I've been working on this FOR DAYS with no luck and I'm not happy with the results, so I'm asking for help.
I'm attempting to use $xoopsDB (as provided in class\database\mysqldatabase.php - the typical database class XOOPS uses to access mySQL) to insert a row into a table I created named XOOPS_registration, and if a duplicate would be the result of the insert report an error that I can trap so I can update the existing row of data with any changes.
When you look at the code I added to include/functions.php (see below), you'll see that I populate the variable $sql with the SQL statement used to attempt to insert data into XOOPS_registration. Pretty standard stuff. What I expect to happen when I use $result = $xoopsDB->query($sql); is for $result to contain FALSE (because the SQL failed due to the duplicate key) and $xoopsDB->error() to display "Duplicate entry
'duplicate@x.com' for key 2" and $xoopsDB->errno() to display 1062. I get those results using $result = mysql_query($sql); but not when I use $result = $xoopsDB->query($sql); When I use $result = $xoopsDB->query($sql); I get blank for $xoopsDB->error() and 0 for $xoopsDB->errno().
Using queryF (as in $xoopsDB->queryF($sql);) changes nothing.
phpMyAdmin reports: MySQL 4.0.24-log running on localhost as kcoug@localhost
phpinfo() reports:
PHP Version: 4.3.11
SERVER["SERVER_SOFTWARE"]: Apache/1.3.33 (Unix) mod_fastcgi/2.4.2 FrontPage/5.0.2.2635 mod_jk/1.2.5
Here is how my function is called in modules/reg/reg4.php:
insert_or_update_registration($_POST);
The data in $_POST comes from my form in modules/reg/reg3.php, which is set up as follows:
echo "
