Reputation: 825
I have this table called employeetimesheets:
empsheet_id|employee_id|timesheet_status|last_update
The table allows the manager to have access to all the employee time sheets. One employee can have several time sheets. I would like to display only the most recent entry per employee. I read in the manual I have to write a groupwise-maximum subquery and left join
with inner join
but I'm unsure how to go about it here.
So far this is the query I have:
$sqlempsheets="SELECT * FROM employeetimesheets JOIN employees ON employeetimesheets.employee_id=employees.employee_id WHERE employeetimesheets.timesheet_status='Pending Approval'";
$resultempsheets=mysqli_query($db,$sqlempsheets);
Upvotes: 0
Views: 346
Reputation: 39467
Try this:
select *
from employeetimesheets t
join (
select employee_id,
max(empsheet_id) as empsheet_id
from employeetimesheets
group by employee_id
) t2 on t.employee_id = t2.employee_id
and t.empsheet_id = t2.empsheet_id
join employees e on t.employee_id = e.employee_id
where t.timesheet_status = 'Pending Approval';
Or using left join
:
select t.*, e.*
from employeetimesheets t
left join employeetimesheets t2 on t.employee_id = t2.employee_id
and t.empsheet_id < t2.empsheet_id
join employees e on t.employee_id = e.employee_id
where t.timesheet_status = 'Pending Approval'
and t2.employee_id is null;
Upvotes: 2