同一列上的多個計數與哪里 - MySQL

Question

我試圖在同一列上對不同的where子句進行多次計數。

我已經找出了每個like子句的基本查詢，但是我需要將它們組合起來產生一個結果。

select system_user, COUNT(details) from asset_log where details like 'Viewed';
select system_user, COUNT(details) from asset_log where details like 'Viewed Web%';
select system_user, COUNT(details) from asset_log where details like 'ThumbView';
select system_user, COUNT(details) from asset_log where details like 'Exported%';

我確信它可能，我只是不知道該怎么做。 任何幫助，將不勝感激。

提前致謝

更新：

結束了為我工作

select distinct system_user, 
    SUM(CASE WHEN details ='viewed' then 1 Else 0 end) AS viewed_count,
    SUM(CASE WHEN details Like 'Viewed Web%' then 1 Else 0 end) AS Web_count,
    SUM(CASE WHEN details = 'ThumbView' then 1 Else 0 end) AS ThumbView_count,
    SUM(CASE WHEN details Like 'Exported%' then 1 Else 0 end) AS Exported_count
from asset_log GROUP BY system_user; 

謝謝！

Answer 1

您可以使用SUM / Case來“轉動”計數

select system_user, 

    SUM(CASE WHEN details ='viewed' then 1 Else 0 end) viewed_count
    SUM(CASE WHEN details Like 'Viewed Web%' then 1 Else 0 end) Viewed Web_count
    SUM(CASE WHEN details = 'ThumbView' then 1 Else 0 end) ThumbView_count
    SUM(CASE WHEN details Like 'Exported%' then 1 Else 0 end) Exported_count
from asset_log 
where 
    details = 'viewed' or
    details like 'Viewed Web%' or
    details = 'ThumbView' or
    details like 'Exported%' 
 group by 
      system_user

注意：如果沒有外卡，我不會費心使用“贊”

Answer 2

您可以使用SUM / CASE的組合來實現所需的結果，如答案https://stackoverflow.com/a/8870028/625594 。

對於您的情況，確切的查詢將是：

select system_user, 
SUM(CASE WHEN details like 'Viewed' THEN 1 ELSE 0) AS `sum1`,  
SUM(CASE WHEN details like 'Viewed Web%' THEN 1 ELSE 0) AS `sum2`,
SUM(CASE WHEN details like 'ThumbView' THEN 1 ELSE 0) AS `sum3`,
SUM(CASE WHEN details like 'Exported%' THEN 1 ELSE 0) AS `sum4`
from asset_log;

Answer 3

你可以像這樣寫：

SELECT MAX(system_user) AS system_user, COUNT(*)
  FROM asset_log
  WHERE details = 'Viewed'
  OR details like 'Viewed Web%'
  OR details = 'ThumbView'
  OR details like 'Exported%'

由於system_user可能都是相同的，因此MAX()將隨機獲得一個，並且它解決了無法將聚合列與非聚合列組合的問題。