Encountering a "Parse error: syntax error, unexpected T_IF" while trying to use the return statement

I've encountered a parse error in some code I'm working on. When trying to create an if function outside of the $return variable and referencing it with a session, it works fine. However, I know this isn't considered "good" coding practice. How can I resolve this issue or improve my return value structure? Any help would be greatly appreciated.

$return = '';
    $return .='<div id="viewport">'. "\n";
    $return .= $ic . "\n";
    $return .='<div id="' . $wallID . '"></div>'. "\n"; 
    $return .='</div>'. "\n";                                  
    $return .= if( isset($coda) )'<div id="coda' . $wallID . '"></div>';. "\n";
    $return .='</div>'. "\n";                                   
    $return .='<div class="clearfix"></div>'. "\n";

 return $return;
phpparse-errorreturn

Answer №1

To ensure proper execution, always remember to include an if statement:

...
if( isset($coda) ) {
  $output .= '<div id="coda' . $wallID . '"></div>'. "\n";
}
$output .= ...
...

Answer №2

This particular code snippet could pose some issues

$return .= if( isset($coda) )'<div id="coda' . $wallID . '"></div>';. "\n";

it should be 

$return .= ((isset($coda)) ? '<div id="coda' . $wallID . '"></div>' : '')."\n";

To improve clarity, consider utilizing HEREDOC syntax

Answer №3

To improve readability, consider utilizing a ternary operator instead of an incorrectly used if block:

$result .= isset($coda) ? "<div id='coda{$wallID}'></div>\n" : '';

Alternatively, you can opt for a more readable if block as suggested by @Mat.

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