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gristoi

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Everything posted by gristoi

  1. People are here to help you with problems that you may have WITH CODE THAT YOU HAVE TRIED TO WRITE YOURSELF. If you want someone to write it for you from scratch then you need to post this in the freelance section and expect to pay for the work. Other wise you will more than likely be out of luck.
  2. remove "// link to update.php and send value of id" from your script
  3. if (!mysql_query($sql,$con)) should be if (!mysql_query($mysql,$con))
  4. this is nothing to do with your localhost not being able to accept sessions. this error : Warning: Cannot modify header information - headers already sent by ..... means that somewhere in your code you are trying to send a header AFTER something has been sent to the page. There are two basic rules for using headers : [*]Ensure thats there is no whitespace directly after the opening / closing tags [*]Ensure that nothing is printed to the page before any kind of header redirect etc. to help you further we need to see some code
  5. gristoi

    mail

    huh?
  6. turn on your error reporting and see what error it throws: <?php ini_set('display_errors',1); error_reporting(E_ALL); ?> put that at the top of your script, just below the opening php tag. post the errors here
  7. look at using sessions http://php.net/manual/en/ref.session.php. sudo code for your page would look something like this: <?php session_start(); if(the users has a valid session){ profile block } else { dont display block } ?>
  8. where is your createMarker function defined? looking at the google map api you make your own createMarker function. I could be wrong, but it looks like you have to build your own function
  9. $SQL ="SELECT * FROM dictionary where word LIKE 'A%'"
  10. try: <?php $emails = array("name1" => "[email protected]", "name2" => "[email protected]", "name3" => "[email protected]", "name4" => "[email protected]"); switch ($Dept) { case "Customer Service": $email_to = $emails['name1'] . ", " . $emails['name2']. ",".$emails['name3'] ; break; ?>
  11. errrrr??? is your production server being used by users?
  12. What have you tried so far? You cant just ask people to write the upload code for you. if you want someone to do it for you then you need to post this in the freelance section. With that said, look at using fgetcsv. There are examples on how to read csv's. you also need to lookk at how to upload and move files on the server and using functions like fopen
  13. where are you declaring $tbl_name and $tblname2 ???????????
  14. thats an issue between you and your hosing company.
  15. No, you do however need to read up on managing access and permissions in linux. using 777 is a very bad option so dont go using that all of the time. It is basically granting all read and write permissions for that file, and as you can guess can lead to security problems. http://www.linuxforums.org/forum/linux-tutorials-howtos-reference-material/10862-linux-file-permissions.html a starting point
  16. nothing 'mystrious' about it, what permissions are set for the .htm file. run chmod 777 /var/www/html/filename/htm and try again
  17. sorry, remove the 'where user=' part out of the query
  18. $u_count = mysql_result(mysql_query("SELECT COUNT(*) AS 'Total Entries', user FROM directory WHERE user='$u_id' AND status='1' GROUP BY user ORDER BY 'Total Entries' DESC LIMIT 5"),0);
  19. your query can't be TWO dates at the same time. use between: <?php SELECT name, date, comment FROM narvaro_all WHERE date BETWEEN '2011-07-14' AND '2011-07-20' ?>
  20. <?php // offset diefference between you and the server $offset = 6; $now = date('Y-m-d H:i:s', time()+$offset*60*60) ?>
  21. where in the world are you?
  22. sounds like your server is not in the same timezone as you . Is this your own server? you could try passing in the time on the php side: <?php $now = date('Y-m-d H:i:s', time()) ?>
  23. and what time do you think it should be?
  24. INSERT INTO table (datetimeField) values (now())
  25. http://dev.mysql.com/doc/refman/5.0/en/comparison-operators.html#function_in use the IN function in mysql. example: <?php $exampleValues = array(1,2,3,4,5,6,7,; $sql ="update table1 set field1 = 1 where field2 IN($exampleValues)"; ?>
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