如果左连接返回任何行,更好的方法是返回'1'？

Question

我有三张桌子,'A','B'和'C'.我查询了'A'和'B',但是我想添加一个字段,告诉我是否有一个或多个(我不关心多少)'C'是外来键入'A'.

这就是我所拥有的:

SELECT    A.A_id, A.col_2, col_3, B.col_2, A.col_4
          count(C.id) as C_count
FROM      A
JOIN      B ON (A.B_id = B.B_id)
LEFT JOIN C ON (A.A_id = C.A_id)
WHERE     A.A_id = ?
GROUP BY  A.A_id, A.col_2, col_3, B.col_2, A.col_4
ORDER BY  CASE WHEN A.col_2 = ?
               THEN 0
               ELSE 1 
          END, col_3;

这似乎有点低效,因为我必须列出所有这些字段,GROUP BY并且因为我在计算我真正想要的地方是否至少有一个匹配.这可以改善吗？

Answer 1

使用子查询存在而不是......

 Select A.A_id, A.col_2, col_3, 
    B.col_2, A.col_4, 
    Case When Exists (Select * From C
                      Where A_id = A.A_id)
         Then 1 Else 0 End As C_Exists
 From A Join B 
     On (A.B_id = B.B_id) 
 Where A.A_id = ?    
 Order By Case When A.col_2 = ? 
           Then 0 Else 1 End, col_3;