警告：mysql_num_rows（）期望参数1为资源，第27行的C：\ wamp \ www ovel \ loginadmin.php中给出布尔值[重复]

Question

This question already has an answer here:

• mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc… expects parameter 1 to be resource or result 32 answers

Can anyone help with this error? I would like to make log in session in my system. Here is the coding.

  <?PHP
   error_reporting (E_ALL^E_NOTICE);
   session_start();
   $username =$_POST['username'];
   $password =$_POST['password'];

    if ($username && $password)
     { // if 1

      $server='localhost';
       $userid2='root';
          $password2='';
          $dbname='suhaibah';

       $conn=mysql_connect("$server", "$userid2", "$password2");
       mysql_select_db("$dbname", $conn) or die(mysql_error()); 
         $query =  mysql_query("select * from staff where username = '$username'");

          $numrows = mysql_num_rows($query);

         if ($numrows !=0)
           { // if 2

     while ($row = mysql_fetch_assoc($query))
    { // if 3
    $dbusername = $row['username'];
    $dbpassword = $row['password'];
    } // if 3

      if ($username==$dbusername&&$password==$dbpassword)
       {
        //echo "You're in! <a href='index2.php'>click </a> here to enter";
        $_SESSION['username']=$username;
        header( 'Location: http://localhost/novel/lamanutamaadmin.php' ) ;
       }
       else 
          {echo " Incorrect Password!! <a href='/novel/landingpageadmin.php'>Click here to login</a>";}
   } // if 2    
else 
  die (" That user dosen't exist<a href='/novel/landingpageadmin.php'>Click here to login</a>");

 } // if 1
 else 
 {
  echo "Please enter userid and password!!<a href='/novel/landingpageadmin.php'>Click here to login</a>";
  }

?>


</body>
</html>

</div>

Answer 1

The problem is not in mysql_num_rows, but in mysql_query!

A call from mysql_query can return a resource on success or the boolean false on fail!

http://php.net/mysql_query

When mysql_num_rows invoked, it takes the boolean false (returned from mysql_query) as parameter, showing an indirect failure in mysql_query.

Try to change it for

$query =  mysql_query("select * from staff where username = '$username'") or die(mysql_error());

and you'll surely get the right error.