Smokin Joe Posted May 29, 2009 Share Posted May 29, 2009 It's a simple way of spitting out the list of training videos that my boss has available, and then view whichever one you want. However, I can't really test it that well because I keep getting that blank screen. I've tried some little debug techniques but I've learned that my debugging abilities are somewhat lacking. Would anyone happen to see what I done wrong with my code or perhaps can point me towards some good debugging techniques? <?php include "config.php"; ?> <!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> <link href="../lrg_css_HOME.css" rel="stylesheet" type="text/css"> <meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1" /> <title>Salesperson Training</title> </head> <body> <table width="100%" border="0" cellpadding="0" cellspacing="0"> <tr> <td width="190" valign="bottom"><a href="/index.php"><img src="/img/lrg_logo.gif" width="180" height="90" border="0" align="baseline"></a> </td> <td width="491" valign="bottom" nowrap class="headertext1"> </td> <td valign="bottom" nowrap class="headertext2"> </td> </tr> </table> <br /><br /> <?php //If no previous command if(!isset($cmd)) { mysql_connect($vid_server, $vid_DBusername, $vid_DBpassword) or die ("$DatabaseError - 162"); mysql_select_db($vid_database); //display all $result = mysql_query("SELECT * FROM $vid_table ORDER BY id"); //loop to grab all news while($r=mysql_fetch_array($result)) { //grab the title and the ID of the news $title=$r["vname"];//take out the title $id=$r["id"];//take out the id $url=$r["vaddr"];//take out the address $desc=$r["vdesc"];//take out the description //spit out linked title, and a [Delete] link echo "<a href=\"$url\">$title</a> - [<a href='index.php?cmd=view&id=$id'>View</a>]"; echo "<p>$desc</p>"; echo "<br><br>"; } } //VIEW MOVIE if($_GET["cmd"]=="view") { mysql_connect($vid_server, $vid_DBusername, $vid_DBpassword) or die ("$DatabaseError"); mysql_select_db($vid_database); // open connection and grab the array $resultVid = mysql_query("SELECT vname, vaddr, vdesc FROM $vid_table WHERE id=$id") or die ("Database READ Error"); $r=mysql_fetch_array($result); //grab the title and the ID of the news $title=$r["vname"];//take out the title $id=$r["id"];//take out the id $url=$r["vaddr"];//take out the address $desc=$r["vdesc"];//take out the description echo "<br /><br />"; echo "<B>$title</B><br />"; echo "<table border='0' cellpadding='0' align=\"center\""; echo "<tr><td>"; echo "<OBJECT classid='clsid:02BF25D5-8C17-4B23-BC80-D3488ABDDC6B' width=\"720\" height=\"500\" codebase='http://www.apple.com/qtactivex/qtplugin.cab'>"; echo "<param name='src' value=\"http://www.leiterrealty.com/training/videos/".$url."\">"; echo "<param name='autoplay' value=\"true\">"; echo "<param name='controller' value=\"true\">"; echo "<param name='loop' value=\"true\">"; echo "<EMBED src=\"http://www.leiterrealty.com/training/videos/".$url."\" width=\"720\" height=\"500\" autoplay="true" controller=\"true\" loop=\"true\" pluginspage='http://www.apple.com/quicktime/download/'>"; echo "</EMBED>"; echo "</OBJECT>"; echo "</td></tr>"; echo "<tr><td>"; echo "<p>$vdesc</p></td></tr>"; echo "</table>"; echo "<br /><br /><hr />"; } ?> </body> </html> Quote Link to comment https://forums.phpfreaks.com/topic/160199-phpmysql-syntax-issue-keep-getting-blank-screen/ Share on other sites More sharing options...
lonewolf217 Posted May 29, 2009 Share Posted May 29, 2009 debugging technique #1 - turn on error reporting so you can see what error you are getting also, im not sure what this is <?php mysql_connect($vid_server, $vid_DBusername, $vid_DBpassword) or die ("$DatabaseError - 162"); I think it should be something like this <?php mysql_connect($vid_server, $vid_DBusername, $vid_DBpassword) or die (mysql_error()); Quote Link to comment https://forums.phpfreaks.com/topic/160199-phpmysql-syntax-issue-keep-getting-blank-screen/#findComment-845254 Share on other sites More sharing options...
Maq Posted May 29, 2009 Share Posted May 29, 2009 debugging technique #1 - turn on error reporting so you can see what error you are getting Yes, put this at the top of your script, before the include file but after the opening '<?php' tag. ini_set ("display_errors", "1"); error_reporting(E_ALL); Quote Link to comment https://forums.phpfreaks.com/topic/160199-phpmysql-syntax-issue-keep-getting-blank-screen/#findComment-845260 Share on other sites More sharing options...
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