include 'file.php' is printing my vars before i tell them to.

[acid.waste]

Alrite, i have a configs.php that holds a bunch of information (Login info mostly) and when i do

 

$UsrNme = $_POST["user"];
$Passw = $_POST["pass"];

include 'configs.php'

if ($UsrNme == $UserName && $Passw == $Password) {
setcookie("User", "True", time()+3600);
echo "Welcome {$UsrNme}; You have been successfully logged in!";
} else {
echo "Sorry, the username {$UsrNme} was not successfully logged in.";
}

 

It prints my configs.php instead of using the variables $UserName and $Password to check against UsrNme and Passw... Am i doing something wrong or is this supposed to happen?

[Pikachu2000]

Hard to tell without knowing what's in configs.php.

[acid.waste]

config.php has

 

$UserName = "Username";

$Password = "Password";

$AdminEmail = "[email protected]";

$sqlHOST = "host.host.host";

$sqlUser = "username";

$sqlPass = "Password";

$sqlDatabase = "database";

$ForumName = "Test";

$ForumName = "http://www.lol.com/forum/";

 

[freelance84]

There is no semi colon after include 'configs.php', that may be messing things up a bit

 

 

[BlueSkyIS]

does config.php have opening and closing php tags?

[rwwd]

It will print out the contents if you don't wrap the file in php tags - either that or you left them off the post?

And in your first example your missing the  ;  from the include instruction.

 

Rw

[acid.waste]

Thanks for your help everyone! I forgot to have fwrite add the opening and closing php tags, thanks for helping me see this!