i am getting this error returned but i dont know why :( error: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/pferrie/public_html/vinrev/adm/insert2.php on line 6
php file <?php include("connection.php"); if(!empty($rating)){ $query="SELECT * FROM $tablename WHERE rating = '$rating'"; if(mysql_query($query)) { $myrow = mysql_fetch_array($query);// This line returns an error! $id = $myrow ['id']; echo"Error! There is another row in the database with a rating value of $rating\n You must first edit the row with a rating of $rating ! Row id = $id<br><br>"; echo"<a href='edit2.php?tablename=$tablename&id=$id'>Edit record</a><br>"; } }else { if(!empty($_FILES['userfile']['tmp_name'])) { $name = strtolower(eregi_replace('#| |\?|!', '', $_FILES['userfile']['name'])); $destination = 'img/'.$name; if(move_uploaded_file($_FILES['userfile']['tmp_name'], $destination)) { $picture=$destination; $query = "INSERT INTO $tablename (artist, title, picture, review, label, format, price, rating)VALUES ('$artist', '$title', '$picture', '$review', '$label', '$format', '$price', '$rating')"; if(!mysql_query($query)) { fail("Error updating DB1"); } Header("Location: main.php?success=New row has been inserted into the database"); } } if(empty($userfile)) { $query2 = "INSERT INTO $tablename (artist, title, picture, review, label, format, price, rating)VALUES ('$artist', '$title', '$picture', '$review', '$label', '$format', '$price', '$rating')"; if(!mysql_query($query2)) { fail("Error updating DB2"); } Header("Location: main.php?success=New row has been inserted into the database"); } } ?> I am trying to check for a duplicate entry for rating and if one is found grab the row ID and then use it to load the edit page for the row. Anyone Cheers Paul -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php