What the code is suppose to do is get data from one table and insert that data into a new table.
Just so you know how the data is stored in the Items table.
Items
ItemID | ItemName | genre
1 | Name1 | 1,2,3
2 | Name2 | 1,3
3 | Name3 | 3
and this is how the new table will be (tblSetToRef)
tblSetToRef
auto_id | ref_item_id | ref_genre_id
1 | 1 | 1
1 | 1 | 2
1 | 1 | 3
1 | 2 | 1
1 | 2 | 3
1 | 3 | 3
the code I have works perfectly, but I keep getting this error:
Warning: Supplied argument is not a valid MySQL result resource in /var/www/test/alter_table_form.php on line 39
I don't know why I get this warning and i want to stop it. what do i have to do?
this is the code I'm using
//using define() and its all correct
function openConn(){
global $dbServerConn;
$dbServerConn = mysql_connect(DB_HOST,DB_USER,DB_PASS) or mysql_error($dbServerConn);
$dbSel = mysql_select_db(DB_NAME, $dbServerConn) or mysql_error($dbSel);
}
openConn();
$getItems = mysql_query("SELECT * FROM Items", $dbServerConn);
while($row = mysql_fetch_array($getItems)){
$item_id = $row['ItemID'];
$i_genre = $row['genre'];
$query = "SELECT * FROM tblGenreList WHERE genre_id in(".$i_genre.")";
$getGenreId = mysql_query($query, $dbServerConn);
echo $query."<br />";
while($row2 = mysql_fetch_array($getGenreId)){
$query1 = "INSERT INTO `tblSetToRef` (`ref_item_id`, `ref_genre_id`) VALUES ('".$item_id."', '".$row2['genre_id']."');";
mysql_query($query1, $dbServerConn);
echo $query."<br />";
}
}
I would appreciate any suggestions.
i must also add, i only get the warning if i do the nested while loops. If i just do a regular while($row = mysql_fetch_array($query)).... i get no warning
Thanks.