hi all,
i have done simple login form with username="admin" and password="admin123".
i am checking for sql injections.i have given "mysql_real_escape_string"
for both username and password fields.
but also it is not working......
if i give username as "admin --" and click the submit button(not giving password also) it is taking to the next page...
tell me whats wrong in my below code.....
Code:<?php session_start(); mysql_connect("localhost","root",""); mysql_select_db("test"); if(isset($_POST['sub'])) { $username=mysql_real_escape_string($_POST['txtuname']); $password=mysql_real_escape_string($_POST['txtpwd']); $check=mysql_query("SELECT DISTINCT `username`,`password` FROM `log` WHERE `username`='$username'") or die("Error: " . mysql_error()); while($find = mysql_fetch_array($check)) { list($username,$output) = $find; } if($password==$output) { $_session['si']=session_id(); echo "<script> location='view1.php'</script>"; } else echo "invalid"; } ?> <table width="200" height="150" bgcolor="lightblue" border="1" align="center"> <tr><td style="font-size:25;color:red" align="center" colspan="2">Login Form </td></tr> <form method="post" action=""> <tr><td align="right" width="100"> Username:</td><td><input type="text" name="txtuname" </td></tr> <tr><td align="right" width="100"> Password:</td><td><input type="password" name="txtpwd" </td></tr> <tr><td align="right" width="100"> <input type="submit" value="login" name="sub" </td></tr> </form> </table>



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