PHP – Nested If Statements Issue

5 comments

1. There’s no such thing as an “if loop”. 🙂

   Since you’re basically only repeating the same condition again and again, you can logically restructure it to this:

   if (function_exists('dynamic_sidebar')) {
       switch ($layout) {
           case 'one' :
               dynamic_sidebar('Sidebar 1');
               break;
           case 'two' :
               dynamic_sidebar('Sidebar 1');
               dynamic_sidebar('Sidebar 2');
               break;
           case ...
       }
   }
   

   This should also take care of the syntax/typo problems that are causing your script to misbehave.

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2. Edit: I just did a test and it looks like this answer is wrong. I was able to mix colon and brace if and while syntax together. However, the manual does state:

   ‘Note: Mixing syntaxes in the same
   control block is not supported.’

   So I’m a bit confused about this.

   ---

   PHP If statements have two types of syntax, one with braces, and one with colons.

   if () {
   
   }
   

   and

   if ():
   
   endif;
   

   There are also similar syntaxes for while, for, foreach etc. But you aren’t allowed to mix the braces with the colon syntaxes. So you need to either change your code to

    if($layout=="one") { 
   
         if ( !function_exists('dynamic_sidebar') || !dynamic_sidebar('Sidebar 1') ) {
         }
   
    } elseif($layout=="two") {
   
         if ( !function_exists('dynamic_sidebar') || !dynamic_sidebar(' Sidebar 1') ) {
         }    
   
         if ( !function_exists('dynamic_sidebar') || !dynamic_sidebar('Sidebar 2') ) {
         }     
   
    } elseif($layout=="three") { (...)
   

   or

    if($layout=="one"):
   
         if ( !function_exists('dynamic_sidebar') || !dynamic_sidebar('Sidebar 1') ) :  
         endif; 
   
    elseif($layout=="two"):
   
         if ( !function_exists('dynamic_sidebar') || !dynamic_sidebar(' Sidebar 1') ) :  
         endif;    
   
         if ( !function_exists('dynamic_sidebar') || !dynamic_sidebar('Sidebar 2') ) :  
         endif;     
   
    elseif($layout=="three"):
     (...)
   
    endif;
   

   You can read more about it at http://php.net/manual/en/control-structures.alternative-syntax.php

   Edit: How silly of me I didn’t notice this. Like the other answers say, you have used a single equals sign instead of two to test for equality.

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3. I agree with Michael and also, I’m not sure how your functions are set up but usually I’d see something more similar to

   if($layout == "one" && function_exists('dynamic_sidebar'))
   {
       dynamic_sidebar('Sidebar 1');
   }
   

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4. Looks like you mean to use the equality operator == where you’re using the assignment operator =

   if ($layout == "one")
   // etc
   

   By the way, consider also using the more common if/else PHP syntax. Rather than using the elseif and endif use this syntax. Brace according to your conventions and preferences.

   if (condition)
   {
     // code for condition
   }
   else if (condition)
   {
     // code
   }
   else
   {
     // else case code
   }
   

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5. First, this is wrong:

   if ($layout="one")
   

   Always use == or === in an if statement (same for elseif) because = actually assigns the value. This is probably part of what’s causing your bug.

   Better yet, in this particular situation, use the switch statement

   Third, you don’t have to keep copying !function_exists('dynamic_sidebar') just put it in once. It should look like this:

   if function_exists('dynamic_sidebar')
   {
       switch ($layout)
       {
           case 'one':
           dynamic_sidebar('Sidebar 1');
           break;
   
           case 'two':
           dynamic_sidebar('Sidebar 1');
           dynamic_sidebar('Sidebar 2');
           break;
   
           // ...
       }
   }
   

   Even better still, change the whole thing to a number and just do a for loop:

   if function_exists('dynamic_sidebar')
   {
       for ($i = 0; $i < $sidebarCount; ++$i)
       {
           dynamic_sidebar("Sidebar $i");
       }
   }
   

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