Refresh page with different datas without refreshing browser - php

Question

Helloo, I have been making a project in with I have to refresh the data every 10 seconds, I already created a separate file to load one info and it worked, but my question is, I have 3 sensors data on my database, Will I have to create 3 different refresh files to load each data to my main page?
This is one of my data files:
<?php
session_start();
include_once 'includes/dbh.inc.php';
$id = $_SESSION['userId'];
$dBname = "infosensor";
$conn = mysqli_connect($servername, $dBUsername, $dBPassword, $dBname);
$sql = "SELECT * FROM `$id` ORDER BY id DESC LIMIT 1;";
$result = mysqli_query($conn, $sql);
$resultCheck = mysqli_num_rows($result);
if($resultCheck > 0)
{
while ($row = mysqli_fetch_assoc($result))
{
$ss1 = intval($row['sensor1'] * ($p = pow(10, 2))) / $p;
echo "".$ss1."A";
$s1 = $row['sensor1'];
}
}
?>
this is my update file:
<script type="text/javascript" src="jquery-3.4.1.min.js"></script>
<script type="text/javascript">
$(document).ready(function (){
setInterval(function () {
$('#p0-blocoCorrente').load('load.php')
}, 5000);
});
</script>
this is how i diplay it:
<div class="blocoCorrente" id = "blocoCorrente">
<!-- Imprimir os valore dos sensor 1 -->
<div class="p0-blocoshow">Corrente 1:</div>
<div class="p0-blocoCorrente" id ="p0-blocoCorrente"></div>
</div>
UPDATE:
session_start();
include_once 'includes/dbh.inc.php';
$id = $_SESSION['userId'];
$sensor = isset($_POST['sensor']) ? "sensor" . intval($_POST['sensor']) : "sensor1";
$dBname = "infosensor";
$conn = mysqli_connect($servername, $dBUsername, $dBPassword, $dBname);
$sql = "SELECT $sensor FROM `$id` ORDER BY id DESC LIMIT 1;";
$result = mysqli_query($conn, $sql);
$row = mysqli_fetch_assoc($result);
if($row)
{
$s1 = $row[$sensor];
$ss1 = intval($s1 * ($p = pow(10, 2))) / $p;
echo $ss1;
}
$sensor1 = isset($_POST['sensor']) ? "sensor" . intval($_POST['sensor']) : "sensor2";
$sql = "SELECT $sensor1 FROM `$id` ORDER BY id DESC LIMIT 1;";
$result = mysqli_query($conn, $sql);
$row = mysqli_fetch_assoc($result);
if($row)
{
$s2 = $row[$sensor1];
$ss2 = intval($s1 * ($p = pow(10, 2))) / $p;
echo $ss2;
}

Answer 1

Make the PHP script return all the sensors in a JSON object. Then you can display each of them in an appropriate DIV.
session_start();
include_once 'includes/dbh.inc.php';
$id = $_SESSION['userId'];
$dBname = "infosensor";
$conn = mysqli_connect($servername, $dBUsername, $dBPassword, $dBname);
$sql = "SELECT sensor1, sensor2, sensor3 FROM `$id` ORDER BY id DESC LIMIT 1;";
$result = mysqli_query($conn, $sql);
$row = mysqli_fetch_assoc($result);
$sensors = array();
if($row)
{
for ($i = 1; $i <= 3; $i++) {
$s1 = $row["sensor$i"];
$ss1 = intval($s1 * ($p = pow(10, 2))) / $p;
$sensors["sensor$i"] = $ss1 . "A";
}
echo json_encode($sensors);
} else {
echo json_encode(null);
}
Then change the JavaScript to use $.getJSON().
$(document).ready(function (){
setInterval(function () {
$.getJSON('load.php', function(sensors) {
if (sensors) {
$('#p0-blocoCorrente').text(sensors.sensor1);
$('#p1-blocoCorrente').text(sensors.sensor2);
$('#p2-blocoCorrente').text(sensors.sensor3);
}
});
}, 5000);
});