mysql_num_rows error

icornwal

I am getting a weird error. Let me give some background first. I just had a hard drive fail and so I had to reinstall Redhat 7.3 on a clean hard drive and then move the data from the dying drive to the new one. Everything seems to be ok from the server side. I got the logins and priveleges back in for the mysql stuff but now I get an error saying

"Warning: Supplied argument is not a valid MySQL result resource in /var/www/html/calendar/index.php3 on line 190"

The code is:

if (strlen($db_host)>0) {
$qrsql = "SELECT * FROM $db_table_name WHERE
($db_column_name='$data')";
$qr_ret = mysql_query($qrsql, $db_link);
if (mysql_num_rows($qr_ret)>0) {
.
.
.

the if( mysql_num_rows... is line 190.

What might cause this error. I don't get any errors on connecting and I know that the table exists and contains data.

Like I said this was all working before and I haven't changed this file.

Lars_Berg

Check the error code from mysql_query

$qr_ret = mysql_query($qrsql, $db_link) or die(mysql_error());

icornwal

"No Database Selected" I have a statement at the top of the page:

$db_link = mysql_connect($db_host, $db_username, $db_passwd);

$db = mysql_select_db($db_name, $db_link);

Is there something else I am missing?

icornwal

Actually I just added the die(mysql_error()) to that line and I must have permissions messed up because my user get denied access.

Andy_Frees

Similar to the challenge discussed above I get the following error:

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/serve/fun4nick/PHP/chapter20/authmain.php on line 14

When I run:

<?
session_start();

if ($userid && $password)
{
// if the user has just tried to log in

$db_conn = mysql_connect( 'mysql.serve.com', 'username', 'userpw' );
mysql_select_db( 'auth' , $db_conn);
$query = "select * from auth "
."where name='$userid' "
." and pass=password('$password')";
$result = mysql_query($query, $db_conn);
if (mysql_num_rows($result) >0 )
{
// if they are in the database register the user id
$valid_user = $userid;
session_register("valid_user");
}
}
?>
<html>
<body>
<h1>Home page</h1>
<?

if (session_is_registered("valid_user"))
{
echo "You are logged in as: $valid_user <br>";
echo "<a href=\"logout.php\">Log out</a><br>";
}
else
{
if (isset($userid))
{
// if they've tried and failed to log in
echo "Could not log you in";
}
else
{
// they have not tried to log in yet or have logged out
echo "You are not logged in.<br>";
}

// provide form to log in 
echo "<form method=post action=\"authmain.php\">";
echo "<table>";
echo "<tr><td>Userid:</td>";
echo "<td><input type=text name=userid></td></tr>";
echo "<tr><td>Password:</td>";
echo "<td><input type=password name=password></td></tr>";
echo "<tr><td colspan=2 align=center>";
echo "<input type=submit value=\"Log in\"></td></tr>";
echo "</table></form>";

}
?>
<br>
<a href="members_only.php">Members section</a>
</body>
</html>

This is row 14 if (mysql_num_rows($result) >0 )

Can you see why I am getting this error?

Thanks for your help....

cgraz

When you get bla bla bla is not a valid mysql result resource, it isn't referring to what's on that line, but what that line is calling.

$result = mysql_query($query, $db_conn);

refers to $query, because you're doing a mysql_query on $query.

To fix that, you need to find out what the problem is with your $query.

So change it to

$result = mysql_query($query, $db_conn) or die(mysql_error());

Take a look at mysql_error() at php.net. It's gotta be the best way to fix these kinds of problems. It tells you specifically what the error is.

Cgraz

Andy_Frees

I did what you suggested and got the following message:

No Database Selected

Question:

mysql_select_db( 'auth' , $db_conn);

Should 'auth' be the table name or the name of the database?

If I substitute the db name in place of the table and it drops through to the:

"Could not log you in" if statement

After using valid username and passwords.

Thanks for your help.

Andy_Frees

This script gains access to the db:

<?
if(!isset($name)&&!isset($password))
{
//Visitor needs to enter a name and password
?>
<h1>Please Log In</h1>
This page is secret.
<form method = post action = "secretdb.php">
<table border = 1>
<tr>
<th> Username </th>
<td> <input type = text name = name> </td>
</tr>
<tr>
<th> Password </th>
<td> <input type = password name = password> </td>
</tr>
<tr>
<td colspan =2 align = center>
<input type = submit value = "Log In">
</td>
</tr>
</form>
<?
}
else
{
// connect to mysql
$mysql = mysql_connect( 'mysql.serve.com', 'userid', 'userpw' );
if(!$mysql)
{
echo 'Cannot connect to database.';
exit;
}
// select the appropriate database
$mysql = mysql_select_db( 'fun4nickdb' );
if(!$mysql)
{
echo 'Cannot select database.';
exit;
}

// query the database to see if there is a record which matches
$query = "select count(*) from auth where
          name = '$name' and
          pass = '$password'";

$result = mysql_query( $query );
if(!$result)
{
  echo 'Cannot run query.';
  exit;
}

$count = mysql_result( $result, 0, 0 );

if ( $count > 0 )
{
  // visitor's name and password combination are correct
  echo "<h1>Here it is!</h1>";
  echo "I bet you are glad you you can see this secret page.";
}
else
{
  // visitor's name and password combination are not correct
  echo "<h1>Go Away!</h1>";
  echo "You are not authorized to view this resource.";
}

}
?>

Thanks again for the help

cgraz

mysql_select_db() should have the database name, not table.

$mysql = mysql_connect( 'mysql.serve.com', 'userid', 'userpw' ); 
if(!$mysql) 
{ 
echo 'Cannot connect to database.'; 
exit; 
}

A much easier and efficient way of doing this would be

mysql_connect('mysql.serve.com', 'userid', 'userpw') or die(mysql_error());

// and

mysql_select_db('fun4nickdb') or die(mysql_error());

I don't really understand what you are doign with that form. Someone fills out a form to connect to this database???

Change your code to the above. No need to run the if statement. If there username or password are incorrect, mysql_error will let them know. 😉

Cgraz

Andy_Frees

I did as you suggested and got this message:

Warning: Access denied for user: 'userid@www4.serve.com' (Using password: YES) in /home/serve/fun4nick/PHP/chapter20/authmain.php on line 8

Warning: MySQL Connection Failed: Access denied for user: 'userid@www4.serve.com' (Using password: YES) in /home/serve/fun4nick/PHP/chapter20/authmain.php on line 8
Access denied for user: 'userid@www4.serve.com' (Using password: YES)

I am trying to create a members only session where members userid and password are checked against a table in a MySQL db, but I haven't been able to get passed the verification that they are a member...:rolleyes:

Again, thanks for your help.

wm_chief

I think your sql hava an error on it.
try to
<? echo"$qr_ret";?>
if have not value £¬is because your
"$qrsql = "SELECT * FROM $db_table_name WHERE
($db_column_name='$data')"; "
is not right.

cgraz

Look at the error message:

Warning: Access denied for user: 'userid@www4.serve.com' (Using password: YES) in /home/serve/fun4nick/PHP/chapter20/authmain.php on line 8

It says the userid you are using is userid.

I'm guessing userid and userpw were fake when you posted to the forum. If you just copied and pasted my code, you forgot to replace userid and userpw with your real username and password.

Cgraz