I have one table that holds my ressources:
And a table that holds the associations
How to select the ressources of an Employee that are available, ie not in the association table?
I've tried this, but it's not working:
select r.ress, r.ress_id
FROM Ressource r
LEFT outer JOIN Ressource_Employee_Association a ON r.ress_id = a.ress_id
WHERE a.emp_id = 'ID00163efea66b' and a.ress_id IS NULL
Any ideas?
Thanks Thomas
The WHERE clause is applied after the LEFT JOIN. This means that you are currently trying to get results where there is NO matching record in Ressource_Employee_Association, but where the emp_id
equals 'ID00163efea66b'
.
But if there is no matching record, how can emp_id
be anything other than NULL
?
One option is to move part of the WHERE clause into the join...
SELECT
r.ress, r.ress_id
FROM
Ressource r
LEFT OUTER JOIN
Ressource_Employee_Association a
ON r.ress_id = a.ress_id
AND a.emp_id = 'ID00163efea66b'
WHERE
a.ress_id IS NULL
This will list all resources that are not associated to employee 'ID00163efea66b'
.
EDIT
Your comment implies that what you want is...
- A view listing all employees
- For each employee list each resource that they DON'T have
This requires an extra table listing all of your employees.
SELECT
*
FROM
Employee
CROSS JOIN
Ressource
WHERE
NOT EXISTS (SELECT * FROM Ressource_Employee_Association
WHERE emp_id = Employee.id
AND ress_id = Ressource.id)
Does this work?
select r.ress, r.ress_id
from resource r
where not exists
(
select 1 from ressource_emplyee_association a
where a.emp_id = '...' and a.ress_id = r.ress_id
)
EDIT
Before that I had the following, but changed it according to the comments below:
select r.ress, r.ress_id
from resource r
where not exists
(
select top 1 1 from ressource_emplyee_association a
where a.emp_id = '...' and a.ress_id = r.ress_id
)
SELECT *
FROM Ressource
WHERE ress_id IN (
SELECT ress_id,
FROM Ressource
MINUS
SELECT ress_id
FROM Ressource_Employee_Association
WHERE emp_id = 'ID00163efea66b'
);
After writing my above comments, and looking at the proposed solutions: I think I've got some more understanding of what you are trying to do.
Assuming you have unlimited quantity of resources in your resources table, you want to select the un-assigned resources per employee (based on their non-existence for any specific employee in the resource association table).
In order to accomplish this (and get a comprehensive list of employees) you'll need a 3rd table, in order to reference the complete list of employees. You'll also need to CROSS JOIN
all of the resources onto the list of employees (assuming every employee has access to every resource), then you LEFT JOIN
( LEFT OUTER JOIN
whatever) your association list onto the query where the resource_id
and employee_id
match the resource_id
in the resources
table, and the employee_id
in the employees
table (respectively). THEN you add your where clause that filters out all the records that assign an employee to a resource. This leaves you with the resources that are available to the employee, which they also do not have signed out. This is convoluted to say, so hopefully the query sheds more light:
SELECT e.employee_id, e.employee, r.res_id, r.res
FROM employees e
CROSS JOIN resources r
LEFT JOIN assigned_resources ar
ON ar.employee_id = e.employee_id AND r.res_id = ar.res_id
WHERE ar.res_id IS NULL
If you don't have an employees
table, you can accomplish the same by using the assigned resources table, but you will be limited to selecting employees who already have some resources allocated. You'll need to add a GROUP BY
query because of the possible existence of multiple employee definitions in the association table. Here's what that query would look like:
SELECT e.employee_id, r.res_id, r.res
FROM assigned_resources e
CROSS JOIN resources r
LEFT JOIN assigned_resources ar
ON ar.employee_id = e.employee_id AND r.res_id = ar.res_id
WHERE ar.res_id IS NULL
GROUP BY e.employee_id, r.res_id
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.