simcoweb Posted August 22, 2006 Share Posted August 22, 2006 Ok, this should be simple. I've seen it a dozen times but can't find the right reference to it. I want to display a message if no results show from a database query. So.. 'IF' no results then 'display this message' ... 'ELSE' display this HTML for the results.Here's my code snippet for this part of the script:[code]mysql_connect($dbhost, $dbuser, $dbpass) or die('Database has gone bye bye');mysql_select_db($dbname) or die('Where oh where is that database');$sql="SELECT * FROM 'members' WHERE 'name' = '$name'";$results = mysql_query($sql);$row = mysql_fetch_array($results);//$num = mysql_num_rows($results);$num_rows = mysql_result($results, 0, 0);//start HTML code for display of profileif ($num_rows == '0') {echo "<font class='bodytext'>'We are sorry. That profile is unavailable at this time. Please select another.<br />";} else {echo <<<HTML[/code]Getting this error:[quote]Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home2/wwwxxxx/public_html/profile.php on line 10Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /home2/wwwxxxx/public_html/profile.php on line 12[/quote] Quote Link to comment Share on other sites More sharing options...
Corona4456 Posted August 22, 2006 Share Posted August 22, 2006 OK... I'm confused as to why you aren't using mysql_num_rows().so you would do:[code]mysql_connect($dbhost, $dbuser, $dbpass) or die('Database has gone bye bye');mysql_select_db($dbname) or die('Where oh where is that database');$sql="SELECT * FROM 'members' WHERE 'name' = '$name'";$results = mysql_query($sql);$row = mysql_fetch_array($results);// Change is here$num_rows = mysql_num_rows($results);//start HTML code for display of profileif ($num_rows == 0) {echo "<font class='bodytext'>'We are sorry. That profile is unavailable at this time. Please select another.<br />";} else {echo <<<HTML[/code] Quote Link to comment Share on other sites More sharing options...
simcoweb Posted August 22, 2006 Author Share Posted August 22, 2006 Hi: I tried that along with a few other combinations. I just took your code and pasted it over what I had. It displays the 'if' echo statement about the profile but also still displays these errors:[quote]Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home2/wwwxxxx/public_html/profile.php on line 10Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home2/wwwxxxx/public_html/profile.php on line 12'We are sorry. That profile is unavailable at this time. Please select another.[/quote]View it at: [url=http://www.plateauprofessionals.com/profile.php]http://www.plateauprofessionals.com/profile.php[/url] Quote Link to comment Share on other sites More sharing options...
Corona4456 Posted August 22, 2006 Share Posted August 22, 2006 Try this for your query instead:[code]$sql="SELECT * FROM `members` WHERE `name` = '$name'";[/code] Quote Link to comment Share on other sites More sharing options...
simcoweb Posted August 22, 2006 Author Share Posted August 22, 2006 Hey, thanks for your posts. I referred to another file in my project and used this line:[code]$sql=("SELECT * FROM members");[/code]and it worked fine. No more error messages. Just a couple of other bugs to work out. Thanks for the help! Quote Link to comment Share on other sites More sharing options...
Corona4456 Posted August 22, 2006 Share Posted August 22, 2006 n/p Quote Link to comment Share on other sites More sharing options...
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