login.php

PHP Code:

<?php
$passwords = file('password.php');
echo "$passwords";
$passwords_array = array($passwords);

if((array_key_exists($_POST['user'], $passwords_array))
        and
    ($passwords_array[$_POST['user']] == $_POST['pass'])){ 
?>
<?php $user =  $_POST['user']; ?>
<form method="post" action="upload.php" enctype="multipart/form-data">
Select a file to upload<br>
<input type="hidden" value="<?php echo "$user"; ?>" name="user" />
<input type="file" name="userfile"/><br>
<input type="submit" value="Upload">
</form>
<?php
}
else{
?>
<center>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST">
<table width="313" border="0">
    <tr>
      <td width="75"><p align="right">Username:</p>
        <p align="right">Password:</p></td>
      <td width="228"><p>
        <input type="text" name="user" />
      </p>
        <p>
          <input type="password" name="pass" />    
            </p></td>
    </tr>
  </table>
<input type="submit" value="Login" />
</form>
<P>
Uploads files to your folder. Files examined daily.
</center>
<?php

}

?>

password.php

PHP Code:

'abc' => '123'
,'def' => '456'
,'ghi' => '789' 


I can't for the life of me figure out what's wrong.

You can't just pass a string to array() and have it eval'ed as PHP. Either eval it as PHP like so (not desired):

PHP Code:

eval('$passwords_array = array(' . $passwords . ');'); 


Or make password.php an actual array and simply include it:
login.php

PHP Code:

include('password.php'); // now $passwords_array exists
// .. the rest 


passwords.php:

PHP Code:

<?php

$passwords_array('abc' => '123',
'def' => '456',
'ghi' => '789');

?>

it would also help to know the format of the file. perhaps you could use fgetcsv to put the contents of the file into an array for you.

Well, I tried your first suggestion and got

"
Parse error: syntax error, unexpected T_DOUBLE_ARROW in c:\wamp\www\test\password.php on line 2
"