Assigning PHP Variables To Javascript Variables Not Working

Question

I am currently having trouble with this. I would like to make one of my variables in Javascript have a PHP value. Here is what I mean:

<script>
    JSvariable = <?php echo $PHPvariable; ?>;
</script>

For some reason that is not working. Here is my full (snippet) of code:

<script>
    currentreplyid = <?php echo $allpostcomments[$key]['replyid']; ?>;
    $('#parentcommentholder').val(currentreplyid);
</script>

I am sure it is some stupid mistake, but I can not seem to find it! What is the problem? Thank you!

PS #parentcommentholder is an input field, and it just had the value 0 after the field is supposed to of been changed.

Here is some source:

<?php
$postcommentsquery = "SELECT * FROM comments WHERE parent = :parent AND postid = :postid ORDER BY datecreated DESC";
$postcommentsparams = array(':parent' => $allreplies[$key]["postid"],
                            ':postid' => $postid);
try{
    $postcommentsstmt = $connection->prepare($postcommentsquery);
    $postcommentsresult = $postcommentsstmt->execute($postcommentsparams);
}
catch(PDOException $ex){
    echo ("Failed to run query: " . $ex->getMessage());
}
$allpostcomments = $postcommentsstmt->fetchAll();
foreach ($allpostcomments as $key => $value) {
?>
    <script>
    var currentreplyid = <?php echo $allpostcomments[$key]['replyid']; ?>;
    $('#parentcommentholder').val(currentreplyid);
    </script>

    <input id="parentcommentholder"></div>

You forgot to use the var keyword. Are there any errors in the browser's console? — Jay Blanchard
– Jay Blanchard, Commented Jun 23, 2015 at 14:46
Have you made sure $allpostcomments contains the correct array keys (e.g. var_dump($allpostcomments) ) to ensure the keys exists — Danny Broadbent
– Danny Broadbent, Commented Jun 23, 2015 at 14:47
I tried the var keyword, but that did not fix anything. Yes the key is correct, and in the console there are 0 errors. — Michael Jones
– Michael Jones, Commented Jun 23, 2015 at 14:48
what result do you get when you do var_dump($allpostcomments); ? placed after $allpostcomments = $postcommentsstmt->fetchAll(); — Funk Forty Niner
– Funk Forty Niner, Commented Jun 23, 2015 at 15:25

Manwal · Accepted Answer · 2015-06-23 14:54:46Z

3

Don't forgot for give quotes ' or ". Use following:

<script>
    var JSvariable = '<?php echo $PHPvariable; ?>';
    //or
    var JSvariable = "<?php echo $PHPvariable; ?>";
</script>

Reason: If php variable contains string and if while assigning it to javascript variable we shall not give quote like:

<?php $PHPvariable = 'String';?>
var JSvariable = <?php echo $PHPvariable; ?>;

Will transform into :

var JSvariable = String;//which will give error in javascript

But this will work fine if PHP variable contains a numeric value like:

<?php $PHPvariable = 2;?>
var JSvariable = <?php echo $PHPvariable; ?>;

Will transform into :

var JSvariable = 2;//which will work perfect

Complete code should be:

<script>
    var currentreplyid = "<?php echo $allpostcomments[$key]['replyid']; ?>";
    //or if you are sure your variable contains int value
    var currentreplyid = parseInt("<?php echo $allpostcomments[$key]['replyid']; ?>");
    $('#parentcommentholder').val(currentreplyid);
</script>

edited Jun 23, 2015 at 14:54

answered Jun 23, 2015 at 14:46

Manwal

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1 Comment

Martijn Pieters Over a year ago

Comments are not for extended discussion; this conversation has been moved to chat.

Danny Broadbent · Accepted Answer · 2015-06-23 16:01:29Z

0

Try the below instead of using javascript (as I don't think you need it):

<?php
$postcommentsquery = "SELECT * FROM comments WHERE parent = :parent AND postid = :postid ORDER BY datecreated DESC";
$postcommentsparams = array(':parent' => $allreplies[$key]["postid"],
                            ':postid' => $postid);
try{
    $postcommentsstmt = $connection->prepare($postcommentsquery);
    $postcommentsresult = $postcommentsstmt->execute($postcommentsparams);
}
catch(PDOException $ex){
    echo ("Failed to run query: " . $ex->getMessage());
}
$allpostcomments = $postcommentsstmt->fetchAll();
foreach ($allpostcomments as $key => $value) {
?>
<input id="parentcommentholder" value="<?php echo ((int)$allpostcomments[$key]['replyid']>0) ? $allpostcomments[$key]['replyid'] : 0; ?>" />
<?php
}
?>

If your defiantly sure $allpostcomments[$key]['replyid'] is bringing back a value, this should work without any issues.

answered Jun 23, 2015 at 16:01

Danny Broadbent

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4 Comments

Michael Jones Over a year ago

Hello, thank you for the reply! Thing is though, I really do need the variables :)

Danny Broadbent Over a year ago

Have you tried putting script below input and putting a blank value="" on the input

Michael Jones Over a year ago

No, sorry. I need to grab the value in Javascript from PHP because of where I need it. That is sadly not going to work. It is not THAT simple (I wish it was :D).

Danny Broadbent Over a year ago

If this input is also in a foreach statement make sure the id is set as a unique id for each result

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