How to call a php variable, as a jQuery variable?

Question

I want to call the PHP variable "rate number" as "phpvalue" in jQuery. But the alert of "phpvalue" does not give any output and the alert of "num3" gives "NaN" as the output.

Here is my PHP code.

$jsqla = mysql_query("select id,name,rate_score,rate_number,video_image from products where genre='$genre' limit 0,5");

while($jrowa = mysql_fetch_assoc($jsqla)){
    $arate_num = $jrowa['rate_number'];
}

Here is my jQuery code.

$(function(){
    $(document).ready(function(e) {
        var $stars = $('.input-star-rate');

        $stars.bind('change', function() {

            var phpvalue = "<?php echo $jrowa['rate_score']; ?>";
            alert(phpvalue);

            var num3 = parseInt(phpvalue, 10);
            alert(num3);

        });
    });
});

Here is my HTML code.

<input class="rating form-control input-star-rate" id="<?php echo $rateid; ?>" name="rating" value="<?php echo $ratea; ?>" data-min="0" data-max="5" data-step="0.3" data-size="xs" style="display: none; text-align: center;"/>

Is the jQuery code also processed by PHP?

Thilo
– Thilo

2014-11-28 06:28:08 +00:00
Commented Nov 28, 2014 at 6:28 — Thilo
– Thilo, Commented Nov 28, 2014 at 6:28
possible duplicate of Use a PHP variable in JQuery

Nibin
– Nibin

2014-11-28 06:28:10 +00:00
Commented Nov 28, 2014 at 6:28 — Nibin
– Nibin, Commented Nov 28, 2014 at 6:28

srinath madusanka · Accepted Answer · 2014-11-28 06:44:38Z

3

try

  $(function(){
    $(document).ready(function(e) {
        var $stars = $('.input-star-rate');

        $stars.bind('change', function() {

            var phpvalue = "<?php echo $arate_num; ?>";
            alert(phpvalue);

            var num3 = parseInt(phpvalue, 10);
            alert(num3);

        });
    });
});

answered Nov 28, 2014 at 6:44

srinath madusanka

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