SQL multiple test between 3 tables - mysql

Question

I've been struggling with a query that I have to do. I have basic knowledge with SQL.
I have 3 tables, which are users, companies and orders. Here is a sample for each one with the columns that interest us :
Users table is just a table with id, firstname, lastname.
Companies table linked with user_id
Orders table linked with user_id
I need to display two columns, one that will be all the different user_id, and the second one, a "current step" that doesn't exist yet, that will be determined like this :
IF (user has an order with status field at ok - orders table)
current = 8
else if (user has an order with status field empty - orders table)
current = 7
else if (user is linked to a company with current-step field at 0-5 - companies table)
current = 1-6
else if (user has a test company in name field - companies table)
current = 0
The thing is, a user can be related to multiple companies, we have to take the highest current possible, and can be related to multiple orders, and take the one with status = ok if he has one and if not, current will be 7 since he has an order anyway.
I could make the first test work, but then when I have to test between the three tables, I don't know how to put all the tests together, my query ends up in a mess. If you could help me going the right direction, that would be kind.
EDIT:
I forgot something, we're working with PHP on CakePHP framework. Would it be possible to divide the query, and to make the test directly in PHP ?

Answer 1

SQL is not a procedural language. So it does not support if-else loops.
That being said , you would require an API like JDBC or embedded sql to perform such queries. Look up SQL API's and embedded sql to know what they are.
If this is some college assignment, then I don't know how your teacher expects you to do it. But if you are learning about sql and databases out of self interests, Then you have to look into the topics I mentioned above!
You will also need to look into SQL Joins.

Answer 2

Without test data, I can't test it, but try this:
select UserID,
case when oc.Current > cc.Current then oc.Current else cc.Current end as Current
from(
select userID,
IfNull( Max( case when o.Status = 'OK' then 8 else 7 end ), -1 ) Current
from Users u
left join Orders o
on o.UserID = u.UserID
group by u.UserID
) oc
join(
select userID,
IfNull( Max( case when c.CurrentStep between 0 and 5 then c.CurrentStep + 1
when c.Name = 'Test' then 0 else -1 end ), -1 ) as Current
from Users u
left join Companies c
on c.UserID = u.UserID
group by u.UserID
) cc;
This would be a lot neater with CTEs, alas. Also, you don't state the value of Current for those uses who meet none of the requirements, so I used -1.
The first inline view returns an 8 if a linked company has status=ok, 7 if no status=ok and -1 if not linked to any company. The second inline view returns -1 through 6 as appropriate or -1 if not linked with any orders.

Answer 3

Sorry for answering that late.
I proposed your solutions to my team, but a friend of mine succeeded by doing this query :
select u.* from (
SELECT u.*,
c.name,
c.canton,
c.country as 'company_country',
c.id as 'company_id',
c.user_id,
MAX(case
when o.status = 'ok' then 8
when o.status = '' then 7
when c.current_step = 5 then 6
when c.current_step = 4 then 5
when c.current_step = 3 then 4
when c.current_step = 2 then 3
when c.current_step = 1 then 2
when c.current_step = 0 then 1
else
(case
when c.name LIKE '%test%' then 0
else 'Empty'
end )
end) as current
FROM ezycount_users u
LEFT OUTER JOIN ezycount_companies c ON c.user_id = u.id
LEFT OUTER JOIN ezycount_orders o ON o.user_id = u.id
group by u.id ) as u
And by using a function in CakePHP to group by again, and it worked for us !
Thanks all of you !