Re: [PHP] Debugging problems
best bet is to put some error checking in.
function _check_db($query = '', $line = '')
{
if ( mysql_errno() )
{
echo "Error: Problem with DataBase : {$GLOBALS['db_database']}\n";
echo "Error: " . mysql_errno() . ':' . mysql_error() . "\n";
echo "Query: $query\n";
echo "Line : $line\n";
exit();
}
}
after every mysql command put this.
check_db($query, __LINE__);
ie.
this will help alot in finding any mysql errors.
what was the command on line 27 ? email/news groups format posts, therefore
line 27 for you may be line 33 when Im looking at it.
--
Chris Lee
[EMAIL PROTECTED]
""Taline Makssabo"" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> I'm using this as a log on but it gives me this warning message:
>
> Warning: Supplied argument is not a valid MySQL resource
> /httpd/html/php2/login.php on line 27
>
>
>
> function access_denied()
>
> Header("WWW-Authenticate: Basic realm=\"$title\"");
> Header("HTTP/1.0 401 Unauthorized");
> }
>
> function auth_headers($title)
>
> Header("WWW-Authenticate: Basic realm=\"$title\"");
> Header("HTTP/1.0 401 Unauthorized");
> }
> if(!isset($PHP_AUTH_USER))
>
> auth_headers("Members Area");
> access_denied();
> exit;
> }
> else {
>
> $db_hostname = "localhost"; // database host, usually "localhost"
> $db_username = "contact"; // username for the database (make sure it has
> proper privledges)
> $db_password = "cdi986"; // password that cooresponds to DB USERNAME
> $db_database = "contact_informations"; // what database is your member
info
> stored in?
>
> $query = "SELECT username,password FROM contact_informations WHERE
> username='$PHP_AUTH_USER' AND password='$PHP_AUTH_PW'";
> $link = mysql_connect($db_hostname, $db_username, $db_password) or
> die("Unable to connect to database server");
>
> if (mysql_num_rows(mysql_db_query($db_database, $query)) == 0)
>
> auth_headers("Members Area");
> access_denied();
> exit;
> }
>
> mysql_close($link);
> }
> ?>
>
>
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[PHP] Debugging problems
I'm using this as a log on but it gives me this warning message: Warning: Supplied argument is not a valid MySQL resource /httpd/html/php2/login.php on line 27 -- 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]
RE: [PHP] debugging problems
On 18-May-01 Taline Makssabo wrote:
> This is a program i wrote to get some information at keep track with some
> stats. I am very new at this so bare with meI keep on getting this
> Warning:
>
> Warning: Supplied argument is not a valid MySQL result resource in
> /home/virtual/ppcu/home/httpd/html/php2/stat.php on line 13
>
> Warning: Supplied argument is not a valid MySQL-Link resource in
> /home/virtual/ppcu/home/httpd/html/php2/stat.php on line 17
>
> Whats wrong Is it my code? Please help.
>
>
> global $dblink,$PHP_SELF;
> $ip = getenv("REMOTE_ADDR");
> $host = getenv("HTTP_HOST");
> $browser = getenv("HTTP_USER_AGENT");
> $referer = getenv("HTTP_REFERER");
> $sql = "SELECT ipaddress FROM stat WHERE ipaddress = '$ip'";
> $connection = @mysql_connect("localhost", "username", "password") or
> die("Couldn't connect.");
Not too informative:
die ('problems ' .mysql_errno() .''.mysql_error()."\n");
but anyhow, ok, you're connected.
> $db_name = "contact";
> $table_name = "stat";
What are these ?
are you missing mysql_select_db($db_name); ?
> $result = @mysql_query($sql, $dblink);
Test your queries !
if ($result) {
> $num = mysql_num_rows($result);
> if( $num == 0 )
> {
> $sql ="INSERT INTO stat(ipaddress,host,browser,referer,filename)
> VALUES('$ip', '$host', '$browser', '$referer','$PHP_SELF')";
> $result=mysql_query($sql,$dblink);
> $sql1="INSERT INTO hits(ipaddress,count) VALUES('$ip',1)";
> $result1=@mysql_query($sql1,$dblink);
> $text="you had a new visitor from IP: $ip, Host: $host Browser:
> $browser Referer: $referer";
>if ($result)
> {
> mail("[EMAIL PROTECTED]", "You had a new visitor",$text,
"FROM:
> [EMAIL PROTECTED]");
> }
> }else{
> $sql = "UPDATE hits SET count = count + 1 where ipaddress='$ip'";
> $result =@mysql_query($sql, $dblink) ;
> $sql2 = "UPDATE stat SET filename ='$PHP_SELF' where ipaddress='$ip'";
> $result2 =@mysql_query($sql2, $dblink) ;
> }
> ?>
>
>
> --
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--
Don Read [EMAIL PROTECTED]
-- It's always darkest before the dawn. So if you are going to
steal the neighbor's newspaper, that's the time to do it.
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Re: [PHP] debugging problems
You forgot to select the database!
mysql_select_db($db_name);
-elias
http://www.eassoft.cjb.net
""Taline Makssabo"" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> This is a program i wrote to get some information at keep track with some
> stats. I am very new at this so bare with meI keep on getting this
> Warning:
>
> Warning: Supplied argument is not a valid MySQL result resource in
> /home/virtual/ppcu/home/httpd/html/php2/stat.php on line 13
>
> Warning: Supplied argument is not a valid MySQL-Link resource in
> /home/virtual/ppcu/home/httpd/html/php2/stat.php on line 17
>
> Whats wrong Is it my code? Please help.
>
>
> global $dblink,$PHP_SELF;
> $ip = getenv("REMOTE_ADDR");
> $host = getenv("HTTP_HOST");
> $browser = getenv("HTTP_USER_AGENT");
> $referer = getenv("HTTP_REFERER");
> $sql = "SELECT ipaddress FROM stat WHERE ipaddress = '$ip'";
> $connection = @mysql_connect("localhost", "username", "password") or
> die("Couldn't connect.");
> $db_name = "contact";
> $table_name = "stat";
> $result = @mysql_query($sql, $dblink);
> $num = mysql_num_rows($result);
> if( $num == 0 )
> {
> $sql ="INSERT INTO stat(ipaddress,host,browser,referer,filename)
> VALUES('$ip', '$host', '$browser', '$referer','$PHP_SELF')";
> $result=mysql_query($sql,$dblink);
> $sql1="INSERT INTO hits(ipaddress,count) VALUES('$ip',1)";
> $result1=@mysql_query($sql1,$dblink);
> $text="you had a new visitor from IP: $ip, Host: $host Browser:
> $browser Referer: $referer";
> if ($result)
> {
> mail("[EMAIL PROTECTED]", "You had a new visitor",$text, "FROM:
> [EMAIL PROTECTED]");
> }
> }else{
> $sql = "UPDATE hits SET count = count + 1 where ipaddress='$ip'";
> $result =@mysql_query($sql, $dblink) ;
> $sql2 = "UPDATE stat SET filename ='$PHP_SELF' where ipaddress='$ip'";
> $result2 =@mysql_query($sql2, $dblink) ;
> }
> ?>
>
>
> --
> 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]
[PHP] debugging problems
This is a program i wrote to get some information at keep track with some stats. I am very new at this so bare with meI keep on getting this Warning: Warning: Supplied argument is not a valid MySQL result resource in /home/virtual/ppcu/home/httpd/html/php2/stat.php on line 13 Warning: Supplied argument is not a valid MySQL-Link resource in /home/virtual/ppcu/home/httpd/html/php2/stat.php on line 17 Whats wrong Is it my code? Please help. -- 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]
