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PHP Build a Basic PHP Website Building a Media Library in PHP Working with _GET Variables

Posted by Kylian van der Have
Kylian van der Have
10,124 Points

The browser says there is a Parse Error with the code, despite being it exactly as in this video.

The Browser error is this:

Parse error: syntax error, unexpected ';' in /home/treehouse/workspace/catalog.php on line 5

catalog.php code is this:

<?php
$pageTitle = "Full Catalog";

if ($_GET["cat"] == "books" {
  $pageTitle = "Books" ;
} else if ($_GET["cat"] == "movies" {
  $pageTitle = "Movies" ;
} else if ($_GET["cat"] == "music" {
  $pageTitle = "Music" ;
} 

include("inc/header.php"); ?>```

header.php code is adjusted:

```<ul class="nav">
                <li class="books"><a href="catalog.php?cat=books">Books</a></li>
                <li class="movies"><a href="catalog.php?cat=movies">Movies</a></li>
                <li class="music"><a href="catalog.php?cat=music">Music</a></li>
                <li class="suggest"><a href="suggest.php">Suggest</a></li>
            </ul>```

1 Answer

Jennifer Nordell
seal-mask
STAFF
.a{fill-rule:evenodd;}techdegree
Jennifer Nordell
Treehouse Teacher
Jennifer Nordell
.a{fill-rule:evenodd;}techdegree
Jennifer Nordell
Treehouse Teacher

Hi there, Kylian van der Have! It's not exactly like it is in the video. According to the code you've posted it looks like on each if statement you've forgotten the closing parenthesis:

You have:

if ($_GET["cat"] == "books" {
  $pageTitle = "Books" ;
} else if ($_GET["cat"] == "movies" {
  $pageTitle = "Movies" ;
} else if ($_GET["cat"] == "music" {
  $pageTitle = "Music" ;
}

But it needs to be:

if ($_GET["cat"] == "books") { // added ) here
  $pageTitle = "Books" ;
} else if ($_GET["cat"] == "movies") { // added ) here
  $pageTitle = "Movies" ;
} else if ($_GET["cat"] == "music") {  // added a ) here
  $pageTitle = "Music" ;
}

Hope this helps! :sparkles: