foreach statment?

On 19 May 2003, Abdul-wahid Paterson wrote:

Hi,

I have a situation (that I seem to come across often - hence the
question) where I have 3 tables.

items
items_options
options

items has a many-to-many relationship with options so items_options is
merely a link table with an item_id and and option_id.

What I need to do is give a default option to all items that don't have
any options.

I would do something like this:

select i.item_id from items i where (select count(item_id) from
items_options where item_id=i.item_id) = 0;

And then write a script that will go through the outputted list of
item_id's and for each one do an insert statement like:

insert into items_options values ($item_id, $n);

Maybe something like (not really tested)

insert into items_options select item_id, $n from
items i where not exists (select item_id from item_options where
i.item_id=item_id);

Abdul-wahid Paterson <aw@lintrix.net> writes:

I would do something like this:

select i.item_id from items i where (select count(item_id) from
items_options where item_id=i.item_id) = 0;

And then write a script that will go through the outputted list of
item_id's and for each one do an insert statement like:

insert into items_options values ($item_id, $n);

Where $n is the number of my default option.

Use INSERT/SELECT:

insert into items_options
select i.item_id, $n
from items i where (select count(item_id) from
items_options where item_id=i.item_id) = 0;

BTW, I'd think about a NOT EXISTS instead of a COUNT test in the
WHERE ... should run faster ...

regards, tom lane