Well, first, are the top

Greg K — Sun, 30 Sep 2012 11:22:16 +0000

Well, first, are the top section of code and the bottom section supposed to be the same script?? This is confusing as to what you are trying to accomplish

Best guess with limited information:

$_POST['txtadd'] contains a single quote, which is breaking the query. (Any variables to be using in a query for data should be wrapped with mysql_real_escape_string() to protect it from breaking due to this (and to minimize SQL injection. Think if for $_POST['txtadd'] someone put my remark'); drop table _kursus; // you would loose your table!)

If it is not an issue with the slashes, then time to track what is wrong. You say "the INSERT query does not function". Have you verified that no data has actually be added to the database. (some people just don't "see" in the web page what they expect, and say it doesn't work")

If there is no record even added to the database, modify it to:

$sql = 'INSERT INTO `_kursus` (`remark`) VALUES ("'.mysql_real_escape_string($txtadd).'")';
$result = mysql_query($sql) or die ('SQL FAILED:<br>'.$sql."<br>Error: 'mysql_error());

is it generating an error? How does the SQL statement look? (sometimes in code it is difficult to see a typo, but when you look at the final SQL statement you can see it easier).

Need more exact answer that what mysql_error() gave? Try manually executing the query on the database.

Don't forget, when you are done debugging, take out the die() statement that displays the actual SQL statement. On a production environment, you never want to show your SQL to the user on an error, (honestly, you shouldn't directly echo out mysql_error() either, why give a hacker "inside information" on how the site is set up)

-Greg