Why this PHP is not working in Javascript?

Question

I am trying to set cookies using javascript. but PHP is not working in javascript code. I can alert first value compprd but evrything else is just not working.

<th><input type="button" onClick="setting_my_first_cookie(<?php echo $compprd; ?>)" value="compare">

On click I run this Javascript function:

<script type="text/javascript"> 

function setting_my_first_cookie(compprd) {
alert("welcome"+compprd);
<?php 
    $compcount = 5;

    if (!isset($_COOKIE['compareid_1'])){$comparid = 'compareid_1'; $compcount = 1;}
    elseif (!isset($_COOKIE['compareid_2'])){$comparid = 'compareid_2';$compcount = 2;  }
    elseif (!isset($_COOKIE['compareid_3'])){$comparid = 'compareid_3';$compcount = 3;  }
    elseif(!isset($_COOKIE['compareid_4'])){$comparid = 'compareid_4'; $compcount = 4;  }

    if($compcount <= 4){
    ?>
var compid = <?php echo $comparid;?>
    //alert("create cookie"+compid);
var d = new Date();
    d.setTime(d.getTime() + (30*24*60*60*1000));
    var expires = "expires="+d.toUTCString();   
        document.cookie = compid + "=" + compprd + "; " + "expires; path=/; domain=.domain.com";
    <?php } ?>
}
</script>

Whats the problem here? How to use PHP in javascript?

UPDATE:

here is the code right now I am seeing using View Source:

<script type="text/javascript"> 
function setting_my_first_cookie(compprd) {
alert("welcome"+compprd);
var compid = compareid_1;
    alert("create cookie"+compid);
var d = new Date();
    d.setTime(d.getTime() + (30*24*60*60*1000));
    var expires = "expires="+d.toUTCString();   
        document.cookie = compid + "=" + compprd + "; " + "expires; path=/; domain=.mywebsite.com";
    }
</script>

You can't use php in js like that, only case is passing variables like you did. Jquery is executing in browser and php on server. — Sojtin, Jan 18 '16 at 08:46
@Sojtin php can be used to control your javascript. In above code Rock Sena is trying to control the 'compid' value using php which is still executen in server. — FrozenFire, Jan 18 '16 at 08:55
@Rocky Sena are you trying to get values from javascript? If you are retrieving any variable from javascript in php, you need to use AJAX or make xmlHttpRequest or similar methods. — FrozenFire, Jan 18 '16 at 09:04
I fetch data using PHP and then I pass them to function using onclick. I dont know how to use AJAX to do it. — Rocky Sena, Jan 18 '16 at 09:07
It's trying to run some PHP code to read cookies and output some data to the browser when a button is clicked. It is a duplicate of http://stackoverflow.com/questions/13840429/what-is-the-difference-between-client-side-and-server-side-programming (the fact there is also a typo which means the PHP wouldn't output the right data is a side issue). — Quentin, Jan 18 '16 at 09:20
I don't care about votes... I just found answer. Thank You Marcos Pérez Gude and FrozenFire — Rocky Sena, Jan 18 '16 at 09:20

score 1 · Answer 1 · answered Jan 18 '16 at 09:15

Take a look at this line:

 var compid = <?php echo $comparid;?>

It will render as this:

 var compid = compareid_1

But you need to render this:

 var compid = "compareid_1";

So you need to change the line to this:

 var compid = "<?php echo $comparid;?>";

Note the ; and the quotes "

FrozenFire · Accepted Answer · 2016-01-18T09:14:41.810

0

You've missed the (;) in var compid = <?php echo $comparid;?> it should var compid = <?php echo $comparid;?>;

Edit:

You need to change var compid = compareid_1; to var compid = 'compareid_1'; in php it will be var compid = '<?php echo $comparid;?>';

edited Jan 18 '16 at 09:14

answered Jan 18 '16 at 08:52

FrozenFire

671
14
28

I am getting this error in console: `Uncaught ReferenceError: compareid_1 is not definedsetting_my_first_cookie @ (index):665onclick @ (index):802 Navigated to http://mywebsite.com/websiteurl/` – Rocky Sena Jan 18 '16 at 09:00
Can you show your js code using view page source? Since php will modify it when retrieved on browser. – FrozenFire Jan 18 '16 at 09:07
I update the code. Please check it – Rocky Sena Jan 18 '16 at 09:12

Why this PHP is not working in Javascript?

2 Answers2