Tell me the difference between sql1 and sql2:
sql1:
select count(1)
from table_1 a
inner join table_2 b on a.key = b.key where a.id in (
select id from table_1 group by id having count(1) > 1
)
sql2:
select sum(a) from (
select count(1) as a
from table_1 a
inner join table_2 b on a.key = b.key group by a.id having count(1) > 1
)
Why is the output not the same?
The queries are not even similar. They are very different. Let's check the first one:
select count(1)
from table_1 a
inner join table_2 b
on a.key = b.key
where a.id in (
select id from table_1 group by id having count(1) > 1
) ;
You are first making an inner join:
select count(1)
from table_1 a
inner join table_2 b
on a.key = b.key
In this case, you can use count(1), count(id), count(*), it's equivalent. You are counting the common elements in both tables: those ones that have in common the key field.
After that, you are enforcing this:
where a.id in (
select id from table_1 group by id having count(1) > 1
)
In other words, that every "id" of the table_1 must be at least two times in the table_1 table.
And lastly, you are doing this:
select count(1)
In other words, counting those elements. So, translated into english you have done this:
Let's see what happens with the second query:
select sum(a) from (
select count(1) as a
from table_1 a
inner join table_2 b
on a.key = b.key
group by a.id
having count(1) > 1
);
You are making the same inner join:
select count(1) as a
from table_1 a
inner join table_2 b
on a.key = b.key
but, you are grouping it by the id of the table:
group by a.id
and then filtering out only those elements who appear more than one time:
having count(1) > 1
The result so far are a set of records that have in common the key field in both tables, but grouped by the id: this means that only those fields that are at leas two times in the table_b are outputed of this join. After that, you group by id, collapsing those results into the table_1.id field and counting the result. I presume that very few records will match this strict criteria.
And lastly, you sum all those set.