显示具有重复某些列的所有行

Question

假设我有以下 sql 表

    objid  firstname lastname active
     1       test      test     0
     2       test      test     1
     3       test1     test1    1
     4       test2     test2    0
     5       test2     test2    0
     6       test3     test3    1

现在，我感兴趣的结果如下：

     objid  firstname lastname active
     1       test      test     0
     2       test      test     1
     4       test2     test2    0
     5       test2     test2    0

我怎样才能做到这一点？ 我尝试了以下查询，

select firstname,lastname from table
group by firstname,lastname
having count(*) > 1

但是这个查询给出的结果类似于

    firstname  lastname
     test        test
     test2       test2

Answer 1

您已找到重复记录，但您有兴趣获取附加到它们的所有信息。您需要 join 将您的副本发送到您的主表以获取该信息。

select *
  from my_table a
  join ( select firstname, lastname 
           from my_table 
          group by firstname, lastname 
         having count(*) > 1 ) b
    on a.firstname = b.firstname
   and a.lastname = b.lastname

这与inner join 相同，意味着对于您的子查询中的每条记录，您可以从主表中找到具有相同 firstseen 和 lastseen 组合的所有记录。

您也可以使用though you should test the difference：

select *
  from my_table a
 where ( firstname, lastname ) in   
       ( select firstname, lastname 
           from my_table 
          group by firstname, lastname 
         having count(*) > 1 )

延伸阅读：

• A visual representation of joins 来自 Coding Horror
• Join explanation 来自维基百科

Answer 2

SELECT DISTINCT t1.*
FROM myTable AS t1
INNER JOIN myTable AS t2
  ON t1.firstname = t2.firstname
  AND t1.lastname = t2.lastname
  AND t1.objid <> t2.objid

这将根据firstname 和lastname 输出具有重复项的每一行。

Answer 3

这里有一个更清晰的方式来做 Ben 的第一个答案：

WITH duplicates AS (
   select    firstname, lastname
   from      my_table
   group by  firstname, lastname
   having    count(*) > 1
)
SELECT    a.*
FROM      my_table   a
JOIN      duplicates b ON (a.firstname = b.firstname and a.lastname = b.lastname)

Answer 4

SELECT user_name,email_ID 
FROM User_Master WHERE 
email_ID 
in (SELECT email_ID 
FROM User_Master GROUP BY 
email_ID HAVING COUNT(*)>1) 

Answer 5

不错的选项从表中获取所有重复值

 select * from Employee where Name in (select Name from Employee group by Name having COUNT(*)>1)

Answer 6

这是最简单的方法：

SELECT * FROM yourtable a WHERE EXISTS (SELECT * FROM yourtable b WHERE a.firstname = b.firstname AND a.secondname = b.secondname AND a.objid <> b.objid)

Answer 7

如果要打印表中所有重复的 ID：

select * from table where id in (select id from table group By id having count(id)>1)

Answer 8

我很惊讶使用 Window function 没有答案。我刚刚遇到了这个用例，这对我有帮助。

select t.objid, t.firstname, t.lastname, t.active
from
(
select t.*, count(*) over (partition by firstname, lastname) as cnt
from my_table t
) t
where t.cnt > 1;

小提琴 - https://dbfiddle.uk/?rdbms=sqlserver_2017&fiddle=c0cc3b679df63c4d7d632cbb83a9ef13

---

格式如下

select
    tbl.relevantColumns
from
(
    select t.*, count(*) over (partition by key_columns) as cnt
    from desiredTable t
) as tbl
where tbl.cnt > 1;

此格式从表中选择您需要的任何列（有时是所有列），其中count > 1 用于标识重复行。 key_columns 可以是任意数量的列。

Answer 9

这个答案可能不是很好，但我认为它很容易理解。

SELECT * FROM table1 WHERE (firstname, lastname) IN ( SELECT firstname, lastname FROM table1 GROUP BY firstname, lastname having count() > 1);

Answer 10

此查询返回重复项

SELECT * FROM (
  SELECT  a.* 
    FROM table a 
    WHERE (`firstname`,`lastname`) IN (
        SELECT `firstname`,`lastname` FROM table 
        GROUP BY `firstname`,`lastname` HAVING COUNT(*)>1       
        )  
    )z WHERE z.`objid` NOT IN (
        SELECT MIN(`objid`) FROM table 
        GROUP BY `firstname`,`lastname` HAVING COUNT(*)>1
        )                                         

Answer 11

请尝试

WITH cteTemp AS (
  SELECT EmployeeID, JoinDT,
     row_number() OVER(PARTITION BY EmployeeID, JoinDT ORDER BY EmployeeID) AS [RowFound]
  FROM dbo.Employee 
)
SELECT * FROM cteTemp WHERE [RowFound] > 1 ORDER BY JoinDT