Question

Empid   Deptid  Email
39  42  uni@mail.com
38  43  uni@mail.com
37  44  uni@mail.com
25  35  unique@maill.com
26  36  second@maill.com

如何找到唯一的记录？ empid - ＆gt; 25和26。

我尝试了以下查询。

select empid,deptid,row_number() over(PARTITION BY lower(email) Order By empid desc) from emp 
group by email,empid

Answer 1

这个怎么样：

select *
from emp
where Email =any (select Email from emp group by Email having count(*) = 1);

我还发现了没有子查询或自联接的版本。至少从理论上来说，它应该具备所有人的最佳表现。

SELECT 
    MAX(Empid) KEEP (DENSE_RANK FIRST ORDER BY NULL) AS Empid, 
    MAX(Deptid) KEEP (DENSE_RANK FIRST ORDER BY NULL) AS Deptid,
    Email
FROM emp
GROUP BY Email
HAVING COUNT(*) = 1;

Answer 2

您可以使用COUNT分析功能。

SQL Fiddle

Oracle 11g R2架构设置：

CREATE TABLE emp 
    (Empid int, Deptid int, Email varchar2(16))
;

INSERT ALL 
    INTO emp  (Empid, Deptid, Email)
         VALUES (39, 42, 'uni@mail.com')
    INTO emp  (Empid, Deptid, Email)
         VALUES (38, 43, 'uni@mail.com')
    INTO emp  (Empid, Deptid, Email)
         VALUES (37, 44, 'uni@mail.com')
    INTO emp  (Empid, Deptid, Email)
         VALUES (25, 35, 'unique@maill.com')
    INTO emp  (Empid, Deptid, Email)
         VALUES (26, 36, 'second@maill.com')
SELECT * FROM dual
;

查询1 ：

SELECT Empid, Deptid, Email
  FROM (SELECT e.*, COUNT (Empid) OVER (PARTITION BY Email) ct
          FROM emp e)
 WHERE ct = 1

<强> Results ：

| EMPID | DEPTID |            EMAIL |
|-------|--------|------------------|
|    26 |     36 | second@maill.com |
|    25 |     35 | unique@maill.com |

Answer 3

您可以使用not exists查询来执行此操作：

select e1.*
from emp e1
where not exists (select *  
                  from emp e2
                  where e2.empid <> e1.empid
                    and lower(e2.email) = lower(e1.email));

Answer 4

另一种选择：

select empid
from emp
where email in (select email 
                from emp
                group by email
                having count(*) = 1);

Answer 5

SELECT *
    FROM emp
    WHERE email IN (SELECT email FROM emp GROUP BY email HAVING COUNT(*)=1)

在sql中查找唯一的emai id记录？

5 个答案: