Formatting the result of an SQL query into JSON - python-3.x

Question

I have a small database as defined in the code snippet below. I want to query this to get all of the information and send it to an vue app via a JSON file via a Flask API. At the moment the query that I am using is
SELECT tbl_room.room, tbl_room.room_id, tbl_device.name, tbl_display.display, tbl_function.function, tbl_device.format
FROM tbl_device
INNER JOIN tbl_room ON tbl_room.id = tbl_device.room_id
INNER JOIN tbl_display ON tbl_display.id = tbl_device.display_id
INNER JOIN tbl_function ON tbl_function.id = tbl_device.function_id
ORDER BY tbl_room.room_id;
this gives me an output like:
Bedroom (Main) bedroom_main bme280/1 gauge temperature {"min": 0, "max": 50, "dp": 1, "units": "°C"}
Bedroom (Main) bedroom_main bme280/1 gauge humidity {"min": 0, "max": 100, "dp": 1, "units": "%"}
Bedroom (Main) bedroom_main bme280/1 gauge pressure {"min": 0, "max": 1100, "dp": 1, "units": "hPa"}
Front Room front_room ds18b20/heater gauge temperature {"min": 0, "max": 50, "dp": 1, "units": "°C"}
I would like to get it into a JSON file so that it is arranged as:
[
{ "name": "Office",
"id": "office",
"devices": []
},
{ "name": "Front Room",
"id": "front_room",
"devices": []
}
}
]
Can this be done in a single sql query? Or do I have to do a query for each room in a loop? Or is it more efficient to dump the whole dataset out in one query and process it in pyhton afterwards? This is a small dataset but I'm interested to know which is the most efficient method.
Thank you in advance,
Martyn
Here is my table structure:
-- Table: tbl_device
CREATE TABLE tbl_device (
name VARCHAR NOT NULL ON CONFLICT ROLLBACK,
room_id INTEGER CONSTRAINT fk_room REFERENCES tbl_room (id)
NOT NULL,
function_id INTEGER CONSTRAINT fk_function REFERENCES tbl_function (id)
NOT NULL ON CONFLICT ROLLBACK,
display_id INTEGER CONSTRAINT fk_display REFERENCES tbl_display (id)
NOT NULL ON CONFLICT ROLLBACK,
format VARCHAR NOT NULL ON CONFLICT ROLLBACK
DEFAULT [default],
UNIQUE (
name,
room_id,
function_id,
display_id
)
ON CONFLICT ROLLBACK
);
-- Table: tbl_display
CREATE TABLE tbl_display (
id INTEGER PRIMARY KEY AUTOINCREMENT,
display VARCHAR NOT NULL ON CONFLICT ROLLBACK
UNIQUE ON CONFLICT ROLLBACK
);
-- Table: tbl_function
CREATE TABLE tbl_function (
id INTEGER PRIMARY KEY AUTOINCREMENT,
function VARCHAR NOT NULL ON CONFLICT ROLLBACK
UNIQUE ON CONFLICT ROLLBACK,
control BOOLEAN NOT NULL
DEFAULT (0)
);
-- Table: tbl_room
CREATE TABLE tbl_room (
id INTEGER PRIMARY KEY AUTOINCREMENT,
room_id VARCHAR NOT NULL
UNIQUE ON CONFLICT ROLLBACK,
room VARCHAR NOT NULL ON CONFLICT ROLLBACK
);

Answer 1

First, There is no way to directly feed JSON response from MySQL database to VueJS or any other App. VueJS App is the fronted of your application. You have to create a Backend which connects to MySQL database, fetch necessary data from MYSQL Database, Convert them to JSON and send to Vue App.
To Develop a backend, you may use languages such as PHP, Python, Java, NodeJS etc.
If you can continue with PHP, it is very easy to fetch data and convert to JSON.
But If you still need to continue with Python, you have to use Flask or any other python web framework to do that.
Here is the sample php code
<?php
$dbhost = 'hostname';
$dbuser = 'username';
$dbpass = 'password';
$dbname = 'database';
$db = new mysqli($dbhost, $dbuser, $dbpass, $dbname);
if ($db->connect_errno) {
printf("Failed to connect to database");
exit();
}
$result = $db->query("SELECT * FROM "); // Your SQL query
$data = array();
while ( $row = $result->fetch_assoc()) {
$data[]=$row;
}
echo json_encode($data);
?>

Answer 2

If your version of sqlite was compiled with the JSON1 extension, enabled, something like:
SELECT json_group_array(json_object('name', tbl_room.name,
'id', tbl_room.room_id,
'devices', json_array()))
FROM tbl_room
GROUP BY tbl_room.name, tbl_room.room_id;