Output formatted version of variable if it has a value, blank otherwise

I'm looking to add an if statement to display a title if something is in a variable but not 100% sure how to go about it.

My current coding s开发者_运维问答hows:

<?php echo $quote->getmessage(); ?>

But I would like it to show a title and the content of the message if there is content in the variable. If there is nothing within the variable I don't want to show anything.

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IMHO it's better to be as verbose as necessary so that the code is easily readable and can be extended without totally rewriting it:

<?php 
$message = $quote->getmessage();
if (!empty($message)) {
    echo "Title!";
    echo htmlspecialchars($message);
}
?>

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Use php's isset function to check if a variable is set, and empty to see if it's empty.

For instance

<?php $a = $quote->getmessage();if (!empty($a)) echo $a; ?>

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<?php echo ($quote->getmessage() == "") ? "" : "Title <br />".$quote->getmessage(); ?>

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Use a ternary operator:

<?php 
   $quote_var = $quote->getmessage(); 
   echo ($quote_var != null)?$quote_var:'NOTHING!'; 
   //displays 'NOTHING' if the variable is null
?>

---

<?php 
$mssg = $quote->getmessage();
echo (!empty($mssg))?$mssg:'';
?>

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If you want your line to be concise, then I would advise this syntax:

<?php  $msg = $quote->getmessage()  AND  print "<h6>title</h6>$msg";  ?>

The AND has a lower precedence than the assignment (but extra whitespace or braces make that more readable). And the second part only gets executed if the $msg variable receives any content. And print can be used in this exporession context instead of echo.

---

<?php 
     if(isset($quote->getmessage())
     {
         echo "My Title";
     }
?>