instruction if toujours afficher vrai même l'entrée est faux

Question

<?php 

    include("connection2.php");  

    if(!empty($_POST['lrn']) && !empty($_POST['password'])){ 

     $lrn = $_POST['lrn']; 
     $pass = $_POST['password']; 
     $check = "SELECT * from studentinf where LRN = '".$lrn."'"; 
     $qry = mysql_query($check); 
     $num_rows = mysql_num_rows($qry); 
     $check2 = "SELECT * from studentinf where Password = '".$pass."'"; 
     $qry2 = mysql_query($check2); 
     $num_rows2 = mysql_num_rows($qry2); 

     if($num_rows > 0){ 
      echo "<center>LRN is valid</center> "; 
     } 

     else 
      echo "<font color='#FF0000'><center>Your LRN is Invalid. <br>Please Contact your Web Administrator for your LRN.</center></font>"; 


     if($num_rows2 != "") 
     { 
      echo "account already registered";    
     } 

     else if ($num_rows2 == "") 
     { 
      echo "account not yet registered"; 
      $query = "UPDATE studentinf SET Password = '$pass' where LRN = '$lrn'"; 
      mysql_query($query); 

      echo "Thank You for Registration."; 
      echo '<br><a href="studentlogin.php">Click Here</a> to login your account.'; 
     } 
    } 
?> 

Answer 1

Si je devine juste dont if déclaration que vous voulez dire, essayez de remplacer:

if($num_rows2 != "") 
{ 
    echo "account already registered"; 
} 
else 
if ($num_rows2 == "") 
{ 
    // ... 
} 

avec:

if ($num_rows2 > 0) 
{ 
    echo "account already registered"; 
} 
else 
{ 
    // ... 
} 

$num_rows2 est numérique comparez-le à un nombre, pas à une chaîne.