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mysqljsonapacheraspberry-pi3google-gauges

how to get google gauges work with mysql database from pi3 lamp server

I am working with a raspberry pi3 lamp server to displaying the data from two temperatures sensors AM2302

I am following a tutorial to display the data in a gauge format, here is the link to the video https://www.youtube.com/watch?v=WyUM--RGLH0 ( it is in Spanish, I think that an expert with this stuff will understate the video without a problem. But I don't know too much about this stuff, I am amateur and really noob about coding.

I am using two PHP files one to read the data and display in the JSON format a tne another one to show it in the gauges. Right now I am able to read the data but the gauges don't work. they stay at zero enter image description here

I think it is how the data json is show it: my result is:

{"temperature":"27.8","humidity":"61.2"}

but in the tutorial the result format is

[{"temperature":"27.8","humidity":"61.2"}]

I guess is that my JSON is not displaying the square brackets as the video show at: the 1:07 Mins from the video

here is my index.php file

<!DOCTYPE html>
<html xmlns="http://www.w3.org/1999/xhtml">
<head runat="server">
<meta http-equiv="Content-Type" content="text/html; charset=utf-8"/>
<title>Medidor Temperatura, Humedad</title>
<script 
src="https://ajax.googleapis.com/ajax/libs/jquery/2.2.0/jquery.min.js"> 
</script>

<script type="text/javascript" 
src="https://www.gstatic.com/charts/loader.js"></script>
<script type="text/javascript">
  google.charts.load('current', {'packages':['gauge']});
  google.charts.setOnLoadCallback(drawChart);
  function drawChart() {

    var data = google.visualization.arrayToDataTable([
      ['Label', 'Value'],
      ['humidity', 0],
      ['temperature', 0]
    ]);

    var options = {
      width: 400, height: 400,
      redFrom: 90, redTo: 100,
      yellowFrom:75, yellowTo: 90,
      minorTicks: 5
    };

    var chart = new 
google.visualization.Gauge(document.getElementById("Medidores"));

    chart.draw(data, options);

    setInterval(function() {
        var JSON=$.ajax({
            url:"http:192.168.0.115/sensores.php",
dataType: 'json',
            async: false}).responseText;
        var Respuesta=jQuery.parseJSON(JSON);

      data.setValue(0, 1,Respuesta[0].humidity);
      data.setValue(1, 1,Respuesta[0].temperature);
      chart.draw(data, options);
    }, 1300);

  }
</script>
</head>
<body>
   <div id="Medidores" ></div>

here is the file that read the data is called: sensores.php ( the one that collects the data from the sensors).

<?php
// Settings
// host, user and password settings
$host = "localhost";
$user = "logger";
$password = "123456";
$database = "temperatures";

//how many hours backwards do you want results to be shown in web page.
$hours = 24;

// make connection to database
$connectdb = mysqli_connect($host,$user,$password)
or die ("Cannot reach database");

// select db
mysqli_select_db($connectdb,$database)
or die ("Cannot select database");

// sql command that selects all entires from current time and X hours backward
$sql="SELECT temperature, humidity FROM temperaturedata where sensor = 'Exterior' and dateandtime >= (NOW() - INTERVAL $hours HOUR) order by dateandtime desc LIMIT 1";

// set query to variable
$temperatures = mysqli_query($connectdb,$sql);

// create content to web page
?>

<?php
// loop all the results that were read from database and "draw" to web page

while($temperature=mysqli_fetch_assoc($temperatures))

$json=json_encode($temperature);
echo $json;

?>

I wanna to display the data from the two sensors in the index.php file but I even with one I will feel ok

Solution

  • In the while loop, build an array the [] indicators in JSON mean array.

    It makes sence also because you could have more than one result from the query anyway.

    $temps = [];
while($temperature = mysqli_fetch_assoc($temperatures)) {
    $temps[] = $temperature;
}

echo json_encode($temps);;