etherboo

I tried that at first, it didn't work.

I'm sorry for my long reply. My classwork really picked up, and I've had to lay off this for a bit. I plan on picking it up while I have more free time. Your solution resolved my error, but I'm still having another problem. Thank you again.

Hello, I'm currently in the process of trying to learn php. I've been working on an issue with something I am trying to code from a book I purchased "PHP and MySQL Web Development 5th Ed." by Luke Welling and Laura Thomson. So far the book has been great, but I'm trying to figure out using MySQL and php. When I try to run this particular script for a query for a book store, I get the error: Notice: Trying to get property of non-object in C:\php-book\ch11\results.php on line 37. This is the code: $query = "select * from books where ".$searchtype."like '%".$searchterm."%'"; $result = $db->query($query); $num_results = $result->num_rows; echo "<p>Number of books found: ".$num_results."</p>"; for($i=0; $i<$num_results; $i++) { $row = $result->fetch_assoc(); echo "<p><strong>".($i+1).". Title: "; echo htmlspecialchars(stripslashes($row['title'])); echo "</strong><br />Author: "; echo stripslashes($row['author']); echo "<br /> ISBN: "; echo stripslashes($row['isbn']); echo "<br />Price: "; echo stripslashes($row['price']); echo "</p>"; } $result->free(); $db->close(); Line 37 is specifically: $num_results = $result->num_rows; I've been bashing my head for a couple of hours here, thank you in advance to whoever can offer any help.