Question

I have two tables as given below

    Table A                Table B             Table C              
  =============          ==============        =========
    Id    Name           Id   AId  CId         Id   Name 
    1     A              1    1    1           1    x
    2     B              2    1    1           2    y
    3     C              3    2    1           3    z 
    4     D              4    2    3           4    w
    5     E              5    3    2           5    v

Now I want all the records of Table A with matching Id column CId from Table B where CId = 1.

So the output should be like below :

Id Name CId
1   A   1 
2   B   1
3   C   1
4   D   Null
5   E   Null

Can anyone help me please?

Answer 1

This does what you want:

SELECT
A.Id,
A.Name,
CASE B.CId WHEN 1 THEN 1 ELSE NULL END AS CId
FROM
A LEFT JOIN B ON A.Id = B.Id

This is not about LEFT JOINing. You could as well do it with an INNER JOIN. When you don’t want the 3 and 2 of column CId to appear you would still have to filter with WHERE and therefore the rows with Id 4 and 5 would not appear, which is not what you want.

EDIT:

Given this test data:

create table A (Id int, Name varchar(5));
insert into A values 
(1, 'A'),
(2, 'B'),
(3, 'C'),
(4, 'D'),
(5, 'E');

create table B (Id int, AId int, CId int);
insert into B values
(1,1,1),
(2,1,1),
(3,2,1),
(4,2,3),
(5,3,2);

my query does not give a cartesian product. Read and try before downvoting. Anyway, it was not clear to me what you want to achieve, now I’ve joined on AId column and with this query:

SELECT DISTINCT
A.Id,
A.Name
, CASE 
WHEN B.CId > 1 THEN 1 
WHEN B.CId = 1 THEN 1
ELSE NULL END AS CId
FROM
A LEFT JOIN B ON A.Id = B.AId

and it also gives the right output, like the first before. If this is still not what you want, your test data is wrong or I absolutely don’t get it.