How to equiry data using sql when facing condition? - mysql

Question

type cost
A 10
A 11
A 12
B 10
B 10
I have this small sample table. I want to select data where the cost of the same type is different.So the expected outcome should be:
type cost
A 10
A 11
A 12
The cost for A is different so I need to select these "A" out.
So what is the "select" sentence?
Thanks for the replies. Actually my table is little more complex like this
type cost people
A 10 jack
A 11 frank
A 12 lucy
B 10 amy
B 10 tom
I need to select the data meet one of the requirements below:
Same type with different cost
Same type with people "amy"
So the outcome should be like :
type cost people
A 10 jack
A 11 frank
A 12 lucy
B 10 amy
B 10 tom
Select all of type A because the cost is different
Select all of type B because the people has "amy"
I have firgure out how to select for amy like this:
select type, cost, people
from table
where type in
(select type from table where people = 'amy')
I don't know how to combine these conditions.
SQL Fiddle

Answer 1

You can use EXISTS to look for another row with same type but other cost:
select t1.type, t1.cost
from tablename t1
where exists (select * from tablename t2
where t2.type = t1.type
and t2.cost <> t1.cost)
Or have a sub-query that returns type values having different costs, and join with that result:
select t1.type, t1.cost
from tablename t1
join (select type
from tablename
group by type
having max(cost) <> min(cost)) t2
on t1.type = t2.type

Answer 2

Another way is:
select
t.type, t.cost
from t
left join t t1 on t.type = t1.type
and (t.cost <> t1.cost or t1.people = 'amy')
where
not t1.cost is null
group by
t.type, t.cost;
[SQL Fiddle Demo]
Or also:
select *
from t
where exists (
select 1
from t t1
where t1.type = t.type
group by t1.type
having count(distinct t1.cost) > 1
-- below code added your new criteria
union all
select 1
from t t2
where t2.people = 'amy'
);
[SQL Fiddle Demo]