আমি একাধিক কলামে নকলগুলি কীভাবে খুঁজে পাব?

Question

সুতরাং আমি নীচে এই এসকিএল কোড এর মতো কিছু করতে চাই:

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

To produce the following, (but ignore where only name or only city match, it has to be on both columns):

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

Answer 1

Duplicated id for pairs name and 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

Comments

Note that if either name or city contain null, then they will fail to be reported in the outer query, but will be matched in the inner query.

— Adam Parkin

If the values can possibly contain null then (unless I'm missing something) you need to change it to a CROSS JOIN (full Cartesian product) and then add a WHERE clause such as: WHERE ((s.name = t.name) OR (s.name is null and t.name is null)) AND ((s.city = t.city) OR (s.city is null and t.city is null))

— Adam Parkin

Answer 2

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

Comments

The cleanest answer.

— Repcak

With that, you can't know the id of each line.

— Juan.Queiroz

Answer 3

Something like this will do the trick. Don't know about performance, so do make some tests.

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

Using count(*) over(partition by...) provides a simple and efficient means to locate unwanted repetition, whilst also list all affected rows and all wanted columns:

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

While most recent RDBMS versions support count(*) over(partition by...) MySQL V 8.0 introduced "window functions", as seen below (in MySQL 8.0)

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<>fiddle here

Window functions. MySQL now supports window functions that, for each row from a query, perform a calculation using rows related to that row. These include functions such as RANK(), LAG(), and NTILE(). In addition, several existing aggregate functions now can be used as window functions; for example, SUM() and AVG(). For more information, see Section 12.21, “Window Functions”.

Answer 5

A little late to the game on this post, but I found this way to be pretty flexible / efficient

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 6

You have to self join stuff and match name and city. Then group by count.

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

Comments

Fails in SQL Server: all non-aggregate columns must be in the GROUP BY

— gbn

Answer 7

Given a staging table with 70 columns and only 4 representing duplicates, this code will return the offending columns:

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

.