Loading PHP Variable Into A Javascript Variable

Question

I'm building a batabase driven website which uses Javascript to play videos on my website using the YouTube API, but I can't load in the data I have extracted via PHP into the Javascript player.

My table is (Example codes only):

ID | One | Two
---------------
01 | 832 | 943

Here's the PHP code I use to get the data:

<?php
$code_sql = "SELECT * FROM codes WHERE id = '$url_id'";
$code_res = mysqli_query($con, $code_sql);
while($code = mysqli_fetch_assoc($code_res)){

    $code_1 = $code["one"];
    $code_2 = $code["two"];

};
?>

And Here's the Javascript I'm trying to load it into:

var $vid1 = <?php echo json_encode("$code_1"); ?>;
var $vid2 = <?php echo json_encode("$code_2"); ?>;
var videoIDs = [
    '$vid1',
    '$vid2'
];

When I load the page it gives me a Invalid Paramiters in the video player. Any ideas where I'm going wrong?

You're constantly overriding $code_1 and $code_2 in your php while loop, making them hold the last result only — baao
– baao, Commented Feb 23, 2017 at 9:51
And variables dont need to be in doublequotes json_encode($code_1) — RiggsFolly
– RiggsFolly, Commented Feb 23, 2017 at 9:52
@baao, cam you please explain what you mena by overriding? And how do I fix that? — Steve Gee
– Steve Gee, Commented Feb 23, 2017 at 9:56
Possible duplicate of Good practice to return DateTime from PHP to JS — Jadeye
– Jadeye, Commented Feb 23, 2017 at 11:15

B. Desai · Accepted Answer · 2017-02-23 10:06:24Z

0

Remove quotes from json_encode. Also remove single quote inside array

var $vid1 = <?php echo json_encode($code_1); ?>;
var $vid2 = <?php echo json_encode($code_2); ?>;
var videoIDs = [
    $vid1,
    $vid2
];

edited Feb 23, 2017 at 10:06

answered Feb 23, 2017 at 10:01

B. Desai

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2 Comments

Steve Gee Over a year ago

Thanks, but when I try this with quotes, it completely blocks out the video player. Any ideas why?

B. Desai Over a year ago

Remove double quotes from outside of php then try. See my edited answer

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