false74 Posted October 24, 2012 Share Posted October 24, 2012 Here is my problem/question: Say I have a class stored in a php file called 'myclass.php' and I have my main index php page 'index.php'. In that index.php page I include the 'myclass.php' file and create a variable as a new object of the class from myclass.php. However in the index page there are multiple php breaks <?php ?> <?php ?>, how can I use the variable that is of my class throughout the entire page? myclass.php: class MyClass { function MyClass() { ... } function foo() { ... } function bar() { ... } } index.php <html> <head> ... </head> <body> <?php include_once('myclass.php'); $thing = new MyClass(); ?> ... <?php $thing->foo(); ?> ... <?php $thing->bar(); ?> </body> </html> When trying to call $thing outside of the first php block I get an 'undefined variable' error. What is the scope of this variable, how can I have it shared amoungst all the php code blocks? Quote Link to comment Share on other sites More sharing options...
ManiacDan Posted October 24, 2012 Share Posted October 24, 2012 There's no reason why this isn't working the way you wrote it here. Are you creating $thing inside a function? Quote Link to comment Share on other sites More sharing options...
false74 Posted October 24, 2012 Author Share Posted October 24, 2012 Are you creating $thing inside a function? $thing is defined inside the first php code block, but cannot be accessed outside of that. Quote Link to comment Share on other sites More sharing options...
akphidelt2007 Posted October 24, 2012 Share Posted October 24, 2012 $thing is defined inside the first php code block, but cannot be accessed outside of that. Can you post your actual code. Because what you posted will work and you can access $thing outside the first code block. So your code must be doing something different. Quote Link to comment Share on other sites More sharing options...
false74 Posted October 24, 2012 Author Share Posted October 24, 2012 (edited) showdb.php <?php class DBManager { var $showthumbfile; var $dblocation; var $db; var $mediaroot; var $showtable; var $epsiodetable; function CityBeatDBManager() { $showthumbrootdir = "thumbs/"; } function connectDB() { try { $db = new PDO('sqlite:' . $dblocation); } catch( PDOException $e ) { die('Error connecting to show database'); } } //QUERIES function queryAllShowRows() { return $db->query('SELECT * FROM shows ORDER BY show_id DESC')->fetchall(PDO::FETCH_ASSOC); } } ?> index.php <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Test</title> <LINK REL=StyleSheet HREF="grey.css" TYPE="text/css" MEDIA=screen> </head> <body> <?php include_once("showdb.php"); $databse = new DBManager(); ?> <ul class="nav"> <li><a id="navbrowse" class="more" href="javascript:void(0);" onclick="toggleBrowse();">Our Programming</a></li> <li><a id="navabout" href="?page=about">About Us</a></li> </ul> </div> <div id="browse" class="close" tabindex=0 onclick="displayBrowse(false);" onblur="displayBrowse(false);"> <h1>Browse our programs</h1> <ul class="shownav"> <?php function listShows() { foreach($database->queryAllShowRows() as $row) { echo " <li> <a href='?show=" . $row['show_mediadir'] . "' class='show'>" . "<img src='" . $row['show_mediadir'] . "/mainthumb.jpg' />" . "<span>" . $row['show_title'] . "</span>" . "</a>" . "</li>"; } } listShows(); ?> </ul> </div> </div> <div id="wrapper"> <h1> <?php echo $databse->$dblocation; ?> </h1> </div> </body> </html> undefined variable when calling $database inside that foreach loop Edited October 24, 2012 by false74 Quote Link to comment Share on other sites More sharing options...
akphidelt2007 Posted October 24, 2012 Share Posted October 24, 2012 You can't access a variable like that within the scope of a function. Why are you wrapping that foreach loop in a function in the first place? Quote Link to comment Share on other sites More sharing options...
ManiacDan Posted October 24, 2012 Share Posted October 24, 2012 You also spelled $database wrong. Read up on PHP's variable scope Quote Link to comment Share on other sites More sharing options...
false74 Posted October 24, 2012 Author Share Posted October 24, 2012 You can't access a variable like that within the scope of a function. Why are you wrapping that foreach loop in a function in the first place? Even when I remove the function and do this: It still is undefined. <?php foreach($database->queryAllShowRows() as $row) { echo " <li> <a href='?show=" . $row['show_mediadir'] . "' class='show'>" . "<img src='" . $row['show_mediadir'] . "/mainthumb.jpg' />" . "<span>" . $row['show_title'] . "</span>" . "</a>" . "</li>"; } ?> Quote Link to comment Share on other sites More sharing options...
false74 Posted October 24, 2012 Author Share Posted October 24, 2012 (edited) Actually. I'm just a dumbass. I can't spell for anything... Sorry for your time. $databse .... $database My mistake Edited October 24, 2012 by false74 Quote Link to comment Share on other sites More sharing options...
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