Ok ok bro,i will improve it..thx again.....
Ok ok bro,i will improve it..thx again.....
thanks broj1...my problem has been resolved...thx a lots....
That error comes that is not sql error,the error is on 39 line,when someone try to login with wrong username....can smeone help me out.
thx for the help bruh....i have improved it little bit...but m getting error...can you please resolve it...
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Latest</title>
</head>
<body>
<?php
include("db.php");
if(isset($_POST["submit"]))
{
$username=$_POST["username"];
$password=$_POST["password"];
$query="select * from latest where username='$username' && password='$password'";
$query1=mysql_query($query);
if(mysql_num_rows($query1)>0)
{?>
<font color="#FF0000" size="+4"><b>
<?php
echo "welcome "."$username";
exit;
}
else
{
$msg="Invalid Username or Password";
}
}
?>
<table align="center" border="2">
<form enctype="multipart/form-data" action="login.php" method="post">
<tr><td> <?php echo($msg);?></td></tr>
<tr><td>Enter Username</td>
<td><input type="text" name="username" />
</td></tr>
<tr><td>Enter Password</td>
<td><input type="password" name="password" /></td>
</tr>
<tr><td>Submit</td><td><input type="submit" value="submit" name="submit"/></td></tr>
</form>
</table>
</center>
</body>
</html>
when i try to access this login page..an empty page open...please help with this coding
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Latest</title>
</head>
<body>
<?php
include("db.php");
if (isset($_POST["submit"]))
{
$username=$_POST["username"];
$password=$_POST["password"];
$query="select * from latest where username='$username' && password='$password'";
$query1=mysql_query($query);
if(mysql_num_rows($query1)>0)
{?>
<font color="#FF0000" size="+4"><b>
<?php
echo "welcome";
echo ($username);
}
else
{
?>
<center>
<table align="center" border="2">
<form enctype="multipart/form-data" action="<?php $_SERVER['PHP_SELF'] ?>" method="get">
<tr><td></td></tr>
<tr><td>Enter Username</td>
<td><input type="text" name="username" />
</td></tr>
<tr><td>Enter Password</td>
<td><input type="password" name="password" /></td>
</tr>
<tr><td>Submit</td><td><input type="submit" value="submit" name="submit"/></td></tr>
</form>
<?php } } ?>
</table>
</center>
</body>
</html>
i m working with sessions..
i have problem with sessions
here is my coding
"
<?php
session_register("admin_id");
if($_SESSION["admin_id"]=="")
{
header("location: login.php");
}
?>
"
i want to register a session
but when i use session_register function i got error ""Deprecated: Function session_register() is deprecated in C:\xampp\htdocs\xampp\c\admin\session_check.php on line 2""
help me..how can i register session without this error...
then how can i fetch another feilds from mysql with image...please help..it urgent....i want to fetch image(blob) and another fields on 1 page...how it can possible?...
i m trying to fetch a image from mysql (blob) with header..here is my coding..."<?php
include("db.php");
$query=mysql_query("select * from table where id='3' ");
$row=mysql_fetch_array($query);
$r=$row;
header("content-type:image");
echo $r;
?>"
i want to fetch another fields from the database....but when i try to echo another fields...the page shows error or it does not echo other fields of database....please help me...how can i resolve it...i want to fetch other fields from database,like'username'password'firstname'lastname and image...thanks in advance...