form output within an IF statement?

[DP23]

having trouble outputting my form within an if statement. what i want to achieve is:

if the cookie is set to 1 - the user is loggin in , so great the user. if they are not logged in print the login form.

please advise

my curent code is:

<?php 
if($_COOKIE['login']==1) 
{ 
print "Hello ".$_COOKIE[firstname]; 
} 
else 
{ 
<FORM METHOD="POST" ACTION="frontpage3.php"> 
<strong>Username:</strong><br> 
<INPUT TYPE="text" NAME="username"> 
<p><strong>Password:</strong><br> 
<INPUT TYPE="password" NAME="password"><br> 
<br><INPUT TYPE="SUBMIT" NAME="submit" VALUE="Login"> 
<p> 
If you are not a registered user please register <a href="index_register2.html">here</a> </p> 
</FORM> 
} 
?> 

[devinemke]

1. you are never setting the cookie (at least not in the code you posted)
2. you are not using [man]echo[/man] to output the form

[Installer]

You can't just go from PHP to HTML like that. Either escape from PHP (with an end tag) before the HTML, or echo the HTML as a string.

<?php
if($_COOKIE['login']==1) 
{ 
print "Hello ".$_COOKIE[firstname]; 
} 
else 
{
?>
<FORM METHOD="POST" ACTION="frontpage3.php"> 
<strong>Username:</strong><br> 
<INPUT TYPE="text" NAME="username"> 
<p><strong>Password:</strong><br> 
<INPUT TYPE="password" NAME="password"><br> 
<br><INPUT TYPE="SUBMIT" NAME="submit" VALUE="Login"> 
<p> 
If you are not a registered user please register <a href="index_register2.html">here</a> </p> 
</FORM>
<?php 
}
?>

or

<?php 
if($_COOKIE['login']==1) 
{ 
print "Hello ".$_COOKIE[firstname]; 
} 
else 
{ 
echo '<FORM METHOD="POST" ACTION="frontpage3.php"> 
<strong>Username:</strong><br> 
<INPUT TYPE="text" NAME="username"> 
<p><strong>Password:</strong><br> 
<INPUT TYPE="password" NAME="password"><br> 
<br><INPUT TYPE="SUBMIT" NAME="submit" VALUE="Login"> 
<p> 
If you are not a registered user please register <a href="index_register2.html">here</a> </p> 
</FORM>';
}
?>

Edit: Put quotes around your array indexes.