Can't address parent field in query with multiple subqueries - mysql

Question

EDIT: Better explanation
I have a page with a job. The job as an idea and three skills (skill_ids) and skill requirements (a user must have at least this skill value to be qualified).
I click on the job to find candidates, so I have the job_id and the three skill_ids and skill_id_requirements. So I can do this so far as the first answer proposed with joins. I find all users who have the three skills. The skills are saved in skill_ratings. So far it works as I use to find the skill_id's only.
But now I want the value and here I have my code where I compute the final value (called rating). The rating respects all given values, but isn't a simple average or the sum of all. That's why I need the long horrible code. In the long horrible code I usually insert a user's ID. But here I need all user_id's who have the skills mentioned above just to calculate if they are qualified. This is dynamic.
I'm having a table where I want to find people who are qualified for a position under some requirements. Here I work with one table called skill_ratings, but (as far as I see) need to add some subqueries. And here I have the problem. There are many subqueries and I've tried to address a parent query field. But it only seems to work in a first-grade subquery to a parent query.
Here's my structure:
SELECT * FROM table t
WHERE EXISTS (SELECT * FROM table d WHERE x > 1
AND b=t.id
AND y <= (SELECT a FROM (MAIN SUBQUERY WITH CALCULATIONS)))
GROUP BY xyx
But the error I get is: #1054 - Unknown column 'skra.usr_id_get' in 'where clause'. skra is the parent table in this case.
I want to get the following (pseudo-sql):
SELECT all FROM table t AS x
WHERE EXISTS (
SELECT all FROM table t AS y
WHERE y.skill_id = 1
AND y.usr_id_get = t.usr_id_get
AND y.value <= (my algorithm)
)
The main subquery is important so far as I want to get a computed number. Elsewhere the code works because I were able to work with predefined PHP-variables for a user's ID. But I can't do this here as I need to find the users within the boundaries of the where-clauses.
How can I solve this? Because addressing a parent-field in a subquery seems to be limited to a first-grade subquery.
EDIT: Code
Code removed due to project status.
Error: #1054 - Unknown column 'c.usr_id_get' in 'where clause'

Answer 1

We want users that have certain skills of certain levels. For example all users that have skill 1 with at least level 20 and skill 2 with at least level 70.
Here is an algorithm:
First of all we must get the skill levels. A user has several skill ratings and the average rating per skill is the level.
Then we want a table of criteria (skill 1 / level 20, skill 2 / level 70 in our example).
We collect all user skill levels that match the criteria (EXISTS clause) and then
keep the users that match all skill levels (count(*) = <desired number of skills>).
The query:
select
sr.usr_id_get
from
(
select usr_id_get, skill_id, avg(value) as level
from skill_ratings
group by usr_id_get, skill_id
) sr
where exists
(
select *
from
(
select 1 as skill_id, 20 as level
union all
select 2 as skill_id, 70 as level
) criteria
where sr.skill_id = criteria.skill_id
and sr.level >= criteria.level
)
group by usr_id_get
having count(*) = 2;
You can also make criteria a real (temporary) table. Then your query stays the same, no matter how many skills are requested. You'd have
where exists
(
select *
from criteria
where sr.skill_id = criteria.skill_id
and sr.level >= criteria.level
)
group by usr_id_get
having count(*) = (select count(*) from criteria);
then.

Answer 2

This looks like it could be done with a simple JOIN:
SELECT T.*
FROM your_table T
JOIN other_table Y ON (
T.usr_id_get = Y.usr_id_get
AND T.skill_id = 1
AND Y.value <= [...]
)
If you need to perform some sort of calculations before the join, then you could join with a subquery:
SELECT T.*
FROM your_table T
JOIN (
SELECT *
FROM other_table Y
WHERE Y.skill_id = 1
AND Y.value = [...]
) Y USING(usr_id_get)

Answer 3

If I understand correctly, you have a user, say user 123, and a skill, say skill 99. Now you want to get the avarage rating for user 123 and skill 99 and then find all users with an equal or better average rating on that skill.
This is how to get the avarage ratings for skill 99 per user:
select usr_id_get, avg(value)
from skill_ratings
where skill_id = 99
group by usr_id_get;
This is how to get all users with an equal or better avarage rating for skill 99 than user 123:
select usr_id_get
from skill_ratings
where skill_id = 99
group by usr_id_get
having avg(value) >=
(select avg(value) from skill_ratings where skill_id = 99 and usr_id_get = 123);
Add to this whatever other criteria you need.