Question

Quick question about the most efficient was to handle this, I have a database which has a tree structure of user types,

system ---  1 to m --- account --- 1 to m --- customer  

All three types can have multiple email addresses, I was wondering what would be the best way to handle this in a database, all user types have auto_id as their primary key so having a table with address identified by there related primary key wont work as there will be potentially 3 of each. e.g. email with key 2 , will link to system with id 2 , account with id 2 , etc.

Should I create a table of email addresses for each type .? Or is there a more elegant solution .?

Thanks

Answer 1

You either have an email table that has a foreign key which is either a system_id, account_id, or customer_id. Then you can have a field specifying the type of that foreign key. Another more complicated strategy would be to have still the email table but no foreign key. Another table which you would call email_relation consisting of the email_id and the foreign key. That way you could use one e-mail address for all three tables.

Example of the use of two tables

system
--------
s1 
s2
s3

account
--------
a1
a2
a3

customer
--------
c1
c2
c3

email
------
e1 [e-mail]
e2 [e-mail]
e3 [e-mail]
e4 [e-mail]

email_relation
---------------
email_id     foreign_id      relation_type
e1           s1              system
e1           a1              account
e1           c1              customer
e2           c1              customer
e3           c2              customer
e4           a3              account
e4           c3              customer

if you want the customer table including the e-mail address

select c.customer_id, e.email
from customer c
left join email_relation r on (r.foreign_id = c.customer_id and relation_type = 'customer')
left join email          e on (e.email_id    = r.email_id)
where r.email_id is not null

If you want all e-mail to a system you could also

select e.email
  from email e
  join email_relation r on (r.email_id = e.email_id and relation_type = "system")
 where r.foreign_id = 1