Shouldn't it be: $result = mysql_query($sql, $link_id);
PHP will be quick to tell you that there is a missing link identifier in the mysql_query() call. It's happened to me plenty of times. HTH! Martin Clifford Homepage: http://www.completesource.net Developer's Forums: http://www.completesource.net/forums/ >>> PHPCoder <[EMAIL PROTECTED]> 07/24/02 02:16PM >>> Yes, what on earth was I thinking! should be: ... $result = mysql_query($sql); if (mysql_num_rows($result)) { $row = mysql_fetch_assoc($result); } else { echo "whatever"; } ... Kevin Stone wrote: >You beat me too the punch and I think you explained it better than me, but >just one minor little thing to note. Where you said.. > >"if ( $result = mysql_query($sql)) " > >This is not a valid way to check if the query has returned anything. >mysql_query() returns FALSE on error. So if there was no error but there >also wasn't anything returned then the object stored in $result wiill more >than likely evaluate to TRUE. For the determining factor you should count >the number of rows with mysql_num_rows($result). If the returned value is >zero then you know it hasn't returned anything. > >-Kevin > >----- Original Message ----- >From: "PHPCoder" <[EMAIL PROTECTED]> >To: "Matthew Bielecki" <[EMAIL PROTECTED]> >Cc: "php-general" <[EMAIL PROTECTED]> >Sent: Wednesday, July 24, 2002 11:50 AM >Subject: Re: [PHP] Help with msql_fetch_array() > > >>I can almost guarantee that it's not the second line that is "failing", >>the problem here is that $result is not containing naything, and that is >>normally due to the fact that you are not connecting to the db, or the >>table "tablename" is not there. >> >>I use the following format as my "standard" MySQL connect and query >> >snippet: > >>$link = @mysql_connect("localhost",$username,$password) or die ('Could >>not connect!'); //@ suppresses the default error message generated by >>this function and the "or die()" bit kills the script right then and >>there should it not be able to connect. >> mysql_select_db("YOUR_DB_NAME",$link); >> $sql = "select * from your_table_name"; >> if ( $result = mysql_query($sql)) { // checks to see if $result >>contains anything before it even tries to fetch an associative array >>from it. >> $row = mysql_fetch_assoc($result); >> } else { >> echo "Empty result set!"; >> >>Note also that I use mysql_fetch_assoc and NOT mysql_fecth_array, as 9 >>out of 10 times, you don't need the array element id's that is returned >>by mysql_fetch_array. >> >>Matthew Bielecki wrote: >> >>>I have a couple of scripts that fail with the error of: >>>Warning: mysql_fetch_array(): supplied argument is not a valid MySQL >>> >result > >>>resource in... >>> >>>I'm new to both SQL and PHP and I'm wondering if I have some setting >>>turned off or what. >>> >>>Here's the piece of code that is failing (the second line fails): >>> >>>$result = mysql_db_query($dbname, "SELECT * FROM tablename ORDER BY id"); >>> $row = mysql_fetch_array($result); >>> >>> >>>Thanks for your help in advance!! >>> >> >> >>-- >>PHP General Mailing List (http://www.php.net/) >>To unsubscribe, visit: http://www.php.net/unsub.php >> > > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
