從每個領域獲得總價值？

Question

如何獲得每個用戶名的是，否，其他字段的總值？

我喜歡加總字段。

SELECT Username, 
    SUM(CASE WHEN type = 'Yes' THEN 1 ELSE NULL END) as Yes, 
    SUM(CASE WHEN type = 'No' THEN 1 ELSE NULL END) as No, 
    SUM(CASE WHEN type = '' THEN 1 ELSE NULL END) as Other 
    //How to get total of Yes/No/Other 
    FROM table 
WHERE source = 'CompanyName' "; 

也是最高的總數在最高的順序。使用

Answer 1

0，而不是NULL，添加缺少的group by和使用COUNT(*)得到總的各組並責令結果：

SELECT Username, 
    SUM(CASE WHEN type = 'Yes' THEN 1 ELSE 0 END) as Yes, 
    SUM(CASE WHEN type = 'No' THEN 1 ELSE 0 END) as No, 
    SUM(CASE WHEN type = '' THEN 1 ELSE 0 END) as Other, 
    COUNT(*) as TOTAL 
    FROM table 
WHERE source = 'CompanyName' 
group by Username 
order by TOTAL desc; 

這假定type只能是「是」，「否」要麼 ''。

Answer 2

使用子查詢來總結你的結果，並添加排序：

select yes, no, other, yes + no + other as Total 
from (

    SELECT Username, 
     SUM(CASE WHEN type = 'Yes' THEN 1 ELSE 0 END) as Yes, 
     SUM(CASE WHEN type = 'No' THEN 1 ELSE 0 END) as No, 
     SUM(CASE WHEN type = '' THEN 1 ELSE 0 END) as Other 
     FROM table 
    WHERE source = 'CompanyName' 
) 
order by (yes + no + other) desc 

Answer 3

不要使用SUM（空）的SUM（1,1,1，空）= NULL，而不是3.

SELECT s.*, s.yes+s.no+s.other as all FROM (
    SELECT Username, 
    SUM(CASE WHEN type = 'Yes' THEN 1 ELSE 0 END) as Yes, 
    SUM(CASE WHEN type = 'No' THEN 1 ELSE 0 END) as No, 
    SUM(CASE WHEN type = '' THEN 1 ELSE 0 END) as Other 
    FROM table 
    WHERE source = 'CompanyName' 
    GROUP BY Username 
) s 
ORDER BY all DESC