Question

我试图找到pageType的百分比评级小于20。

我的表格如下：

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

到目前为止，我试过了：

 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

但是得到错误：

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

Answer 1

是的，您需要将任何非聚合字段添加到逐个字段列表中。虽然MySQL不会给你一个错误，即使你没有。

就此而言，您要么将Top20添加到group by中，要么从select中删除Top20：

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

如何在前20个SQL中获得列的百分比？

1 个答案: