A bit of trouble with mySQL querying - mysql

Question

say I have a table called "users", which houses all the users of a system. Each user has a "role" as defined by the "roles" column in addition to a "name" and "id" column. Each user with the role "worker" is advised by a "supervisor" as defined by the "advised_by" column.
I'd like to query my database so that it returns the both the name of the worker and their supervisor and one line, is that possible without joins?
Thanks for any advise, I've been banging my head against this for a while now
Edit: Thanks to everybody who tried to help. Sorry, I guess that was a pretty vague/awful description in retrospect. I'll try to model the table better here.
Columns: Name, ID, Role, Advised_by
Row 1: Bob, 123, Worker, 321
Row 2: Tom, 321, Supervisor, N/A
I would like to return a result with all the people tagged as "Workers", along with their name and the name of their supervisor, so something like: Bob, Worker, Tom on one line.
I was trying to avoid adding a join because my professor asked us not to use them due to the performance hit, but it doesn't seem like there's a practical way around it, in which case I'll be alright with a join.
Edit 2: Guess I'm just an idiot, I realized after I typed it all out I could just do it pretty easily with aliasing. Sorry for the trouble guys

Answer 1

Some more details would be necessary, but you need to join the table to get both the name of the supervisor and worker in one MySQL sentence.
Assuming the advised_by column contains the ID of the user who advises this worker:
SELECT u.id, u.name, u2.id AS supervisorID, u2.name AS supervisorNAME
FROM users u, users u2
WHERE u.advised_by = u2.id AND u.role = 'worker'

Answer 2

At the risk of repeating the answer provided by EpochGalaxy...
Q: Is that possible without joins?
A: Yes. It is possible. One approach is to use a correlated subquery.
Consider the query you need to run to get the supervisor name...
SELECT s.name
FROM users s
WHERE s.id = 321
What if you ran a query like that for each row you retrieved from users.
Say you ran query like this:
SELECT u.id
, u.name
, u.role
, u.advised_by
FROM users u
ORDER BY u.id
And for each row from that query, right before you returned it, you ran a query to get the name of the supervisor.
We can achieve that by using a subquery in the SELECT list.
As a trivial example of running a subquery in the SELECT list consider this query. (There's no logical reason we'd run a query like this, but as a simple demonstration that we can run a subquery...
SELECT u.id
, u.name
, u.role
, u.advised_by
, ( SELECT 'bar' ) AS foo
FROM users u
ORDER BY u.id
The rows returned by this query will include an additional column, named foo. The value assigned to the column will be provided from the result of the subquery... "SELECT 'bar'".
Now that we know we can run a subquery in the SELECT list, and return a column with that, we can try something else. Like this:
SELECT u.id
, u.name
, u.role
, u.advised_by
, ( SELECT s.name
FROM users s
WHERE s.id = 321
) AS s_name
FROM users u
ORDER BY u.id
With that, for each row returned by the outer query, MySQL will run the subquery in the SELECT list, and take the value returned by the subquery, and put the value returned into a column of the row being returned by the outer query.
That subquery is going to get only the name from the row in users with id = 321.
So that doesn't really get us what we need.
Here's the trick... instead of a literal value 321, we can put there an expression that includes a reference to the table in the outer query.
SELECT u.id
, u.name
, u.role
, u.advised_by
, ( SELECT s.name
FROM users s
WHERE s.id = u.advised_by
) AS s_name
FROM users u
ORDER BY u.id
Aha!
Now, for each row returned by the outer query, MySQL will run the subquery in the SELECT this.
This time, the subquery includes a reference to the advised_by column from the row of the outer query. MySQL will take the value that's in the advised_by column from the row being returned (by the outer query), and then slide that into the subquery before the subquery executes. The return from the subquery goes into the column of the resulset.
With this, the results we get from the sbuquery are determined by values in the outer query. The subquery is related to the outer query.
Some important restrictions to notes: The subquery in a SELECT list of another query must return exactly one column. (It can be an expression, a literal. But the return from the subquery is not allowed to return two or more columns.
The subquery must return no more than one row. If it returns zero rows, a NULL value is assigned to the column in the outer query.
The subquery in the example above satisfies the conditions. There's exactly one expression in the SELECT list, and the id column is unique in the users table, so it can't return more than one row.
The SQL terminology for using a subquery like this is correlated subquery.
So, to answer your question, yes... it is possible to achieve the specified result without a join operation.
Some other notes:
In terms of performance, it's important that the subquery be efficient. It is going to be executed for every row returned by the outer query. If we're returning 10,000 rows, that's 10,000 executions of the subquery.
So, this approach is best suited to cases where the outer query is returning a small number of rows, and the correlated subquery is very efficient.

Answer 3

Use the following query:
SELECT (u.name)workername, (select name from users where id=u.advised_by)supervisor FROM `users` u where u.roles="worker"
Sample table structure as described by you:
--
-- Database: `test`
--
-- --------------------------------------------------------
--
-- Table structure for table `users`
--
CREATE TABLE `users` (
`id` int(11) NOT NULL,
`name` varchar(255) NOT NULL,
`roles` varchar(255) NOT NULL,
`advised_by` int(11) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
--
-- Dumping data for table `users`
--
INSERT INTO `users` (`id`, `name`, `roles`, `advised_by`) VALUES
(1, 'test1', 'supervisor', 0),
(2, 'test2', 'worker', 1),
(3, 'test3', 'worker', 1);