Simonvetterli
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Everything posted by Simonvetterli
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Hi at All I have the problem, that I do not know, how to get the value from the selected option in my dropdown list. Hope someone can help me. I have the follow code: <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>File Uploader</title> <link href="css/reset.css" rel="stylesheet" type="text/css" /> <style type="text/css"> body { margin: 10px; font: 62% Tahoma, Arial, sans-serif; } #main_container{ font-size: 1.4em; } h2 { font-size: 2em; padding-bottom: 20px; } </style> <link href="css/ui-lightness/jquery-ui-1.8.14.custom.css" rel="stylesheet" type="text/css" /> <link href="css/fileUploader.css" rel="stylesheet" type="text/css" /> <script src="js/jquery-1.6.2.min.js" type="text/javascript"></script> <script src="js/jquery-ui-1.8.14.custom.min.js" type="text/javascript"></script> <script src="js/jquery.fileUploader.js" type="text/javascript"></script> </head> <body> <div id="main_container"> <h2>Bilder Girls Uploader</h2> <p> </p> <p> <?php mysql_connect("localhost:8888", "root", "root") or die("Connection Failed"); mysql_select_db("mydb")or die("Connection Failed"); $query = "SELECT idgirls,name FROM girls"; $result = mysql_query($query); ?> <form name="form1" method="POST" action="<?=$PHP_SELF?>"> <select name="select1" onchange="this.form1.submit();" method="post"> <?php while ($line = mysql_fetch_array($result, MYSQL_ASSOC)) { ?> <option value="<?php echo $line['idgirls'];?>"> <?php echo $line['name'];?> </option> <?php } ?> </select> </form> </p> <p> </p> <p> </p> <form action="upload.php" method="post" enctype="multipart/form-data"> <input type="file" name="userfile" class="fileUpload" multiple> <input type="hidden" name='girl' value='<?php echo $_POST['select1'];?>' /> <button id="px-submit" type="submit">Upload</button> <button id="px-clear" type="reset">Clear</button> </form> <script type="text/javascript"> jQuery(function($){ $('.fileUpload').fileUploader(); }); </script> </div> </body> </html>
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$_SERVER['SERVER_NAME'] without the www.
Simonvetterli replied to Simonvetterli's topic in Regex Help
Hi And how I do with my "problem": Here is my code: $lang = $_GET['lang']; $currency = ''; list($subdom,$secdom,$tld) = explode(".",$domain); if (($lang == '') AND ($tld == 'ch')){ $lang = 'ge'; $currency = 'CHF'; } elseif (($lang == 'ge') AND ($tld == 'ch')){ $lang = 'ge'; $currency = 'CHF'; } elseif (($lang == 'en') AND ($tld == 'ch')){ $lang = 'en'; $currency = 'CHF'; } elseif (($lang == '') AND ($tld == 'com')){ $lang = 'en'; $currency = 'THB'; } elseif (($lang == 'ge') AND ($tld == 'com')){ $lang = 'ge'; $currency = 'THB'; } $_GET['lang'] = $lang; $url1 = $_SERVER['PHP_SELF']; $url = basename($url1); How I "include" Your code? Kindly Simon -
I have a simmilar same problem, hope somebody can help me: Depend of the tld I have I choose the language on the site: so I have the follow code: $domain = $_SERVER['SERVER_NAME']; list(subdom,$secdom,$tld) = explode(".",$domain); If I have in the browser www.turtlesdive.com no problem, ich I have www.turtlesdive.ch no problem. But if I have only turtlesdive.com or turtlesdive.ch - I not get any more com or ch as $tld. How can I do? Kindly Simon