Problem getting JSON variable from another file

Question

I'm trying to get an Object from my file called load.php, I'm trying to get the information from load.php every 5s. For some reason, I could make one part of it, but I can't seem to get the other variable.

<script type="text/javascript" src="jquery-3.4.1.min.js"></script>
    <script type="text/javascript">
    $(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);
              //Error here:
              $('#p3-blocoCorrente').text(soma);
          }
        });
      }, 5000);
    });
    </script>

This is my load.php file:

<?php 

  session_start();

  include_once 'includes/dbh.inc.php';

  $soma = 0;
  $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;
          $soma += $row["sensor$i"];
          $sensors["sensor$i"] = $ss1 . "A";
      }
      echo json_encode($sensors);
  } else {
      echo json_encode(null);
  }
?>

Can't see where you output $soma from the php, nor where you would receive it in the ajax success. So when you say you "can't seem to get the other variable" do you mean you can't output the other variable or that it's not in javascript where, for some reason, you're expecting it to be? — fdomn-m
– fdomn-m, Commented Oct 31, 2019 at 11:40
I want to display it, but I just don't know how to get it into the script — Matheus
– Matheus, Commented Oct 31, 2019 at 11:42

RiggsFolly · Accepted Answer · 2019-10-31 11:46:41Z

1

In the PHP add soma to the array before json'ing it

<?php 

session_start();

include_once 'includes/dbh.inc.php';

$soma = 0;
$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;
        $soma += $row["sensor$i"];
        $sensors["sensor$i"] = $ss1 . "A";
    }

    // add soma to the array
    $sensors['soma'] = $soma;
    echo json_encode($sensors);
} else {
    echo json_encode(null);
}
?>

Then in the javascript, get it out to whereever you want to place it

<script type="text/javascript" src="jquery-3.4.1.min.js"></script>
<script type="text/javascript">
$(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);

                $('#p3-blocoCorrente').text(sensors.soma);
            }
        });
    }, 5000);
});
</script>

answered Oct 31, 2019 at 11:46

RiggsFolly

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5 Comments

RiggsFolly Over a year ago

Basically yes, its really quite simple once you get the first couple under your belt isn't it

Matheus Over a year ago

This comes from javascript right? It's because I actually have to get borrow some time to learn decent javascript, I'm really noob

fdomn-m Over a year ago

No - this has nothing to do with javascript / ajax. That part you had fine. It's the server-side php that provides the information that your javascript has access to, if you don't include it in the output, you can't see it in the javascript. This answer shows you how to change the php to include it.

RiggsFolly Over a year ago

@freedomn-m I see where you are coming from there, but the server response IS an integral part of AJAX process

fdomn-m Over a year ago

Well, yes, you need a server response, but it's not part of Asynchronous Javascript. Better to keep the separation of concerns, especially when learning. Javascript makes an ajax request; the server provides a response. When you change the server side, you're not changing the javascri[t/ajax, which is what was being asked with "this comes from javascript, right?". I may have been unclear when saying "nothing to do with javascript" - the change to the server-side/php is nothing to do with javascript. It's important to keep them separate.

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