How to make query to count values from two tables - sql-server

Question

I have three tables main 'maintable' table and two sub tables 'table1' and 'table2' the main table 'maintable' contains tow columns 'ID' and 'name' like that:
ID name
.... ......
1 Khalid
2 Jone
3 Steve
and the first sub table 'table1' contains 't1ID' and 'column' and 'ID' (foreign key) from 'maintable' like that:
t1ID column ID
...... ....... ....
1 Value 1
2 Value 1
3 Value 1
4 Value 2
and the second sub table 'table2' contains 't2ID' and 'column' and 'ID' (foreign key) from 'maintable' like that:
t2ID column ID
...... ....... ....
1 Value 2
2 Value 1
3 Value 1
4 Value 3
I want to make query to find count of (table1.ID) as A and count of (table2.ID) as B like that:
name A B
...... ... ...
khalid 3 2
Jone 1 1
Steve 0 1

Answer 1

Try this :
select name,
(select count(t1.ID) from table1 t1 where t1.ID = main.ID) as A,
(select count(t2.ID) from table2 t2 where t2.ID = main.ID) as B
from maintable main

Answer 2

Try this out:
;with cte1 as (
SELECT ID, COUNT(1) AS Cnt
FROM table1
GROUP BY ID
), cte2 as (
SELECT ID, COUNT(1) AS Cnt
FROM table2
GROUP BY ID
)
SELECT m.name, ISNULL(cte1.Cnt, 0) AS A, ISNULL(cte2.Cnt, 0) AS B
FROM maintable m
LEFT JOIN cte1 ON cte1.ID = m.ID
LEFT JOIN cte2 ON cte2.ID = m.ID
It can also be done with subqueries, but I like CTEs more (query is more readable).

Answer 3

Try this:
with t0_t1 as (
select
t.id,
t.nm,
count(t1.id) as A
from table0 t
left join table1 t1 on t.id = t1.id
group by t.id, t.nm
)
select t.nm, t.A, count(t2.id) as B
from t0_t1 t
left join table2 t2 on t.id = t2.id
group by t.nm, t.A
Example: http://sqlfiddle.com/#!6/341ff/10
create table table0 (id int, nm varchar(20));
insert into table0 values (1,'Khalid'),(2,'Jone'),(3,'Steve');
create table table1 (t1id int, col varchar(20), id int);
insert into table1 values
(1, 'v', 1), (2, 'v', 1), (3, 'v', 1), (4, 'v', 2);
create table table2 (t2id int, col varchar(20), id int);
insert into table2 values
(1, 'v', 2), (2, 'v', 1), (3, 'v', 1), (4, 'v', 3);
Result:
| nm | A | B |
|--------|---|---|
| Steve | 0 | 1 |
| Jone | 1 | 1 |
| Khalid | 3 | 2 |