how to select rows with same column_a but different column_b? - sql

Question

I want to select rows in sql server, there's my questions below:
Table1
--------------------------
| Name | Type |
--------------------------
| A | 1 |
| A | 2 |
| B | 1 |
| B | 3 |
| A | 3 |
| C | 1 |
| C | 3 |
| D | 1 |
| D | 2 |
| D | 3 |
| . | . |
| . | . |
Select rows like below:
Table2
--------------------------
| Name | Type |
--------------------------
| A | 1 |
| A | 2 |
| A | 3 |
| D | 1 |
| D | 2 |
| D | 3 |
| . | . |
| . | . |
The select rules is...
Show Name and Type which Type must have 1,2 and 3.
Example: A had 1,2,3 types,so i would select it.
Example: B only has 1,2 types,so i wouldn't select it.

Answer 1

You can use window functions for this:
select name, type
from (
select
t.*,
sum(case when type in (1, 2, 3) then 1 else 0 end)
over(partition by name) cnt
from mytable t
) t
where cnt = 3
This assumes that each (name, type) tuple occurs only once in the original table, which is consistant with your sample data.
Demo on DB Fiddle:
name | type
:--- | ---:
A | 1
A | 2
A | 3
D | 1
D | 2
D | 3

Answer 2

You could use INNER JOINs on the three Type columns to achieve this:
SELECT Table1.[Name],
Table1.[Type]
FROM Table1
INNER JOIN (
SELECT [Name]
FROM Table1
WHERE ([Type] = 1)
) A ON A.[Name] = Table1.[Name]
INNER JOIN (
SELECT [Name]
FROM Table1
WHERE ([Type] = 2)
) B ON B.[Name] = A.[Name]
INNER JOIN (
SELECT [Name]
FROM Table1
WHERE ([Type] = 3)
) C ON C.[Name] = A.[Name]
This outputs:
Name Type
A 1
A 2
A 3
D 1
D 2
D 3
The matching sqlfiddle.
This works by returning rows that contain [Type] = 1, and then ONLY matching rows where [Type] = 2 and [Type] = 3. Then this is joined back to your main table and the results are returned.

Answer 3

Get the names with group by name and set the condition in the having clause:
select * from Table1
where name in (
select name
from Table1
group by name
having count(distinct type) = 3
)
If there are for the column Type other values than 1, 2, 3 then:
select * from Table1
where type in (1, 2, 3) and name in (
select name
from Table1
where type in (1, 2, 3)
group by name
having count(distinct type) = 3
)
See the demo.
Results:
> Name | Type
> :--- | ---:
> A | 1
> A | 2
> A | 3
> D | 1
> D | 2
> D | 3

Answer 4

you can use string_agg if it is sql server 2017 and above or Azure SQL as below:
Select * from #yourTable yt join (
select [name], string_agg([Type], ',') as st_types
from #YourTable
group by [name] ) a
on yt.name = a.[name] and a.st_types like '%1,2,3%'

Answer 5

I give you this, this will work if you have:
A 1
A 2
A 3
A 2
It will then only give you B.
SELECT *
FROM Table1
WHERE Name in (
SELECT Name from
(
SELECT Name, Type, count(Name) c from Table1 where Type = 1
GROUP BY Name, Type
HAVING count(Name) = 1
UNION
SELECT Name, Type, count(Name) c from Table1 where Type = 2
GROUP by Name, Type
HAVING count(Name) = 1
UNION
SELECT Name, Type, count(Name) c from Table1 where Type = 3
GROUP by Name, Type
HAVING count(Name) = 1) t
GROUP by name
HAVING count(c) = 3)
Here is the DEMO