UNION, SUM and count(*) together in PHP SQL query - mysql

Question

So I have php SQL query that queries the floors and the total workstations on each floor.
config_location_workstation workstation asset_inventory
floor floor floor
workstation_number workstation_number workstation_number
workstation_name assigned_user
(production|non-production) machine_type
I have config_location_workstation table that consist of floor, workstation_number, workstation_name (Production, Non-Production).
Then, I have asset_workstation table that consist of floor, workstation_number, assigned_user, machine_type (Laptop, Desktop). workstation_number has all the users workstation number per floor.
I also have asset_inventory table that also has floor and workstation_number of each machine.
I was able to query the total count of workstations per floor. But I also need to pull up all the workstation_name that has 'Production' in it. I need to combine all the tables to produce one query to get the total count of each workstation, machines deployed to each floor, total head count per floor. I need to get the count of all users workstation number per workstation_name (Production, Non-Production).
This is the current code I have to pull up the floors and the total workstations.
$query=mysqli_query($conn,"SELECT workstation.floor, COUNT(*) TOTAL FROM ( SELECT floor FROM config_location_workstation UNION ALL SELECT floor FROM asset_workstation ) AS workstation GROUP BY floor ORDER BY floor");
I am expecting to output
+-------+-------------+------+------+------+------+-----+
| Floor | Head Count | Workstations| Machines Deployed |
+-------+-------------+------+------+------+------+-----+
| 6TH | 100 | 150 | 140 |
| 18TH | 200 | 250 | 200 |
| 19TH | 300 | 320 | 310 |
+-------+-------------+------+------+------+------+-----+
How am I going to do this in one query.

Answer 1

Your tables are not normalized: there appears to be duplicate information in all 3 tables. (unless workstation_number is not unique and floor/workstation_number is a composite key. I would advise against that...)
I would suggest a schema similar to this (assuming a many-to-many relationship between users and workstations):
Workstation User_Workstation User
id <- - workstation_id
workstation_number user_id - -> id
machine_type name
name
floor
Each of the fields in Workstation describe an attribute of a workstation. Any information you may have gathered from the other two tables is now in one place. Don't Repeat Yourself! Duplicate data is evil.
The goal is
I need to combine all the tables to produce one query to get the
total count of each workstation,
machines deployed to each floor,
total head count per floor.
I need to get the count of all users workstation number per workstation_name (Production, Non-Production).
Individually, these are a piece of cake. (And, if you're running a script to get your results, it's all you need)
select count(id) number_of_workstations
from Workstation;
select count(id) workstations_per_floor, floor
from Workstation
group by floor
select count(u.id) users_per_floor, floor
from Workstation w
inner join User_Workstation uw on w.id = uw.workstation_id
inner join User u on u.id = uw.user_id
-- optionally filter by production or non-production
-- where w.name = ?
group by w.floor
But your desired result seems to be something different:
+-------+-------------+------+------+------+------+-----+
| Floor | Head Count | Workstations| Machines Deployed |
+-------+-------------+------+------+------+------+-----+
| 6TH | 100 | 150 | 140 |
| 18TH | 200 | 250 | 200 |
| 19TH | 300 | 320 | 310 |
+-------+-------------+------+------+------+------+-----+
I'm guessing that Head Count is number of users per floor, Workstations is total workstations per floor, and Machines Deployed is... I have no idea. Production?
select floor, count(u.id) Head_Count, count(distinct w.id) Workstations, -- ??? as Machines_Deployed
from workstation w
left join user_workstation uw on w.id = uw.workstation_id
left join user on u.id = uw.user_id
group by w.floor
(I'll modify this based on feedback on what Machines Deployed is)