Can we put php in jquery code?

Question

I want to use php script in jquery code. Here I put php script in my jquery but my jquery function is not working then. Is it right way to put php script in a jquery function.

<script type="text/javascript">

    $(function() {

        var count2=1;
        $('a#addTelefono').click(function() {
            count2 +=1;

            $('<p><div style="float:left;width: 100%;"><select name="product_id[]" ><?php $sql=mysql_query("SELECT * FROM tbl_product"); while($res=mysql_fetch_array($sql)){ echo "<option value='".$res['id']."'>".$res['product_name']."</option>";}?></select><input type="text" name="discount[]" placeholder="discount ' + count2 + '" style="margin-left: 8px;" id="discount_' + count2 + '" />%<a href="#" class="remove" id="elimina"><img src="images/cross.png"></a></div></p>').fadeIn("slow").appendTo('#extendTelefono'); 

                i++;    
                return false;           
        });
        //fadeout selected item and remove
        $('.remove').live('click', function() {
            $(this).parent().fadeOut(300, function(){ 
                $(this).remove();
                return false;
            });
        });

    });
    </script>

If you intend to run the script on client side (what I assume you want to), it cannot work, because PHP is a server-side language, not a client-side language. — Abrixas2
– Abrixas2, Commented Jun 16, 2014 at 7:12
use jquery .post.. here is the link api.jquery.com/jquery.post — Eddie
– Eddie, Commented Jun 16, 2014 at 7:13
@Abrixas2: You are partially right. This script will work if he have this JS code inside a PHP as inline JS. If the script is placed in an external JavaScript file, with .js extension, it will not work. — Sarvap Praharanayuthan
– Sarvap Praharanayuthan, Commented Jun 16, 2014 at 7:13
PHP inside a JS file (not parsed by PHP engine) will not execute. — Sarvap Praharanayuthan
– Sarvap Praharanayuthan, Commented Jun 16, 2014 at 7:16

fdehanne · Accepted Answer · 2014-06-16 07:16:38Z

2

You have to make it in two steps and separate your php code from jQuery for readability :

$sql = mysql_query("SELECT * FROM tbl_product");
while($res=mysql_fetch_array($sql))
{
    $contents .= "<option value='".$res['id']."'>".$res['product_name']."</option>";
}


$('<select name="product_id[]" ><?php echo($contents); ?></select> .....');

Also, be carreful of caracters you have in $res['product_name'] : It can fail if you have a quote in your product name (so you must escape it).

Moreover, mysql_* is deprecated, see mysqli_* or PDO.

answered Jun 16, 2014 at 7:16

fdehanne

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fdehanne Over a year ago

Use console to see errors (All browser has one). In chrome : developer.chrome.com/devtools/docs/console In Firefox : developer.mozilla.org/fr/docs/Outils/Web_Console

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