Create . JSON file based on the result of the SQL query

Question

I am trying to send data from my database to a . json file but I am not able to because it creates the file and nothing appears, only the empty array.

my code is as follows::

<?php
$connect = mysqli_connect("localhost","root","admin","opcuaserver");
$query = "SELECT * FROM sensores (Timestamp, Metros, LM335, LM35)";
$result = mysqli_query($connect, $query);

$data = array();
$data[$result] = array();


//format the data
$formattedData = json_encode($data);

//set the filename
$filename = 'basedadossensores.json';

//open or create the file
$handle = fopen($filename,'w+');

//write the data into the file
fwrite($handle,$formattedData);

//close the file
fclose($handle);

?>

and the result when I run the code

{"Values->":[]}

Answer 1

I made him walk the $result to generate a vector to be converted by the function json_encode(); work:

<?php
$connect = mysqli_connect("localhost","root","admin","opcuaserver");
$query = "SELECT Timestamp, Metros, LM335, LM35 FROM sensores";
$result = mysqli_query($connect, $query);
if ($result) {
    $data = array();
    while($row = mysqli_fetch_array($result,  MYSQLI_ASSOC)) {
            $data[] = $row;
    }
} else  {
    echo "erro na query";
}

//format the data
$formattedData = json_encode($data);

//set the filename
$filename = 'basedadossensores.json';

//open or create the file
$handle = fopen($filename,'w+');

//write the data into the file
fwrite($handle,$formattedData);

//close the file
fclose($handle);

?>

Test there and see if it will work as you need it. I followed this reference.