[RESOLVED] Unexpected T_IF in... Please Help!

[JackRolfe]

Hi.

I am trying to insert data from a register form into my database but I keep getting this error.

Parse error: syntax error, unexpected T_IF in /home/a2046892/public_html/login1.php on line 49

I can't figure out why I have the error for the life of me!

Here is the code

if(mysql_num_rows($res) == 0) {
			$curnum ++;
			echo $curnum . ". The username '<b>".$username."</b>' does not exist!<br>\n";

	}

The first line of the code is line 49.

This is the second time I have had to work around php and the first time I have ever had this error.

Thanks.

[djjjozsi]

Lets post the previous/following 3-4 lines from your code.

This IF is not in the right logic:

If you have no result why you want to fetch the results?
A (while) cycle is missing from this program.

[JackRolfe]

	if ($_POST['submit']) {

	$curnum = 0;

	$username = $_POST['username'];
	$password = $_POST['password'];

	$sql = "SELECT * FROM users WHERE username='".$username."'";
	$res = mysql_query($sql) or die(mysql_error)

	if(mysql_num_rows($res) == 0) {
		$curnum ++;
		echo $curnum . ". The username '<b>".$username."</b>' does not exist!<br>\n";

	}

}

That alright mate?

[djjjozsi]

You forget to close this line by a semicolumn: ;

$res = mysql_query($sql) or die(mysql_error() );

mysql_error is a constant , mysql_error() is a function name. Be careful.

You expect only one user with a unique username, so you should apply a LIMIT 0,1 into your query. If the mysql_num_rows($resul) is 1 , you fetch the user data, and test the password,
in other case you say there is no username.

[JackRolfe]

Aahh I see..

I said I was inserting data from a registration form into my database.

What I should've said was that I was making a login form.

I just finished a registration page.

Sorry for the inconvenience.

[djjjozsi]

Oh, i see. Now, you can get the username is in used or not.
There is a simpliest way to do it:

<?php
    if ( mysql_result(
            mysql_query( sprintf( "select COUNT(*) from `users` WHERE username='%s' LIMIT 1",
                    mysql_real_escape_string( $_POST["username"] )
                        )
                    ), 0 ) > 0 )
    die("Username is in use");
else
{
/*  registration ...*/
}
?>

[JackRolfe]

Thanks djjjozsi

I tweaked my code so that it was similar to yours.

However, now I have experienced another problem!

Parse error: syntax error, unexpected $end in /home/a2046892/public_html/login1.php on line 41

I find this very confusing. Line 41 is the last line in the following code.

<?php
    if ( mysql_result(
            mysql_query( sprintf( "select COUNT(*) from `users` WHERE username='&#37;s' LIMIT 1",
                    mysql_real_escape_string( $_POST["username"] )
                        )
                    ), 0 ) > 0 )
    die("Username is in use");
else
{
/*  registration ...*/
}
?> 

[laserlight]

Check that every opening brace is matched by a closing brace in the right place.

[JackRolfe]

They are :S

[djjjozsi]

Post the code you have.

[JackRolfe]

Ok here it comes.

<?php

mysql_connect("host","user","pass") or die(mysql_error());
mysql_select_db("Login") or die(mysql_error());

if ($_SESSION['uid']) {

echo "You are already logged in!<br>\n";

} else {

?>
<form method="post" action="login1.php">
<table border="0" style="font-size: 15px; font=family: Tahoma; border: 1px solid black;">
<tr align="right">
  <td><span>Username:</span></td>
  <td><input name="username" type="text" id="uname"></td>
</tr>
<tr align="right">        
  <td><span>Password:</span></td>         
  <td><input id="pword" name="password" type="password"></td> 
	<tr>
		<td colspan="2" align="center">
			<input type="submit" name="submit" value="Login">
		</td>
	</tr>
</table>
</form>

<?php
    if ( mysql_result(
            mysql_query( sprintf( "select COUNT(*) from `users` WHERE username='%s' LIMIT 1",
                    mysql_real_escape_string( $_POST["username"] )
                        )
                    ), 0 ) > 0 )
    die("Username is in use");
else
{
/*  registration ...*/
}
?> 

And I get this error with it..

Parse error: syntax error, unexpected $end in /home/a2046892/public_html/login1.php on line 41

[djjjozsi]

• remove your passwords from your code before post.

You have this:

} else {

You did not close that else section.

[JackRolfe]

Ah crap!

Forgot about the passwords :/

I have closed the bracket and get the same error :S

[djjjozsi]

See this: Here

[JackRolfe]

This is doing my head in!

With the code that you showed me

<?php

mysql_connect( "host", "user", "pass" ) or die( mysql_error() );
mysql_select_db( "Login" ) or die( mysql_error() );

if ( $_SESSION['uid'] )
{
    echo "You are already logged in!<br>\n";
}
else
{
    if ( isset( $_POST ) )
    {
        if ( empty( $_POST["username"] ) )
            die( "username is empty" );

    if ( empty( $_POST["password"] ) )
        die( "password is empty" );

    if ( mysql_result(
            mysql_query( sprintf( "select COUNT(*) from `users` WHERE username='%s' LIMIT 1",
                    mysql_real_escape_string( $_POST["username"] )
                        )
                    ), 0 ) > 0 )
        die( "Username is in use" );
    else
    {
        echo "User is available... continue the registration...";
        /* registration here...*/
    }
}

?>
<form method="post" action="login1.php">
<table border="0" style="font-size: 15px; font=family: Tahoma; border: 1px solid black;">
<tr align="right">
  <td><span>Username:</span></td>
  <td><input name="username" type="text" id="uname"></td>
</tr>
<tr align="right">
  <td><span>Password:</span></td>
  <td><input id="pword" name="password" type="password"></td>
    <tr>
        <td colspan="2" align="center">
            <input type="submit" name="submit" value="Login">
        </td>
    </tr>
</table>
</form>

<?php

}

?>

I get this error:

Warning: mysql_connect() [function.mysql-connect]: Host '64.191.87.134' is not allowed to connect to this MySQL server in /home/a2046892/public_html/login.php on line 3

Host '64.191.87.134' is not allowed to connect to this MySQL server

Please Help!?

[bradgrafelman]

The error message pretty clearly explains the problem - your webserver isn't allowed to connect to the remote MySQL server.

I note that the address you're trying to connect to resolves to the same IP as the one listed in your error message. This likely means that you should be connecting via "localhost" rather than a remote address.

EDIT: Note that I said "likely" - if that doesn't work (sharing an external IP address doesn't mean that the two services are located on the same server), find out how your host expects you to connect to the SQL server.

[JackRolfe]

Anyone have any ideas?

[JackRolfe]

I downloaded wampserver 2.0 before, I looked at a few tutorials and every time I tried setting up localhost, the IIS homepage came up. When I tried looking for one of my folders I got this error.

Server Error in Application "DEFAULT WEB SITE"

I am now using the details of the mysql database I created within my webhost.

The mysql error didn't use to come up.

Any ideas?

[halojoy]

I agree with bradgrafelman
try connect with 'localhost' first, instead of IPnumber, if your mysql is on same PC as webserver.

mysql_connect( "localhost", "xxxx", "xxxx" ) or die( mysql_error() );

If this does not work, then I do not know.
Because 90% this will work when you run PHP, MySQL on your own machine.

When we have our stuff elsewhere, at some web hosting,
then we often use some special link for 'host' to the database server