Question

所以我想做下面的SQL代码：

select s.id, s.name,s.city 
from stuff s
group by s.name having count(where city and name are identical) > 1

要生成以下内容，（但忽略只有名称或仅匹配城市的位置，它必须位于两列上）：

id      name  city   
904834  jim   London  
904835  jim   London  
90145   Fred  Paris   
90132   Fred  Paris
90133   Fred  Paris

Answer 1

对id和name对<{1}}重复city。

select s.id, t.* 
from [stuff] s
join (
    select name, city, count(*) as qty
    from [stuff]
    group by name, city
    having count(*) > 1
) t on s.name = t.name and s.city = t.city

Answer 2

 SELECT name, city, count(*) as qty 
 FROM stuff 
 GROUP BY name, city HAVING count(*)> 1

Answer 3

这样的事情可以解决问题。不知道性能，所以做一些测试。

select
  id, name, city
from
  [stuff] s
where
1 < (select count(*) from [stuff] i where i.city = s.city and i.name = s.name)

Answer 4

使用count(*) over(partition by...)提供了一种简单有效的方法来查找不必要的重复，同时还列出了所有受影响的行和所有需要的列：

SELECT
    t.*
FROM (
    SELECT
        s.*
      , COUNT(*) OVER (PARTITION BY s.name, s.city) AS qty
    FROM stuff s
    ) t
WHERE t.qty > 1
ORDER BY t.name, t.city

最新的RDBMS版本支持count(*) over(partition by...) MySQL V 8.0引入的“窗口函数”，如下所示（在MySQL 8.0中）

CREATE TABLE stuff(
   id   INTEGER  NOT NULL
  ,name VARCHAR(60) NOT NULL
  ,city VARCHAR(60) NOT NULL
);

INSERT INTO stuff(id,name,city) VALUES 
  (904834,'jim','London')
, (904835,'jim','London')
, (90145,'Fred','Paris')
, (90132,'Fred','Paris')
, (90133,'Fred','Paris')

, (923457,'Barney','New York') # not expected in result
;

SELECT
    t.*
FROM (
    SELECT
        s.*
      , COUNT(*) OVER (PARTITION BY s.name, s.city) AS qty
    FROM stuff s
    ) t
WHERE t.qty > 1
ORDER BY t.name, t.city

    id | name | city   | qty
-----: | :--- | :----- | --:
 90145 | Fred | Paris  |   3
 90132 | Fred | Paris  |   3
 90133 | Fred | Paris  |   3
904834 | jim  | London |   2
904835 | jim  | London |   2

db <>提琴here

窗口函数。 MySQL现在支持窗口函数，对于查询中的每一行，都使用与那排。这些包括诸如RANK（），LAG（）和NTILE（）之类的函数。此外，现在可以将几个现有的聚合函数用作窗口功能；例如SUM（）和AVG（）。了解更多信息，参见Section 12.21, “Window Functions”。

Answer 5

你必须自己加入东西并匹配名称和城市。然后按计数分组。

select 
   s.id, s.name, s.city 
from stuff s join stuff p ON (
   s.name = p.city OR s.city = p.name
)
group by s.name having count(s.name) > 1

Answer 6

给定一个包含70列的登台表，只有4列表示重复，此代码将返回有问题的列：

SELECT 
    COUNT(*)
    ,LTRIM(RTRIM(S.TransactionDate)) 
    ,LTRIM(RTRIM(S.TransactionTime))
    ,LTRIM(RTRIM(S.TransactionTicketNumber)) 
    ,LTRIM(RTRIM(GrossCost)) 
FROM Staging.dbo.Stage S
GROUP BY 
    LTRIM(RTRIM(S.TransactionDate)) 
    ,LTRIM(RTRIM(S.TransactionTime))
    ,LTRIM(RTRIM(S.TransactionTicketNumber)) 
    ,LTRIM(RTRIM(GrossCost)) 
HAVING COUNT(*) > 1

Answer 7

这篇文章对游戏来说有点晚了，但是我发现这种方式非常灵活/高效

select 
    s1.id
    ,s1.name
    ,s1.city 
from 
    stuff s1
    ,stuff s2
Where
    s1.id <> s2.id
    and s1.name = s2.name
    and s1.city = s2.city

Answer 8

SELECT Feild1, Feild2, COUNT(*)
FROM table name
GROUP BY Feild1, Feild2
HAVING COUNT(*)>1

这将为您提供所有答案。

如何在多列中找到重复项？

8 个答案: