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PHP

error message after commenting

My code :

<?php

  // This is my name 
  $name = "Eph ";

  // $location = "Orlando, FL";
  $full_name = "Mike The Frog";
  $name = $full_name;


?>

<!DOCTYPE html>
<html>
  <head>
    <meta charset=utf-8>
    <title><?php echo $name ?>| Treehouse Profile</title>
    <link href="css/style.css" rel="stylesheet" />
  </head>

  <body>
    <section class="sidebar text-center">
      <div class="avatar">
        <img src="img/avatar.png" alt="Mike The Frog" title="<?php echo $name ?>">
      </div>
      <h1><?php echo $name ?></h1>
      <p><? echo $location ?></p>
      <hr />
      <p>Welcome to PHP Basics!</p>
      <hr />
      <ul class="social">
        <li><a href=""><span class="icon twitter"></span></a></li>
      </ul>
    </section>
    <section class="main">
      <p>Let's Get Started!</p>
      <p><?php echo "Hello World" ?></p>
    </section>
  </body>
</html>

After I commented the $location variable this error message shows up

Notice: Undefined variable: location in /home/treehouse/workspace/index.php on line 27

2 Answers

Erik S.
Erik S.
9,789 Points

This line is missing a php:

<p><? echo $location ?></p>

<p><?php echo $location ?></p>
Jeff Lemay
Jeff Lemay
14,267 Points

You're trying to echo a variable that does not exist. You can write a conditional to check if the variable exists and then echo it if so:

<?php
<p><?php if($location) { echo $location; } ?></p>
?>

You'd have to use isset() for that.

if(isset($location)) { echo $location; }
Jeff Lemay
Jeff Lemay
14,267 Points

Thank you for that, Jason. You also wouldn't wrap the paragraph in php tags...

<p><?php if(isset($location)) { echo $location; } ?></p>