将两个选择合并为一个

Question

I have a couple of tables. I'll try to make it simple: Table_1 ID is unique. Table_2 ID is not unique. The table_2 stores the ID from a row on table_1 and a value, resulting in, for instance, this:

table_1
ID | A
------
1  | a
2  | b
3  | c

table_2
ID | B
------
1  | x
1  | y
3  | z

I want to count how many of each ID is there on table_2, so I do

select t1.id, count(*)
from table_1 t1
group by t1.id

id | count
----------
1  | 2
3  | 1

And I want to list every row on table_2 and its corresponding value on table_1.A, so I do

select t1.id, t1.A, t2.B
from table_2 t2
left join table_1 t1
on t1.id = t2.id

ID | A | B
----------
1  | a | x
1  | a | y
3  | c | z

Is there a way to combine those 2 selections into one, to get a result like this?

ID | A | B | count
------------------
1  | a | x | 2
1  | a | y | 2
3  | c | z | 1

Answer 1

You can combine the results by joining the count result.

Fiddle with sample data

select t1.id, t1.A, t2.B, x.cnt as count
from  t2
left join t1
on t1.id = t2.id
join (select t2.id, count(*) as cnt
from  t2
group by t2.id
) x
on x.id = t1.id

Answer 2

Besides the generic Derived Table solution posted by @vkp you might also utilize a Scalar Subquery to return a single value:

select t1.id, t1.A, t2.B, 
 (select count(*)
  from  t2 as x
  where x.id = t2.id
) cnt
from  t2

http://www.sqlfiddle.com/#!9/07a3a/6