i dont know why this error cuming. can any one plz

Question

niths -1 Posting Whiz in Training

14 Years Ago

--You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'Resource id #10' at line 1--

<html>
<body>
<form action="login.php" method="post">
<div>
<table width="100%">
<tr>
<td><img src="Logofinalcopy.gif"></td>
</tr>
<tr>
<td bgcolor="aqua"><h2>Login</h2></td>
</tr></table>
<table align="right" style="width:40%">
<br>
<tr>
<td>Username:</td>
<td><input type="text" name="username"></td>
</tr>
<tr>
<td>Password:</td>
<td><input type="password" name="password"></td>
</tr>
<tr>
<td align="CENTER" COLSPAN="4">
<input TYPE="SUBMIT" name="submit" value="Login">
<input TYPE="reset" name="submit" value="clear"></td>
</tr>
</table>
</div>
</form>
</body>
</html>
<?php
if(isset($_POST['submit']))
{
    $username=$_POST['username'];
    $password=$_POST['password'];
    $username = stripslashes($username);
    $password = stripslashes($password);
    
    if(strlen($username) < 1)
    {
        $msg=$msg."Please enter the username<br>";
        $flag="NOTOK";
    }
    else if(strlen($password) < 1)
    {
        $msg=$msg."Please enter the password<br>";
        $flag="NOTOK";
    }
    else
    {
        $flag="OK";
        $msg="";
    }



    if($flag != "OK")
    {
        echo "<br><br><br><br><br><strong style='color:#FF3333'>"."<right>Please enter Username/Password <br></right>"."</strong>";
    }
    else
    {
        mysql_connect("10.70.1.50","invensis","invensis") or die('Could not connect: ' . mysql_error());                
        mysql_select_db("database_myproject") or die('Could not connect DB: ' . mysql_error());
        $username = mysql_real_escape_string($username);
        $password = mysql_real_escape_string($password);


        $sql = "SELECT count(*) as count FROM Users WHERE username = '$username' and `password` = '$password'";
        $result = mysql_query($sql) or die(mysql_error());
        
        $row_count = mysql_fetch_array($result);
        
        $count = $row_count['count'];
        
        if($count == 1)
        {
            $sql = "SELECT role FROM Users WHERE username = '$username' and password = '$password'";
            $result = mysql_query($sql);
            $info = mysql_fetch_array($result);


            if($info['Role']== '0')
            {
                @header("location: http://localhost/Project/Superadmin.php");
            }
            else if($info['Role']== '1')
            {
                @header("location: http://localhost/Project/Admin.php");
            }
            else if($info['Role']== '2')
            {
                @header("location: http://localhost/Project/User.php");
            }
        }
        else
        {
            echo "<br><br><br><br><br><strong style='color:#FF3333'><left>Incorrect UserName OR Password</left></strong>";
        }
    }
}
?>

php

Edited 14 Years Ago by Ezzaral because: Added code tags. Please use them to format any code that you post.

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Latest Post 14 Years Ago Latest Post by ko ko

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Arsench 0 Light Poster

14 Years Ago

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'Resource id #10' at line 1

<html>
<body>
<form action="login.php" method="post">
<div>
<table width="100%">
<tr>
<td><img src="Logofinalcopy.gif"></td>
</tr>
<tr>
<td bgcolor="aqua"><h2>Login</h2></td>
</tr></table>
<table align="right" style="width:40%">
<br>
<tr>
<td>Username:</td>
<td><input type="text" name="username"></td>
</tr>
<tr>
<td>Password:</td>
<td><input type="password" name="password"></td>
</tr>
<tr>
<td align="CENTER" COLSPAN="4">
<input TYPE="SUBMIT" name="submit" value="Login">
<input TYPE="reset" name="submit" value="clear"></td>
</tr>
</table>
</div>
</form>
</body>
</html>
<?php
if(isset($_POST['submit']))
{
    $username=$_POST['username'];
    $password=$_POST['password'];
    $username = stripslashes($username);
    $password = stripslashes($password);

    if(strlen($username) < 1)
    {
        $msg=$msg."Please enter the username<br>";
        $flag="NOTOK";
    }
    else if(strlen($password) < 1)
    {
        $msg=$msg."Please enter the password<br>";
        $flag="NOTOK";
    }
    else
    {
        $flag="OK";
        $msg="";
    }



    if($flag != "OK")
    {
        echo "<br><br><br><br><br><strong style='color:#FF3333'>"."<right>Please enter Username/Password <br></right>"."</strong>";
    }
    else
    {
        mysql_connect("10.70.1.50","invensis","invensis") or die('Could not connect: ' . mysql_error());                
        mysql_select_db("database_myproject") or die('Could not connect DB: ' . mysql_error());
        $username = mysql_real_escape_string($username);
        $password = mysql_real_escape_string($password);


        $sql = "SELECT count(*) as count FROM Users WHERE username = '$username' and `password` = '$password'";
        $result = mysql_query($sql) or die(mysql_error());

        $row_count = mysql_fetch_array($result);

        $count = $row_count['count'];

        if($count == 1)
        {
            $sql = "SELECT role FROM Users WHERE username = '$username' and password = '$password'";
            $result = mysql_query($sql);
            $info = mysql_fetch_array($result);


            if($info['Role']== '0')
            {
                @header("location: http://localhost/Project/Superadmin.php");
            }
            else if($info['Role']== '1')
            {
                @header("location: http://localhost/Project/Admin.php");
            }
            else if($info['Role']== '2')
            {
                @header("location: http://localhost/Project/User.php");
            }
        }
        else
        {
            echo "<br><br><br><br><br><strong style='color:#FF3333'><left>Incorrect UserName OR Password</left></strong>";
        }
    }
}
?>

end quote.

Hi, check you Superadmin.php and Admin.php and User.php files, also, the files normally creating with low cases.

regards

Edited 11 Years Ago by mike_2000_17 because: Fixed formatting

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ko ko 97 Practically a Master Poster · Answer 1 · 2010-03-19T22:44:11+00:00

Remove single quote from the red color text.

$sql = "SELECT count(*) as count FROM Users WHERE username = '$username' and `password` = '$password'";

i dont know why this error cuming. can any one plz

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