CODEWITHSUNDEEP
phpjavascript
Kichu
Kichu

Reputation: 3267

How can i call a javascript function using php code with php variable passing

This is my php code:

<?php
  $typeid = 65;
  $subledgerid = 'subledger'.$typeid;
  $loadledger = 'loadledgers';
   ?>
   <script type="text/javascript">view_subledger('<?php echo $subledgerid;?>',<?php echo $typeid;?>,'<?php echo $loadledger;?>');</script>

This is not calling the view_subledger().

But when i call something like this it will work fine:

<script type="text/javascript">view_subledger('subledger65',65,'loadledgers');</script>

How can i call this?

Upvotes: 1

Views: 104

Answers (3)

ioseb
ioseb

Reputation: 16951

Here is the safest way to do this:

<?php

$typeid = 65;
$subledgerid = 'subledger'.$typeid;
$loadledger = 'loadledgers';

?>

<script type="text/javascript">
view_subledger.apply(window, <?php print json_encode(array(
  $typeid,
  $subledgerid,
  $loadledger
)); ?>);
</script>

Which generates following code:

<script type="text/javascript">
view_subledger.apply(window, [65,"subledger65","loadledgers"]);
</script>

json_encode() will ensure that variables are escaped properly and with .apply() method you can pass parameter array to JS function.

Another suggested version:

<?php
$typeid = 65;
$subledgerid = 'subledger'.$typeid;
$loadledger = 'loadledgers';
$param_str = implode(', ', array_map('json_encode', array(
  $typeid,
  $subledgerid,
  $loadledger
)));
?>

<script type="text/javascript">
view_subledger(<?php print $param_str; ?>);
</script>

Generates following:

<script type="text/javascript">
view_subledger(65, "subledger65", "loadledgers");
</script>

Upvotes: 2

Alexander Taver
Alexander Taver

Reputation: 474

This is all the code you have in the file? All this looks OK and must work as far as $subledgerid, $typeid,$loadledger are correct. But what about this line?

  <?php } ?>

It must crash the script

Upvotes: 0

Adam
Adam

Reputation: 1264

Why is <?php } ?> after the script? Should be like:

<?php
$typeid = 65;
$subledgerid = 'subledger'.$typeid;
$loadledger = 'loadledgers';
?>
<script type="text/javascript">view_subledger('<?php echo $subledgerid;?>',<?php echo $typeid;?>,'<?php echo $loadledger;?>');</script>

Upvotes: 0

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