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DeanWhitehouse

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Everything posted by DeanWhitehouse

  1. are you storing the message in this var $message?? mail($to, $subject, $message, $headers);
  2. You arnt storing this to a var
  3. thats what this means
  4. whats the error and the corresponding code
  5. or you could just show us the $message , and then we can solve this
  6. add error_reporting(E_ALL); to the top of both php pages
  7. where are you setting $message
  8. change the value of it or unset it
  9. this line $query = "SELECT comments.comment_image, album.image_id FROM album, comments WHERE comment_image = image_id ORDER BY comment_id"; shoud be $query = "SELECT * FROM album, comments WHERE comment_image = image_id ORDER BY comment_id";
  10. put the echos in the while loop
  11. Are you sure the hostname is correct?
  12. it should work on any i think, just need the correct details
  13. well i am to tired to write the code out , but the logic is , have a set of tags (div for me) that have the image inside , and use JS to hide the rest of the page (that needs to be in tags, div) e.g. <html> <script type="text/javascript"> document.getElementById('main').style.visibility="hidden"; function loading() { document.getElementById("main").style.visibility="visible"; document.getElementById('loader').style.visibility="hidden"; document.getElementById('loader').style.marginTop="0%"; document.body.style.backgroundColor="#CCCCCC"; document.getElementById('loader').innerHTML = ' '; } </script> <style type="text/css"> body { visibility:hidden; background-color:#000000; } </style> <body onLoad="loading()"> <div id="loader" align="center" class="test"> <h1>Loading</h1> <br /><img src="loading.gif" alt="loading" align="absmiddle"> </div> <div align="center" id="main"> Main stuff </div> </body> </html>
  14. do you want something like this http://djw-webdesign.awardspace.com/Mail%20App/
  15. have a look at innerHTML that should do it
  16. then its either the wrong username or hostname or password
  17. why not try elseif? it will do the same thing
  18. yes, it will just show what the error is
  19. thnks DW im still right though
  20. try an elseif elseif(mysql_num_rows($links) >= 1)
  21. THIS IS PHP HELP NOT JS
  22. change this $db = mysql_connect($hostname, $username, $password); to $db = mysql_connect($hostname, $username, $password) or die (mysql_error());
  23. yer i just saw, i agree with DW , i have been doin this a few months and never seen a syntax error for no reason.
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