New print question:
<?php
$db_host = 'myhost';
$db_user = 'user';
$db_pwd = 'password';
$database = 'survey';
$table = 'january08_practices';
if (!mysql_connect($db_host, $db_user, $db_pwd))
die("Can't connect to database");
if (!mysql_select_db($database))
die("Can't select database");
$query="select COUNT(*) from january08_practices
where Practices = 'yes'";
$rt=mysql_query($query);
if($rt){echo " Command is successful ";}
else {echo "Command is not successful ";}
print mysql_error();
print_r ($rt);
?>
The results I get on the webpage are:
Command is successful Resource id #2
I want the value to be "30" which is the number of "yes" responses
and when I run the query in CocoaMySQL, select COUNT(*) from
january08_practices
where Practices = 'yes'; I get an answer of 30.
How do I print "30" on my webpage?
Thanks,
Grant
On Jan 7, 2008, at 8:57 AM, Allen D. Tate wrote:
> Good deal. I think that there are security issues with the echo
> command
> so this is the better solution anyway. :)
[Non-text portions of this message have been removed]
