Thanks so much guys, It was my user permissions all along. Funnily enough the book gives you guidelines on how to set them up, and sets up a fair few... but for some reason they selected the wrong user to use.
Thanks very much for your help David and Brian- much appreciated. Yours, GF ----- Original Message ----- From: David Jackson To: [EMAIL PROTECTED] Cc: [EMAIL PROTECTED] ; [EMAIL PROTECTED] Sent: Monday, December 31, 2001 2:41 AM Subject: Re: [PHP] MySQL problem I almost forgot add a or mysql_error() for each line like this: <?php // database connect script $dbhostname = "localhost"; $dbuser = "db_user_name"; $dbpasswd = "db_password"; $dbname= "db_name"; $link = mysql_connect("$dbhostname", "$dbuser", "$dbpasswd") echo mysql_error(); mysql_select_db("$dbname") echo mysql_error(); ?> This will return "human readable error messges" Can you onto mysql database from command. mysql -u root -p mysql or mysql -u root mysql # A root password isn't usally get during install. if so: >select user,host,password from user; then: select user,host,db from db; My quesss you don't have any permission for the databases or to connect to local host. If this is correct do: GRANT ALL on db_name.* TO you@local host idendified by 'your_password'; > Hiya, > > Thanks for the quick reply. I used the PHP manual example and it > connects to the database successfully but cannot select the database. > > I'm not sure why this is? I've looked hard at it and I cannot see where > I have gone wrong. > > Thanks. > > GF. > ----- Original Message ----- > From: David Jackson > To: [EMAIL PROTECTED] > Cc: [EMAIL PROTECTED] > Sent: Monday, December 31, 2001 1:48 AM > Subject: Re: [PHP] MySQL problem > > > Here's the example from the PHP manual: > The tutorial here are very helpfull: > http://www.melonfire.com/community/columns/trog/ > > -- David > > <?php > // Connecting, selecting database > $link = mysql_connect("mysql_host", "mysql_login", "mysql_password") > or die("Could not connect"); > print "Connected successfully"; > mysql_select_db("my_database") > or die("Could not select database"); > > // Performing SQL query > $query = "SELECT * FROM my_table"; > $result = mysql_query($query) > or die("Query failed"); > > // Printing results in HTML > print "<table>\n"; > while ($line = mysql_fetch_array($result, MYSQL_ASSOC)) { > print "\t<tr>\n"; > foreach ($line as $col_value) { > print "\t\t<td>$col_value</td>\n"; > } > print "\t</tr>\n"; > } > print "</table>\n"; > > // Closing connection > mysql_close($link); > ?> > > > > > > Hello, > > > > I am extremely new to MySQL and have never managed to get working > > smoothly with PHP before. I am trying really hard to understand how > > to work it, and am almost there. > > > > I have a problem which I do not know how to resolve and was > > wondering if anybody could help me. I have no idea what is wrong > > with the code and why I am getting the error message; > > > > Warning: Supplied argument is not a valid MySQL result resource in > > C:\apache\htdocs\sams\chapter10\results.php on line 47 > > > > I am currently using a book to aid me with MySQL, and this is an > > example from the book. It does not seem to work and I have no idea > > what I may have done wrong to obtain this warning. > > > > I have changed my login and password to question marks. > > > > <? > > > > if (!$searchtype || !$searchterm) > > > > { > > echo "You have not entered search details. Please go back and > > try > > again."; > > > > exit; > > > > } > > > > > > $searchtype = addslashes($searchtype); > > > > $searchterm = addslashes($searchterm); > > > > @ $db = mysql_pconnect("mesh", "bookorama", "bookorama123"); > > > > if (!$db) > > > > { > > echo "Error: Could not connect to database. Please try again > > later."; > > > > exit; > > > > } > > > > mysql_select_db("booktest"); > > > > $query = "select * from booktest where ".$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>"; > > > > } > > > > ?> > > > > The problem seems to be around the lines of code; > > > > $result = mysql_query($query); > > > > $num_results = mysql_num_rows($result); > > > > Any assistance is appreciated. > > > > Yours, > > > > GF. > > > > _________________________________________________________________ > > Chat with friends online, try MSN Messenger: > > http://messenger.msn.com > > > > > > -- > > PHP General Mailing List (http://www.php.net/) > > To unsubscribe, e-mail: [EMAIL PROTECTED] > > For additional commands, e-mail: [EMAIL PROTECTED] To > > contact the list administrators, e-mail: > > [EMAIL PROTECTED] > > > -- -- -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]