SELECT * FROM TABLE, TABLE depending on another query - SQL - mysql

Question

Here's what I'm trying to achieve :
Select a person's or firm's name depending on the contact type.
Tables :
CONTACTS : PEOPLE FIRMS
- id - contact_id - contact_id
- type (person/firm) - name - name
I would like this kind of magic :
SELECT type
FROM contacts
WHERE contacts.id = 42;
SELECT model.name
FROM contacts LEFT JOIN model ON model.contact_id = contacts.id
WHERE contacts_id = 42
AND model = CASE (WHEN type = "person" THEN "people"
WHEN type = "firm" THEN "Firms" END); (Not sure this one works)
All in one query. Thanks !

Answer 1

You could do like below, join two tables and then decide which column to use according to contacts.type.
SELECT
(CASE WHEN t1.type = 'person' THEN t2.name WHEN t1.type = 'firm' THEN t3.name) AS contact_name
FROM contacts t1
LEFT JOIN people t2 ON t1.id = t2.contact_id
LEFT JOIN firms t3 ON t1.id = t3.contact_id

Answer 2

Simply left join both tables; one of them will match id and type. Use COALESCE then to display the one name found.
select coalesce(p.name, f.name) as name
from contacts c
left join people p on p.contacts_id = c.id and c.type = 'person'
left join firms f on f.contacts_id = c.id and c.type = 'firm'
where c.id = 42;

Answer 3

You can use dynamic sql like this:
declare
#table_name nvarchar(20),
#query nvarchar(max)
SELECT
#table_name =
case type
when 'person' then 'people'
when 'firm' then 'firms'
end
FROM contacts
WHERE contacts.id = 42;
set #query = N'SELECT ' + #table_name + '.name ' +
'FROM contacts LEFT JOIN model ON ' + #table_name + '.contact_id = contacts.id
WHERE contacts_id = 42 '
exec sp_executesql #query

Answer 4

Here is an alternative way without a join:
select
case when c.type = 'person' then
(select name from people p where p.contacts_id = c.id)
else
(select name from firms f on f.contacts_id = c.id)
end as name
from contacts c
where c.id = 42;
You see, there are many ways to address the problem. Just pick the one you find most readable.