PHP mysql_num_rows() missing some parameter

Question

i am new in PHP world. anyhow i am trying to make logging form which authenticate the entered data from a database (that data will be entered from sing-up form which is working fine) but its given me the following error.

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in \wamp\www\hightech\login_form.php on line 23

and my code is:

HTML form: index.php

<form id="login" action="login_form.php" method="post">
   <table style="color:white;border-color:white">
            <tr>
                 <td align="center" width="300" colspan="2">
                     <h3>Log In</h3>
                 </td>
            </tr>
            <tr>
                 <td align="right" width="200" id="username">
                      User&nbspName
                 </td>
                 <td>
                    <input type="text" name="user_name" placeholder=" Type user name" style="width:170;color:blue;text-align:center;border-top-left-radius:25;border-bottom-right-radius:25;"/>
                 </td>

            </tr>
            <tr>
                 <td align="right" id="passd">
                     Password
                 </td>
                 <td>
                     <input type="password" name="pass" placeholder="Type  password" style="width:170;color:blue;text-align:center;border-top-left-radius:25;border-bottom-right-radius:25;"/>
                 </td>
            </tr>
            <tr>
                 <td width="300" colspan="2" align="right">
                     <input type="submit" value="Log In" onclick="chack_info()" style="color:blue;border-top-left-radius:25;border-bottom-right-radius:25"/>
                 </td>
             </tr>
       </table>
    </form>

loging_form.php

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "hightech"; 
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

   $user_name= filter_input(INPUT_POST, 'user_name');
   $password=  filter_input(INPUT_POST, 'pass');
   $result=mysql_query("SELECT * FROM accounts WHERE  user_name='$user_name' and password='$password'");

   $count=mysql_num_rows($result);
   if($count==0){
      echo'data found';
   } else {
      echo 'Wrong Username or Password!'; 
   }

Plz help me as soon as possible.

In your question you mention mysqli_fetch_array but in your code you're using mysql_*. You can't mix the two. Now that I see your edit, I have to say, why are you using mysql_*? You should be using mysqli — Zarathuztra, May 23 '15 at 15:38
You need to show more code. Has a connection been established? Have you verified the connection was succsesful? — Zarathuztra, May 23 '15 at 15:40
What are you using? In case of MySQLi you cannot use MySQL. How do you initialize the database connection? — Richard, May 23 '15 at 15:40
@Zarathuztra i have that connection part too now in my post. and guys thanks for replay. — Hacan Champ, May 23 '15 at 15:43
You're using a mysqli connection, but then mysql functions to do the work. You need to use the mysqli api — Zarathuztra, May 23 '15 at 15:44

score 1 · Accepted Answer · answered May 23 '15 at 15:43

1

Use this:

$result=mysqli_query($conn, "SELECT * FROM accounts WHERE  user_name='$user_name' and password='$password'");

$count=$result->num_rows;
if($count==0){
   echo'data found';
} else {
   echo 'Wrong Username or Password!'; 
}

You cannot mix MySQLi and MySQL because these are two different database drivers which other architecture (OOP vs. Non-OOP).

answered May 23 '15 at 15:43

Richard

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@HacanChamp: No problem, you can vote up my answer and mark it as solution if you want so. – Richard May 23 '15 at 16:52
actually i have less reputation points – Hacan Champ May 23 '15 at 17:47

PHP mysql_num_rows() missing some parameter

1 Answers1