MySQL COUNT DISTINCT

Question

我想在昨天的cp中收集不同的訪問次數，然後數一下。

SELECT 
    DISTINCT `user_id` as user, 
    `site_id` as site, 
    `ts` as time 
FROM 
    `cp_visits` 
WHERE 
    ts >= DATE_SUB(NOW(), INTERVAL 1 DAY) 

由於某種原因，這是拉多個結果與相同的網站ID ....我怎麼只拉和計數獨特的site_id cp登錄？

Answer 1

Select 
    Count(Distinct user_id) As countUsers 
    , Count(site_id) As countVisits 
    , site_id As site 
From cp_visits 
Where ts >= DATE_SUB(NOW(), INTERVAL 1 DAY) 
Group By site_id 

Answer 2

您需要使用group by子句。

SELECT site_id, MAX(ts) as TIME, count(*) group by site_id 

Answer 3

總體

SELECT 
     COUNT(DISTINCT `site_id`) as distinct_sites 
    FROM `cp_visits` 
WHERE ts >= DATE_SUB(NOW(), INTERVAL 1 DAY) 

或站點

SELECT 
     `site_id` as site, 
     COUNT(DISTINCT `user_id`) as distinct_users_per_site 
    FROM `cp_visits` 
    WHERE ts >= DATE_SUB(NOW(), INTERVAL 1 DAY) 
GROUP BY `site_id` 

有結果中的time列沒有意義 - 因爲你是聚集行，表現出一個特別是time是無關緊要的，除非它是min或你在追求210。