SQL query with two EXISTS statements behaving differently than expected - sql

Question

The following SQL query is intended to find label_item_lists which have label_items with given names.
SELECT lils.id FROM label_item_lists AS lils
INNER JOIN label_items AS items ON lils.id = items.label_item_list_id
WHERE EXISTS(SELECT * FROM label_item_lists WHERE items.name=?)
OR EXISTS(SELECT * FROM label_item_lists WHERE items.name=?)
It properly returns the ids of label_item_lists having an item with either name. However, the same query using the AND operator rather than OR returns no results.
SELECT lils.id FROM label_item_lists AS lils
INNER JOIN label_items AS items ON lils.id = items.label_item_list_id
WHERE EXISTS(SELECT * FROM label_item_lists WHERE items.name=?)
AND EXISTS(SELECT * FROM label_item_lists WHERE items.name=?)
There are definitely label_item_list entries that have label_items matching both names provided. In fact the OR SQL query returns the id twice for these entries, but the AND query returns no results. For this reason I think I might be missing an important piece of info on how these JOINed queries with EXISTS work. Thanks for any assistance!
----------------------------------------------------------------
| label_items | id | name | label_item_list_id |
----------------------------------------------------------------
| Row1 | 1 | foo | 1 |
----------------------------------------------------------------
| Row2 | 2 | bar | 1 |
----------------------------------------------------------------
| Row3 | 3 | bar | 2 |
----------------------------------------------------------------
--------------------------------
| label_item_lists | id |
--------------------------------
| Row1 | 1 |
--------------------------------
| Row2 | 2 |
--------------------------------
I want to return the first label_item_list but not the second, as it only has one of the two names I am searching for, 'foo' and 'bar'.

Answer 1

try changing the where condition from
WHERE EXISTS(SELECT * FROM label_item_lists WHERE items.name=?)
AND EXISTS(SELECT * FROM label_item_lists WHERE items.name=?)
to
WHERE EXISTS(SELECT * FROM label_item_lists lst WHERE lst.name=?)
AND EXISTS(SELECT * FROM label_item_lists lst WHERE lst.name=?)

Answer 2

In your query AND will not return anything because on same output row it will apply both filters which will never happen hence it is giving blank output.
And Or operator will never check condition after OR operator until first condition is false.
Try something like this, # is just a place holder to distinguish between two searches:
select * from label_items lil
where label_item_list_id
in (
select li.label_item_list_id from
label_items li
inner join label_items l1
on li.label_item_list_id = l1.label_item_list_id
and li.name <> l1.name
where concat(li.name,'#',l1.name) = 'foo#bar')

Answer 3

This is what I eventually came up with! I'm not 100% confident yet, but it has worked so far. I added a bit of functionality in Ruby and ActiveRecord to allow for as many necessary matches as desired and to return only those which match exactly (without any extra names not in the list).
items = ["foo", "bar"]
db = ActiveRecord::Base.connection
query = <<-EOS
SELECT lils.id FROM label_item_lists AS lils
JOIN label_items AS items ON items.label_item_list_id = lils.id
WHERE lils.id IN (
SELECT label_item_list_id FROM label_items AS items
WHERE items.name IN (#{(['?'] * items.length).join(',')})
) AND lils.id NOT IN (
SELECT label_item_list_id FROM label_items AS items
WHERE items.name NOT IN (#{(['?'] * items.length).join(',')})
)
GROUP BY lils.id
HAVING COUNT(DISTINCT items.name) = #{items.length}
EOS
query = ActiveRecord::Base.send(:sanitize_sql_array, [query, *(items*2)])
Basically it checks that a name is both IN the list of given names (items array) AND it also checks that the name IS NOT outside (or NOT IN) the list of given names. Any list of label_items that has a non-matching name is excluded by the latter IN query. This is helpful so that a label_item_list with both "foo" and "bar" but also "lorem" is not included.