2013-12-18 46 views
-2

任何人都可以幫助解決此錯誤嗎?我想在我的系統中登錄會話。這是編碼。警告:mysql_num_rows()期望參數1是資源,在第27行C: wamp www novel loginadmin.php中給出的布爾值

<?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> 
+0

請參閱[本答案](http://stackoverflow.com/a/11674313/250259)瞭解如何解決此問題。 –

+0

您對「//評論」的使用乍一看實在令人困惑。我想知道你爲什麼要放'if 1' – Sir

回答

0

問題不在mysql_num_rows中,而是在mysql_query中!

從的mysql_query的呼叫可以成功或失敗上的布爾返回資源

http://php.net/mysql_query

當mysql_num_rows調用時,它需要布爾(來自的mysql_query返回)作爲參數,示出的mysql_query間接故障。

試圖改變它的

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

,你一定會得到正確的錯誤。

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