Opening a window in Javascript with the value of a PHP variable

Question

I would like to open a window with a 5 second delay when a button is clicked. I'm trying:

<script type="text/javascript">
function sample() {
   setTimeout(function() {
       window.open('<?php echo esc_attr(wpcoupon_coupon()->get_go_out_url()); ?>', '_self');
   }, 5000); 
}
</script>

which I call in the onclick attribute of the <button>:

<button class="ui right labeled icon button btn btn_secondary" onClick="sample();">
    <i class="copy icon"></i>
    <span><?php esc_html_e('Copy', 'wp-coupon'); ?></span>
</button>

The problem is that <?php echo esc_attr(wpcoupon_coupon()->get_go_out_url()); ?> doesn't return the correct value, and correct URL doesn't open.

What could be going wrong?

So what is the value written when the page loads, that is the value. View the page source. Issue would be with the PHP code. — epascarello
– epascarello, Commented Jun 21, 2018 at 11:55
This is mixture of JS and PHP; such a thing is not possible. You should try another approach. — ata
– ata, Commented Jun 21, 2018 at 12:00
Of course is such mixing possible, but it won't work as expected sometimes. In this case, you should explain which URL is opened and what you expect — Nico Haase
– Nico Haase, Commented Jun 21, 2018 at 12:03

Álvaro Romero · Accepted Answer · 2018-06-21 11:54:54Z

1

You need ajax to read that php variable asynchronously from javascript. Otherwise i think your question is better answered here:

Get variable from PHP file using JQuery/AJAX

answered Jun 21, 2018 at 11:54

Álvaro Romero

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1 Comment

Pauet Over a year ago

No knowledge about that.

mscdeveloper · Accepted Answer · 2018-06-21 14:16:43Z

0

Is the page address correct?

esc_attr(wpcoupon_coupon()->get_go_out_url())

For test your script in php file use construction:

<?php
   //php code
?>
<script>
function sample() {
   setTimeout(function() {
       window.open('<?php echo('https://google.com'); ?>', '_self');
   }, 5000);
}
</script>
<button class="ui right labeled icon button btn btn_secondary" onClick="sample();">test</button>
<?php
  //php code
?>

answered Jun 21, 2018 at 14:16

mscdeveloper

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3 Comments

Pauet Over a year ago

yes, if i echo esc_attr(wpcoupon_coupon()->get_go_out_url()) it shows example.com/go-to-store/1198 so is correct, but with the function it shows another url example.com/go-to-store/202.

mscdeveloper Over a year ago

Maybe use: var url='<? echo esc_attr(wpcoupon_coupon()->get_go_out_url()); ?>'; and : window.open(url, '_self');

Pauet Over a year ago

think php problem because with google.com it works correctly

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