PHP nested Ifs are not working

Question

I'm checking if Facebook users don't like a page. If so, the code displays the user's name and an invitation message to "Like it"; if not, it displays the "incentive landing page". However, I just get the first part: the code doesn't show the "incentive" when users hit the Like button. The code looks like this:

<?php if(!$like_status): ?> 
    <?php if ($me): ?>
      <div id="name">Hola <?php echo $me['first_name']; ?></div> //Everything works fine.  
    <?php endif; ?>
     // Display graphics... Everything works fine.
<?php else : ?>
     //Page with the incentive... This does not work when the nested if is present.
<?php endif; ?>

If I get rid of the nested if (the one that displays the user's name), everything works fine. Any ideas?

@Charles Sprayberry : is perfectly fine. @Byktor what is the value of the like_status; — Ibu
– Ibu, Commented Jul 19, 2011 at 3:19
LOL @ Charles :) Read this bro: php.net/manual/en/control-structures.alternative-syntax.php — SeanCannon
– SeanCannon, Commented Jul 19, 2011 at 3:20
Yes, I got to the alternative syntax. I still say the : should be {}. To me, that looks hideous :) — Charles Sprayberry
– Charles Sprayberry, Commented Jul 19, 2011 at 3:21
This is the recommended syntax for PHP when code is placed within HTML syntax in a view in most MVC frameworks. — SeanCannon
– SeanCannon, Commented Jul 19, 2011 at 3:23

SeanCannon · Accepted Answer · 2011-07-19 03:19:27Z

1

You're missing your semicolon after the endif's

<?php if(!$like_status): ?>   
    <?php if ($me): ?>
      // Display user's name... Everything works fine.  
    <?php endif; ?>
     // Display graphics... Everything works fine.
<?php else: ?>
     //Page with the incentive... This does not work when the nested if is present.
<?php endif; ?>

answered Jul 19, 2011 at 3:19

SeanCannon

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5 Comments

Byktor Over a year ago

I already tried the semicolon after the endif... didn't work. I'll recheck it, just in case, and let you know.... Rechecked it and it didn't work.

SeanCannon Over a year ago

I'm guessing there is an error in that code you're not showing us ;) Code works fine, just tested it:

$like_status = true;         if(empty($like_status)):             if(!empty($me)):                 echo 'name';             endif;             echo 'graphics';         else:             echo 'incentive';         endif;

outputs "incentive"

Byktor Over a year ago

I added the three critical lines.

Byktor Over a year ago

No spamimng, @Ibu... I just wanted @AlienWebguy to see the result. I deleted the link a few minutes later.

Byktor Over a year ago

Thank you everybody. I just rewrote the code no nested if and that solved the problem.

Byktor · Accepted Answer · 2011-07-26 00:48:52Z

0

In case somebody is facing the same problem, I found the solution: for some reason IE 7-8 doesn't like the variable $me and the call to the API('/me') in the same context. I just changed the name of the variable and everything works fine in FF, Chrome, and IE 7-9.

answered Jul 26, 2011 at 0:48

Byktor

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