I'm trying to pass a PHP $_SESSION variable to javascript

Question

So my ultimate goal is for my database to inform my site when something is sold so that it will darken out and so a hypothetical customer couldn't buy something that was already sold. I have the my database setting php session variables but I'm having trouble transferring that information into javascript/jquery. I'm not sure if it's because I'm doing it wrong or because there is something logically wrong with my approach. Thank you! Here is my code: PHP:

<?php
$customer = new order;
$customer->item1Avail();
$customer->item2Avail();
$customer->item3Avail();
$customer->item4Avail();
var_dump($_SESSION);

$item1Status = $_SESSION['item1'];
$item2Status = $_SESSION['item2'];
$item3Status = $_SESSION['item3'];
$item4Status = $_SESSION['item4'];


?>

Jquery:

var item1Status = <?php echo json_encode($item1Status); ?>;
var item2Status = <?php echo json_encode($item2Status); ?>;
var item3Status = <?php echo json_encode($item3Status); ?>;
var item4Status = <?php echo json_encode($item4Status); ?>;


if(item1Status == "sold"){
$("#picture1").css("opacity", "0.5");
$("#button1").css("visibility", "hidden");
$("#item1").css("opacity", "0.5"); 
} 

if(item2Status == "sold"){
$("#picture2").css("opacity", "0.5");
$("#button2").css("visibility", "hidden");
$("#item2").css("opacity", "0.5"); 
} 

if(item3Status == "sold"){
$("#picture3").css("opacity", "0.5");
$("#button3").css("visibility", "hidden");
$("#item3").css("opacity", "0.5"); 
} 

if(item4Status == "sold"){
$("#picture4").css("opacity", "0.5");
$("#button4").css("visibility", "hidden");
$("#item4").css("opacity", "0.5"); 
}

var item1Status = "<?php echo json_encode($item1Status); ?>"; — Flakes
– Flakes, Commented Jul 27, 2015 at 14:01

Juan Serrats · Accepted Answer · 2015-07-27 14:04:20Z

2

Nothing logically wrong, you just need to wrap those php echos with quotes:

var item1Status = "<?php echo json_encode($item1Status); ?>";
var item2Status = "<?php echo json_encode($item2Status); ?>";
var item3Status = "<?php echo json_encode($item3Status); ?>";
var item4Status = "<?php echo json_encode($item4Status); ?>";

answered Jul 27, 2015 at 14:04

Juan Serrats

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4 Comments

random_user_name Over a year ago

Not actually true. You don't need quotes, AND at least part of the issue is that the JSON does not get parsed before evaluated. Why bother json_encoding if you aren't going to $.parseJSON, but attempt to access directly?

Mitchell J Gardepe Over a year ago

I'm honestly not sure why I'm using json. I'm tried just echoing and ran into the same issue. Also tried with the quotes. Not sure what I'm doing wrong...

Mitchell J Gardepe Over a year ago

I'm checking it in firebug and the js variables seem to be being set. The css effect I set in the jquery still isn't happening. Am I somehow writing that part wrong?

Juan Serrats Over a year ago

The structure you're using seems to be fine: $("selector").css("property", "value"); Make sure you're entering those if statements and that you're using #id/.class correctly on the selector

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