日期？

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

獲取最新的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 2

Select min(date_col) from table_name 

This文件可以證明是有益的搜索

Answer 3

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

Answer 4

既然你已經編組，採取viewed的min

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;