Parse error that i cannot make sense of

Now bear with me, the page itself probably wont work. It is probably wrong. But you have to learn some way right?..anyway above is the error and below is the entire page. Print seems fine. must be something above. i perfer print over echo

I just threw the page together based on what i learned so far. Soooo if it makes no sense thats ok...I just cannot see this error. helpage?

thanks = )

<?
session_start();

//required pages
require "connect.php";

// make connect global on the page
global $connect;


// makes the ip a variable
$ip = $_SERVER['REMOTE_ADDR'];


// If username is posted it sets it to a var
if($_POST['username'])
{
$username=$_POST['username'];
}


// queries db to see if the username already exists
$userquery=mysql_query("SELECT * FROM users WHERE loginname='{$username}'",$connect);



if(mysql_num_rows($userquery))
{
Print "Username currently in use. Choose something else.";
}


//checks to see if the password and confirm passwords dont match
else if($_POST['password'] != $_POST['confirmpassword'])
{
print "Passwords did not match";
}

else
{

// inserts into the database
mysql_query("INSERT INTO users (loginname, loginpassword, ip) VALUES ( '{$username}',
'{$_POST[password]}', '{$ip}');

}





else
{

print "
<center><table>
<tr>
<td><form action=register.php method=post>Username:</td>
<td><input type=text name=username></td>
</tr>
<tr>
<td>Password: </td>
<td><input type=password name=password></td>
</tr>
<tr>
<td>Confirm Password:</td>
<td><input type=password name=confirmpassword></td>
</tr>
<td>Email:</td>
<td><input type=text name=email></td>
</tr>
<td>Promo Code:</td>
<td><input type=text name=promo></td>
</table>


<input type=submit value=Submit></form>";
}
print "</body></html>";
?>

PHP:

 

// inserts into the database
mysql_query("INSERT INTO users (loginname, loginpassword, ip) VALUES ( '{$username}',
'{$_POST['password']}', '{$ip}'");

PHP:

You need the quotation mark to end your query, and the apostrophes around password.

 

^^ thanks for resolving my first issue. 1+ for you. now i came into a second problem. I cannot insert into the database anything. i checked the field names in the db. they match up. and the table name is users so thats right too. there is already one user in there i manually put in myself to start the auto increment of a field called id.
I also assigned that query to $query and tried to print it and nothing happened(blank page).

ps, i recycled this thread as not to spam up your forum = )

can someone please tell me whats wrong?
excuse the notes im learning = )

<?session_start();

//required pages
require "connect.php";

// makes the ip a varialbe
$ip = $_SERVER['REMOTE_ADDR'];


// If username is set it sets it to a var
//first if
if($_POST['username'])
{
$username=$_POST['username'];
// queries db to see if the username already exists
$userquery=mysql_query("SELECT * FROM users WHERE loginname='{$username}'",$connect) or die(mysql_error());

//nested if in the first if
if(mysql_num_rows($userquery))
{

Print "Username currently in use. Choose something else.";
}//close of nested if


//checks to see if the password and confirm passwords dont match
//still in first if
elseif($_POST['password'] != $_POST['confirmpassword'])
{
print "Passwords did not match";
}//close of elseif

//checks to see if username box or password box are blank. it can be "or" because it only takes 1 blank to kill the signup
elseif($_POST['password']=='' || $_POST['username']=='')
{
print "Your username or password was blank";
}//close of elseif

/*
After checking for existing username, matching passwords
and blank username/password fields and all that passes inspection this next part
inserts the information from the page into the db
*/

else
{

// inserts into the database

mysql_query("INSERT INTO users (loginname, loginpassword, ip) VALUES ( '{$username}',
'{$_POST['password']}', '{$ip}'")or print "POOOF didnt work";


}
}//close of first if to bunch it all together


//what happens if nothing above is happening.pretty much the signup page
else{




print "
<center><table>
<tr>
<td><form action=register.php method=post>Username:</td>
<td><input type=text name=username></td>
</tr>
<tr>
<td>Password: </td>
<td><input type=password name=password></td>
</tr>
<tr>
<td>Confirm Password:</td>
<td><input type=password name=confirmpassword></td>
</tr>

</table>


<input type=submit value=Submit></form>";}

print "</body></html>";?>

PHP:

 

if(mysql_num_rows($userquery))

this doesnt make sense for a start..

You will need to set it to apply to something

if(mysql_num_rows($userquery) == "1") for example

 

Try replacing the first line with this and let me know what happens.

<?
error_reporting(E_ALL);
ini_set('display_errors','1');
session_start();

PHP:

 

thank you both for your replies but this fixed it.
first i printed out what i wanted to be inserted to see if it was actually grabbing it. and it was. so then i knew it was the insert. after messing around with it i came up with this and it worked

mysql_query("INSERT INTO users (loginname, loginpassword, ip) VALUES ( '{$username}',
'{$_POST['password']}', '{$ip}')")or die(mysql_error());

PHP:

I didnt seem to close ("

as far as

if(mysql_num_rows($userquery))
{

Print "Username currently in use. Choose something else.";

PHP:

I see where your coming from. but the code works anyway. ill put a notation to look at it later but for now its not an issue. both 1+ for helping me if you never got 1+ from me in the past and im denied it.

thanks again