Undefined Variable in PHP

Question

I keep getting an undefined variable error on lines 14 and 28 of this code (The $pageTitle = $cfstory["location"] and the <h3 id=blog-top-h3"> lines): I have attached the 2 relevant php files below. If anyone could help me understand why this variable is undefined I'd greatly appreciate it.

This file is campfire-blog.php in case it matters:

<?php include("inc\campfire_arr.php");

  if (isset($_GET["location"])){
    $cfstoryname = $_GET["location"];
    if (isset($cfstories[$cfstoryname])){
      $cfstory = $cfstories[$cfstoryname];
    }
  }
  // if(!isset($cfstory)){
  //   header("Location:camp-fire.php");
  //   exit();
  // }

$pageTitle = $cfstory["location"];
?>

<!DOCTYPE html>
<html>
<?php include('inc/header.php');?>
    <header class="contact-header">
      <?php include("inc/name_title_header.php");?>
      <?php include("inc/menu.php");?>
    </header>
    <div id="wrapper">
      <!-- <div class="breadcrumb"><a href="camp-fire.php">Campfire Stories</a> &gt; <?php echo $cfstory ["name"]; ?></div> -->
      <div id="main-left">
        <div class="campfire">
          <h3 id="blog-top-h3"> <?php echo $cfstory["location"];?> </h3>

        </div>
      </div>
    </div>
    <footer class="contact-footer">
      <?php include("inc/footer.php"); ?>
    </footer>
    <?php include('inc/js_scripts.php');?>

This one is camp-fire.php:

<?php include("inc/campfire_arr.php");?>

<!DOCTYPE html>
<html>
<?php
  $pageTitle="Campfire Stories";
  include("inc/header.php");
?>
  <header class="contact-header">
      <?php include("inc/name_title_header.php");?>
      <?php include("inc/menu.php");?>
  </header>
    <div id="wrapper">
      <div id="main-left">
        <div class="campfire">
          <h3>Campfire Stories</h3>
          <ul class="cts">
            <?php foreach($cfstories as $cfstoryname => $cfstory)
              echo display_campfire_html($cfstoryname, $cfstory);
            ?>
          </ul>
        </div>
      </div>
    </div>
    <footer class="contact-footer">
      <?php include("inc/footer.php"); ?>
    </footer>
    <?php include('inc/js_scripts.php');?>

This one is inc/campfire_arr.php:

<?php
  function display_campfire_html($cfstoryname,$cfstory){
    //build HTML output here
    $output = "";
    $output = $output . "<li>";
    $output = $output .'<a href="campfire-blog.php?php='.$cfstoryname.'">';
    $output = $output ."<figure>";
    $output = $output .'<img src="'.$cfstory["img"] .'" alt="'.$cfstory["name"].'">';
    $output = $output ."<figcaption>".$cfstory["location"]."</figcaption>";
    $output = $output ."</figure>";
    $output = $output ."</a>";
    $output = $output ."</li>";

    return $output;
  }

  $cfstories = array();
  $cfstories["clearwater-lake"] = array(
    "img" => "../img/cl-onf.jpg",
    "name" => "Clearwater Lake, Ocala National Forest",
    "location" => "Clearwater Lake"
  );
  $cfstories["denali"] = array(
    "img" => "../img/denali.jpg",
    "name" => "Denali National Park, Alaska",
    "location" => "Denali National Park"
  );
  $cfstories["arches"] = array(
    "img" => "../img/arch.jpg",
    "name" => "Clearwater Lake, Ocala National Forest",
    "location" => "Clearwater Lake"
  );
  $cfstories["yosemite"] = array(
    "img" => "../img/yosemite.jpg",
    "name" => "Yosemite National Park, California",
    "location" => "Yosemite National Park"
  );
?>

Answer 1 · 2015-05-16T08:10:01Z

May 16, 2015 8:10am

Hi John,

Your code was in need of some markdown love so I fixed that up for you, if in the future you need to post code again I recommend you read Posting Code to the Forum which explains how to use markdown.

The reason you're getting a undefined variable message is because the link you have setup within your display_campfire_html function isn't correct, at the moment your query string is php but your code is looking for location. Simply change this and your code will behave as expected for known values in the $cfstories array.

$output = $output .'<a href="campfire-blog.php?location='.$cfstoryname.'">';

On another note your technique for concatenating the $output string is verbose, what I mean by that is PHP offers a helpful shortcut in which we can use .= to append new string data onto an existing string without the need to call the variable first and assign it back to the same variable which can get messy, instead we can do the below.

<?php

function display_campfire_html($cfstoryname, $cfstory) {
  $output  = "";
  $output .= "<li>";
  $output .= '<a href="campfire-blog.php?location='.$cfstoryname.'">';
  $output .= "<figure>";
  $output .= '<img src="'.$cfstory["img"] .'" alt="'.$cfstory["name"].'">';
  $output .= "<figcaption>".$cfstory["location"]."</figcaption>";
  $output .= "</figure>";
  $output .= "</a>";
  $output .= "</li>";

  return $output;
}

Hope that helps.

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John McKinney

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Undefined Variable in PHP

1 Answer

Chris Shaw

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John McKinney

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