Oracle SQL 包含要显示的列而不将其添加到 GROUP BY

Question

我正在尝试在用户表上查找所有案例变体重复项：

SELECT LOWER(EMAIL), COUNT(EMAIL) 
    FROM USERS
    GROUP BY LOWER(EMAIL)
    HAVING COUNT (LOWER(EMAIL)) >= 3;

结果类似于：

Emails                   Count (number of duplicates)
bob@example.com          3   
john.smith@example.com   3 
blah@example.com         4 
james.smith@example.com  3 

问题是我需要每封电子邮件的 ID，我该如何实现？由于 GROUP BY，我不能简单地将其添加到 SELECT 语句中：

SELECT **ID**, LOWER(EMAIL), COUNT(EMAIL) 
    FROM USERS
    GROUP BY **ID**, LOWER(EMAIL)
    HAVING COUNT (LOWER(EMAIL)) >= 3;

上面会寻找重复的电子邮件和 ID，这不是我需要的。

Answer 1

您可以使用analytic count() 在内联视图中：

SELECT ID, EMAIL, LOWER(EMAIL), HOW_MANY
FROM (
  SELECT ID, EMAIL, COUNT(*) OVER (PARTITION BY LOWER(EMAIL)) AS HOW_MANY
  FROM USERS
)
WHERE HOW_MANY >= 3
ORDER BY ID;

        ID EMAIL                          LOWER(EMAIL)                     HOW_MANY
---------- ------------------------------ ------------------------------ ----------
         1 bob@example.com                bob@example.com                         3 
         2 Bob@example.com                bob@example.com                         3 
         3 BOB@example.com                bob@example.com                         3 
         4 john.smith@example.com         john.smith@example.com                  3 
         5 John.smith@example.com         john.smith@example.com                  3 
         6 JOHN.smith@example.com         john.smith@example.com                  3 
         7 blah@example.com               blah@example.com                        4 
         8 BLAH@example.com               blah@example.com                        4 
         9 blAH@example.com               blah@example.com                        4 
        10 BLah@example.com               blah@example.com                        4 
        11 james.smith@example.com        james.smith@example.com                 3 
        12 James.smith@example.com        james.smith@example.com                 3 
        13 JAMES.smith@example.com        james.smith@example.com                 3 

SQL Fiddle。分析的好处之一是它只需要提交一次。

Answer 2

试试这个：

SELECT U.*, COUNT(LOWER(EMAIL)) OVER (PARTITION BY (LOWER(EMAIL))) 
    FROM USERS U WHERE LOWER(EMAIL) IN (SELECT LOWER(EMAIL)
    FROM USERS
    GROUP BY LOWER(EMAIL)
    HAVING COUNT (LOWER(EMAIL)) >= 3);