SQL: How to find all parents where parents have children with given id





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I have a question about a parent / child join table. I have a join table called parent_children. The table contains a parent_id and child_id. So like this:



parent_id  | child_id
1 1
1 2
1 3
2 1
2 3
3 1
3 4


What I want is to find all parents who have children in a certain list. So let's say the list contains 1 and 3 then I want parent_id 1 and 2. If the list contains 4 I want parent_id 3. How do I do this?










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  • Child 1 has three parents?

    – jarlh
    Nov 16 '18 at 15:53


















0















I have a question about a parent / child join table. I have a join table called parent_children. The table contains a parent_id and child_id. So like this:



parent_id  | child_id
1 1
1 2
1 3
2 1
2 3
3 1
3 4


What I want is to find all parents who have children in a certain list. So let's say the list contains 1 and 3 then I want parent_id 1 and 2. If the list contains 4 I want parent_id 3. How do I do this?










share|improve this question

























  • Child 1 has three parents?

    – jarlh
    Nov 16 '18 at 15:53














0












0








0








I have a question about a parent / child join table. I have a join table called parent_children. The table contains a parent_id and child_id. So like this:



parent_id  | child_id
1 1
1 2
1 3
2 1
2 3
3 1
3 4


What I want is to find all parents who have children in a certain list. So let's say the list contains 1 and 3 then I want parent_id 1 and 2. If the list contains 4 I want parent_id 3. How do I do this?










share|improve this question
















I have a question about a parent / child join table. I have a join table called parent_children. The table contains a parent_id and child_id. So like this:



parent_id  | child_id
1 1
1 2
1 3
2 1
2 3
3 1
3 4


What I want is to find all parents who have children in a certain list. So let's say the list contains 1 and 3 then I want parent_id 1 and 2. If the list contains 4 I want parent_id 3. How do I do this?







sql join






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share|improve this question













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edited Nov 16 '18 at 15:52









melpomene

62.5k55195




62.5k55195










asked Nov 16 '18 at 15:50









Martijn HiemstraMartijn Hiemstra

127211




127211













  • Child 1 has three parents?

    – jarlh
    Nov 16 '18 at 15:53



















  • Child 1 has three parents?

    – jarlh
    Nov 16 '18 at 15:53

















Child 1 has three parents?

– jarlh
Nov 16 '18 at 15:53





Child 1 has three parents?

– jarlh
Nov 16 '18 at 15:53












2 Answers
2






active

oldest

votes


















2














You can use group by, where, and having:



select parent_id
from t
where child_id in (1, 3)
group by parent_id
having count(*) = 2; -- "2" is the number of items in the list


This assumes no duplicate parent/child rows in the table. If this is possible, then use count(distinct child_id) = 2.






share|improve this answer
























  • This is brilliant. Thanks for the answer. I don't know how on earth you came up with that.

    – Martijn Hiemstra
    Nov 16 '18 at 19:00



















0














So you basically need to count the number of occurrences in the IN list of your input.
The following can help.



create table parent_children(parent_id int, child_id int)


insert into parent_children values(1,1)
insert into parent_children values(1,2)
insert into parent_children values(1,3)
insert into parent_children values(2,1)
insert into parent_children values(2,3)
insert into parent_children values(3,1)
insert into parent_children values(3,4)



with list_data
as(select *
from (values(1),(3))as t(x)
)
select a.x
from list_data a
left join parent_children b
on a.x=b.child_id
group by a.x
having count(*) = count(distinct b.parent_id)





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    You can use group by, where, and having:



    select parent_id
    from t
    where child_id in (1, 3)
    group by parent_id
    having count(*) = 2; -- "2" is the number of items in the list


    This assumes no duplicate parent/child rows in the table. If this is possible, then use count(distinct child_id) = 2.






    share|improve this answer
























    • This is brilliant. Thanks for the answer. I don't know how on earth you came up with that.

      – Martijn Hiemstra
      Nov 16 '18 at 19:00
















    2














    You can use group by, where, and having:



    select parent_id
    from t
    where child_id in (1, 3)
    group by parent_id
    having count(*) = 2; -- "2" is the number of items in the list


    This assumes no duplicate parent/child rows in the table. If this is possible, then use count(distinct child_id) = 2.






    share|improve this answer
























    • This is brilliant. Thanks for the answer. I don't know how on earth you came up with that.

      – Martijn Hiemstra
      Nov 16 '18 at 19:00














    2












    2








    2







    You can use group by, where, and having:



    select parent_id
    from t
    where child_id in (1, 3)
    group by parent_id
    having count(*) = 2; -- "2" is the number of items in the list


    This assumes no duplicate parent/child rows in the table. If this is possible, then use count(distinct child_id) = 2.






    share|improve this answer













    You can use group by, where, and having:



    select parent_id
    from t
    where child_id in (1, 3)
    group by parent_id
    having count(*) = 2; -- "2" is the number of items in the list


    This assumes no duplicate parent/child rows in the table. If this is possible, then use count(distinct child_id) = 2.







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Nov 16 '18 at 15:52









    Gordon LinoffGordon Linoff

    797k37318423




    797k37318423













    • This is brilliant. Thanks for the answer. I don't know how on earth you came up with that.

      – Martijn Hiemstra
      Nov 16 '18 at 19:00



















    • This is brilliant. Thanks for the answer. I don't know how on earth you came up with that.

      – Martijn Hiemstra
      Nov 16 '18 at 19:00

















    This is brilliant. Thanks for the answer. I don't know how on earth you came up with that.

    – Martijn Hiemstra
    Nov 16 '18 at 19:00





    This is brilliant. Thanks for the answer. I don't know how on earth you came up with that.

    – Martijn Hiemstra
    Nov 16 '18 at 19:00













    0














    So you basically need to count the number of occurrences in the IN list of your input.
    The following can help.



    create table parent_children(parent_id int, child_id int)


    insert into parent_children values(1,1)
    insert into parent_children values(1,2)
    insert into parent_children values(1,3)
    insert into parent_children values(2,1)
    insert into parent_children values(2,3)
    insert into parent_children values(3,1)
    insert into parent_children values(3,4)



    with list_data
    as(select *
    from (values(1),(3))as t(x)
    )
    select a.x
    from list_data a
    left join parent_children b
    on a.x=b.child_id
    group by a.x
    having count(*) = count(distinct b.parent_id)





    share|improve this answer




























      0














      So you basically need to count the number of occurrences in the IN list of your input.
      The following can help.



      create table parent_children(parent_id int, child_id int)


      insert into parent_children values(1,1)
      insert into parent_children values(1,2)
      insert into parent_children values(1,3)
      insert into parent_children values(2,1)
      insert into parent_children values(2,3)
      insert into parent_children values(3,1)
      insert into parent_children values(3,4)



      with list_data
      as(select *
      from (values(1),(3))as t(x)
      )
      select a.x
      from list_data a
      left join parent_children b
      on a.x=b.child_id
      group by a.x
      having count(*) = count(distinct b.parent_id)





      share|improve this answer


























        0












        0








        0







        So you basically need to count the number of occurrences in the IN list of your input.
        The following can help.



        create table parent_children(parent_id int, child_id int)


        insert into parent_children values(1,1)
        insert into parent_children values(1,2)
        insert into parent_children values(1,3)
        insert into parent_children values(2,1)
        insert into parent_children values(2,3)
        insert into parent_children values(3,1)
        insert into parent_children values(3,4)



        with list_data
        as(select *
        from (values(1),(3))as t(x)
        )
        select a.x
        from list_data a
        left join parent_children b
        on a.x=b.child_id
        group by a.x
        having count(*) = count(distinct b.parent_id)





        share|improve this answer













        So you basically need to count the number of occurrences in the IN list of your input.
        The following can help.



        create table parent_children(parent_id int, child_id int)


        insert into parent_children values(1,1)
        insert into parent_children values(1,2)
        insert into parent_children values(1,3)
        insert into parent_children values(2,1)
        insert into parent_children values(2,3)
        insert into parent_children values(3,1)
        insert into parent_children values(3,4)



        with list_data
        as(select *
        from (values(1),(3))as t(x)
        )
        select a.x
        from list_data a
        left join parent_children b
        on a.x=b.child_id
        group by a.x
        having count(*) = count(distinct b.parent_id)






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 16 '18 at 16:01









        George JosephGeorge Joseph

        1,590510




        1,590510






























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