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Comments (33)

  1. Sorry im having an error in the jquery. the error is code is 'document' is not defined. [no undef] pls i need your help.

  2. Hello thenewboston. I am looking for a tutorial that will explain how to run a mysql query on a *.php file and then showing the result on a regular HTML web page. What video/tutorial would you suggest? Or may be there is a better approach that I'm not aware of?
    Excellent tutorial channel that you have!!!

  3. What am i doing wrong?
    When I do this, and save it as "1.php" or "index.php" and tell it to open with Google Chrome, it just shows the code below instead of just the word "test".
    <?php
    echo 'test';
    ?>

  4. what I must change in next php code <?php
    // Main menu items
    $mainMenu['Home']      = 'index.php';
    $mainMenu['Projects']  = 'strana/projects.php';
    $mainMenu['About us']  = '/strana/about.php';

    // Sub menu items
    $subMenu['Projects']['Product-1'] = 'strana/product1.php';
    $subMenu['Projects']['Product-2'] = 'strana/product2.php';

    $subMenu['About us']['Staff-1']   = 'strana/staff1.php';

    ?>

    <?php
    class maxNavigation{
       
       function showMenu(){
          global $mainMenu,$subMenu;
          
          $actualPage = $_SERVER['PHP_SELF'];
          $actualPath = $_SERVER['REQUEST_URI'];
          
          $actualPageName = basename($actualPage);
          //echo $page;
          
          //echo "$actualPage <br/> $actualPath";
          $actMenu = '';
        foreach ($mainMenu as $menu => $link) {
       if ($link == $actualPageName) $actMenu = $menu;
       if (isset($subMenu[$menu])){
          foreach ($subMenu[$menu] as $menuSub => $linkSub) {
            if ($linkSub == $actualPageName) $actMenu = $menu;
          }
       }
       }

       // Now display the menu
       foreach ($mainMenu as $menu => $link) {
         $class = ' class="mainMenuLink" '; 
         if ($actualPageName == $link) $class=' class="mainMenuLinkSelected" '; 
         
      echo '<a'.$class.'href="'.$link.'">'.$menu.'</a>';
      
      if ( ($actMenu == $menu) && (isset($subMenu[$menu])) ){
          foreach ($subMenu[$menu] as $menuSub => $linkSub) {
                     $class = ' class="subMenuLink" '; 
                 if ($actualPageName == $linkSub) $class=' class="subMenuLinkSelected" '; 
            echo '<a'.$class.'href="'.$linkSub.'">'.$menuSub.'</a>';
          }
       }
     }
          
          
       }

    }

    $navi = new maxNavigation();
    $navi->showMenu();

    ?>
    if my php web page example in second dir like is name "strana" where is example  page about.php. Width  this code I get  error 
    The requested URL /strana/about.php was not found on this server.

  5. Say you were going to have a user login control and it held two text boxes and a button, as well as the two labels ofc, do you create this within the php tags or is this created seperate from the php but then the code to make them work (send information ot the database and allow log in or not allow)?

  6. Hi very nice!!
    Can you tell me how to echo PHP, but with design and styles?
    How to add html and styles into my PHP echo?

  7. Thanks you dude.I thinked this will be stupid because it's not bucky,but now I see it's even better than buckys

  8. Well, 10 episodes. I managed to install XAMPP and create a empty file with php extension and with a echo inside.

    That felt productive.

  9. start XAMPP and start apache & mysql. make sure you run localhost/filename.php on your browser. change "filename" to the filename you have. watch videos 3-5

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