Dude - sweet book. I bought it myself when i was starting out. Your problem is due to the PHP error message setting in the php.ini file. Whack this at the beginning of your scripts: // Report simple running errors error_reporting (E_ERROR | E_WARNING | E_PARSE); When you know more - check out the php.ini file and read about the error level.. And this page in the PHP manual: http://www.php.net/manual/en/function.error-reporting.php Cheers, Luke Woollard -----Original Message----- From: Gilles [mailto:webmaster@bionix.net] Sent: Monday, 7 April 2003 11:18 AM To: php-db@lists.php.net Subject: Can't do anything with variables Hi all, Sorry for this really beginner question (PHP-Newbie here). I bought the book "PHP and MySQL web developpement" (Welling & Thompson) and I'm trying to recreate the exemples and it won't work. I installed PHP 4.3.1 and MySQL 4.0.12. The easy part. Here's a simple form page (from the book): SEARCH.HTML <html> <head> <title>Book-O-Rama Catalog Search</title> </head> <body> <h1>Book-O-Rama Catalog Search</h1> <form action="results.php" method="post"> Choose Search Type:<br> <select name="searchtype"> <option value="author">Author <option value="title">Title <option value="isbn">ISBN </select> <br> Enter Search Term:<br> <input name="searchterm" type=text> <br> <input type="submit" value="Search"> </form> </body> </html> And now the processing page: RESULTS.PHP <html> <head> <title>Book-O-Rama Search Results</title> </head> <body> <h1>Book-O-Rama Search Results</h1> <? if (!$searchtype || !$searchterm) { echo "You have not entered search details. Please go back and try again."; exit; } $searchtype = addslashes($searchtype); $searchterm = addslashes($searchterm); @ $db = mysql_pconnect("localhost", "bookorama", "bookorama"); if (!$db) { echo "Error: Could not connect to database. Please try again later."; exit; } mysql_select_db("books"); $query = "select * from books where ".$searchtype." like '%".$searchterm."%'"; $result = mysql_query($query); $num_results = mysql_num_rows($result); echo "<p>Number of books found: ".$num_results."</p>"; for ($i=0; $i <$num_results; $i++) { $row = mysql_fetch_array($result); echo "<p><strong>".($i+1).". Title: "; echo stripslashes($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>"; } ?> </body> </html> The problem. Even if I fill both fields, I always get "Notice: Undefined variable: searchtype" The only way it would work is if I remove all the "if" part and leave the "select * from books" (without the where clause). I don't get it. Any help would be much appreciated. Regards Gilles -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
