如何获得前20个SQL中列的百分比？

Question

I am trying to find percentage of pageType is having rating lesser than 20.

My Table looks like :

requestId   rank    pageType    
MMCVS       0       HOME        
MMCVS       1       MOBILE      
MMCVS       2       HOME 
BBVSS       0       HOME        
BBVSS       11      MOBILE      
BBVSS       12      HOME 

so far, I tried :

 select pageType, Top20, Top20/count(DISTINCT requestId) as Percentage
    from (
     SELECT 
        pageType, requestId,
        SUM(CASE WHEN rank <= 20 THEN 1 ELSE 0 END) as Top20  
    FROM 
        tempTable 
    group by pageType, requestId) tempTable group by pageType

But getting error :

 expression 'Top20' is neither present in the group by, nor is it an aggregate function. 

Answer 1

Yes you need to add any non-aggregated fields into group by fields list. Though MySQL won't give you an error even if you don't.

For this matter, you either add Top20 into group by, or remove Top20 from select:

SELECT pageType, Top20/count(DISTINCT requestId) as Percentage
FROM (
  SELECT pageType, requestId,
    SUM(CASE WHEN rank <= 20 THEN 1 ELSE 0 END) as Top20  
  FROM tempTable 
  GROUP BY pageType, requestId
) tempTable
GROUP BY pageType

http://sqlfiddle.com/#!9/e7eb89/4