Show or Hide Button with PHP condition

PHP
1

Hi guys, I ran into a coding problem. I want to show a button only if my $status variable changes to COMPLETE. I don’t know how to do this. Can anyone help me? Here’s what I’m working on:-

<?php echo $status; ?>

<button name="donwloadbtn" href="www.myweb.com/myfile.jpg" download style="visibility: hidden">
<i class="fa fa-download" aria-hidden="true"></i> 
Download</button>

I want if $status == COMPLETE then show $downloadbtn else hide $downloadbtn

2 
<?php
if ($status == 'COMPLETE'){
echo 'Your button code here';
}
3

You mean like this?

<?php if ($status == 'COMPLETE'){ 
echo '<-button type="button" name="donwloadbtn" href="www.myweb.com/myfile.jpg" download style="visibility: hidden"><-i class="fa fa-download" aria-hidden="true"></-i> Download</-button>';
 }
4

Yes that’s one way, alternatively it can be done like this:-

<?php if ($status == 'COMPLETE') : ?>
<button name="donwloadbtn" href="www.myweb.com/myfile.jpg" download style="visibility: hidden">
<i class="fa fa-download" aria-hidden="true"></i> 
Download</button>
<?php endif ?>

Here there is no echo just normal html because you exit the php tags, but the html only shows if the if condition that encloses it is true.

Etiher stye may be used depending on preference or where it’s used.

1 Like
5

Thanks. It worked smoothly.

6

I prefer changing the CSS class parameters and passing .hide {display: none;}


<?php
  $hide=" "; // default visible
   if ($status == 'COMPLETE') : 
      $hide="hide";
 ?>
<button
  class="<?= $hide ?>"
   name="donwloadbtn"
  href="www.myweb.com/myfile.jpg"
 >
  <i class="fa fa-download" aria-hidden="true"></i>
   Download
</button>

<?php endif ?>
7

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