id address retailer
1 A 11
2 A 11
3 A 11
4 A 12
5 A 13
6 B 12
7 B 12
8 B 13
My output should be
id address retailer
1 A 11
4 A 12
5 A 13
6 B 12
8 B 13
i.e my query should return id's which have same address but not same retailer.
How toget this?
Try to use group by clause as below:
select min(id), address, retailer
from tab
group by address, retailer
Assuming you're joining on columns with no duplicates, which is by far the most common case:
An inner join of A and B gives the result of A intersect B, i.e. the inner part of a venn diagram intersection.
An outer join of A and B gives the results of A union B, i.e. the outer parts of a venn diagram union.
Suppose you have two Tables, with a single column each, and data as follows:
A B
- -
1 3
2 4
3 5
4 6
Note that (1,2) are unique to A, (3,4) are common, and (5,6) are unique to B.
Inner join:
An inner join using either of the equivalent queries gives the intersection of the two tables, i.e. the two rows they have in common.
select *
from a
INNER JOIN b on a.a = b.b;
select a.*,b.*
from a,b
where a.a = b.b;
a | b
--+--
3 | 3
4 | 4
Left outer join:
A left outer join will give all rows in A, plus any common rows in B.
select *
from a
LEFT OUTER JOIN b on a.a = b.b;
select a.*,b.*
from a,b
where a.a = b.b(+);
a | b
--+-----
1 | null
2 | null
3 | 3
4 | 4
Full outer join:
A full outer join will give you the union of A and B, i.e. All the rows in A and all the rows in B. If something in A doesn't have a corresponding datum in B, then the B portion is null, and vice versa.
select *
from a
FULL OUTER JOIN b on a.a = b.b;
a | b
-----+-----
1 | null
2 | null
3 | 3
4 | 4
null | 6
null | 5
select min(id) as id,address, retailer
from table1
group by address, retailer
order by id
The query you need is:
SELECT min(id), address, retailer
FROM table1 AS t1
group by address, retailer
order by address
Here's the source
Use This: It's working:
SELECT * FROM `sampletable` GROUP BY address, retailer
