Jump to content

MadTechie

Staff Alumni
  • Posts

    9,409
  • Joined

  • Last visited

  • Days Won

    1

Everything posted by MadTechie

  1. WHAT!! The Switch statment was correct,
  2. We have all seen the Web Hosting list, but what about speed i know my host goes nuts if i use more than 0.5% of the processor!!!! also their download rate is kinda weak when the USA wakes up heehee So are what are the best performance wise??
  3. what about Boot from USB or.. Boot from CD and store to USB!
  4. try <?php if(isset($_POST['change'])==TRUE) { $X= $_POST['first']; }else{ $X= 'john'; } echo "<input type='text' name='first' size='10' value='".$X."' />\r\n"; ?>
  5. change <?php } else { echo "Query failed<br />$query<br />. mysql_error(); } ?> to <?php } else { echo "Query failed<br />$query<br />". mysql_error(); } ?>
  6. Cool Oh Please Click Solved
  7. from the code i can't workout what you want to do! ???
  8. you need to use something like <?php mysql_query("INSERT INTO secondtable ($_GET['select']) VALUES ('field')"); ?>
  9. Link to PHP Manual was wrong but i agree with wildteen88 PHP Manual
  10. using insert... what code do you have ?
  11. ??? erm.. ok why don't use just use 'Order By' deleting rows will have no effect then!! maybe some code will help..
  12. you can't in php, try javascript, what are you trying to do ?
  13. Depends... what you trying to do! syntaxs looks ok
  14. Double check the username & password
  15. change CUSTOMER-INSERT-PROCESS.PHP from <?php mysql_query("INSERT INTO customer (name, address) VALUES('$name', '$address' ) ") or die(mysql_error()); ?> to <?php mysql_query("INSERT INTO customer (name, address) VALUES('{$_POST['name']}', '{$_POST['address']}' ) ") or die(mysql_error()); ?>
  16. ~Sighs~ ok .htaccess Options +FollowSymlinks # no hot-linking RewriteEngine On RewriteCond %{HTTP_REFERER} !^$ RewriteCond %{HTTP_REFERER} !^http://(www\.)?phpfreaks\.com/ [nc] RewriteRule .*\.(avi|mpg|mpeg)$ http://phpfreaks.com/movies/dummy.avi [nc]
  17. erm.. a reboot maybe:p
  18. Not exacly you can make this more complex but for a basic example this is one solution user clicks on a link ie video.php?ID=12 this page creates a random number were say 1234 a session is stored with 1234 a record in the database is created with 4 fields VideoID = 12, Session = 1234, active = 1, IP = connections IP, the Player Now starts to play the Movie from another script MoviePlayer.php MoviePlayer.php uses the session to looks into the database and finds the VideoID checks the IP and that active = 1 now it sets the active to 0, and unsets the session and starts to readin the video file.
  19. Erm.. did you read/understand the sessions part ?
  20. Field A and B may need to be Floats, Strange i just created another table and it works but the old table still fails.. i am trying to work out why! Will try to rebuild the old table, if it fails i'll post it OH one other thing, the Fields used in this calculation were Text but i converted them (before trying the calculation)
  21. wrong section i think.. move to MySQL
  22. add a index.php file to the folder that holds the movies create a video player in flash and the link to the files gets returned via a connection to a php file which check things like sessions (created during login) So when i say session i mean session interlinked database Login create a session Video is steamed via a php which will only work after checking session Flash player picks up the video from the php file of course you can add protection here as well as for HotLink protection you add that to the PHP script and also you have one in cPanel i hope that helps
  23. that message will appear on almost every error so try to workout what the real problem is try print_r($_FILES["uploadFile"];
  24. You can't tell a cookie to update if you don't control the site their on..
  25. OK this should be simple but again i have messed up somewhere, i have review my "O'Reilly's MySQL Pocket Referance.. :-\" but its not much help.. OK the problem Table1 Field A Field B Field C All Type Float Now Field A = 140 Field B = 211 Field C = 0 i run the SQL statement UPDATE `Table1` SET `Field C` = ((`Field A`/`Field B`)*100 ); and Field C becomes ...150.714... surely that shoud be 66.666.....? can someone please help me workout what i am doing wrong..
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.