PHP Variable Troubles - $_GET

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  • Cobain
    New Member
    • Mar 2011
    • 5

    PHP Variable Troubles - $_GET

    Hi,

    I am building my new website and I want to confine the menu to one file, but make it so that the user can see what page he/she is on. For this I came up with the following code:
    Code:
    <?php $page = $_GET['page']; ?>
    <?php $page1 = 'page1'; $page2 = 'page2'; $page3 = 'page3'; $page4 = 'page4';?>
    Where $pagenumber = 'inputvalue' in $_GET from the URL.
    The menu is then below, as:
    Code:
    <?php if ($page == $page1) {echo "<div id=\"menuon\"><a href=\"/index.php?page=$page1\">Homepage</a></div>";} else {echo "<div id=\"menuoff\"><a href=\"/index.php?page=$page1\">Homepage</a></div>";} ?>
    <?php if ($page == $page2) {echo "<div id=\"menuon\"><a href=\"/index.php?page=$page2\">Some Page</a></div>";} else {echo "<div id=\"menuoff\"><a href=\"/index.php?page=$page2\">Some Page</a></div>";} ?>
    This works absolutely fine until I go to the top-level domain where there is no "$_GET['page']" input for the script to pick up on. I need it to recognise a non-input as the homepage, so that I don't have to redirect it to " /index.php?page= page1 ".
    I have tried a few wacky and wonderful ways of doing this but none have been successful. Does anybody have any solutions or suggestions?

    Many thanks,
  • abhishekmiet
    New Member
    • Mar 2011
    • 8

    #2
    I hope the following bunch of code will help you...

    Code:
    <?php
    if(!isset($_GET['page'])){
     $page='page1';
    }
    else{
    $page=$_GET['page'];
    }
    ?>
    Last edited by Dormilich; Mar 29 '11, 05:47 AM. Reason: please use [CODE] [/CODE] tags when posting code

    Comment

    • Cobain
      New Member
      • Mar 2011
      • 5

      #3
      Hi abhishekmiet,

      Thanks for that, works a treat! I think that PHP is great, but it is sooo annoying when you forget things!

      Thanks again!

      Comment

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