Parse Error - unexpected $

waterfoul

Oct 29 2005, 04:27 AM

i have this code

CODE

<?php
if($_POST['user']=="" or $_POST['pass']=="" or $_POST['host']=="" or $_POST['root']=="" or !isset($_POST['run'])){
print('<form method=post name=form>');
print('FTP username <input type=\'text\' name=user value=""><br>');
print('FTP password <input type=\'password\' name=pass value=""><br>');
print('FTP Host <input type=\'text\' name=host value=""><br>');
print('FTP Root Directory <input type=\'text\' name=root value=""><br>');
print('<input type=\'hidden\' name=run value="1">');
print('<input type=\'submit\'>');
}else{
$ftp_server = $_POST['host'];
$ftp_user = $_POST['user'];
$ftp_pass = $_POST['pass'];

$conn_id = ftp_connect($ftp_server) or die("Couldn't connect to $ftp_server please refresh and do not click retry");
// try to login
if (@ftp_login($conn_id, $ftp_user, $ftp_pass)) {
echo "Connected as $ftp_user@$ftp_server\n";
if(!isset($_POST['out'])){
ftp_chdir($conn_id,$_POST['root']);
$title="";
$lib=implode(array(''),',');
$topimage='';
$topcolor='';
$sideimage='';
$sidecolor='';
$midimage='';
$midcolor='';
$links=implode(array(''));
$linkimage=implode(array(''));
print('<form method=post name=form>');
print('<input type=\'hidden\' name=user value="' . $_POST['user'] . '"><br>');
print('<input type=\'hidden\' name=pass value="' . $_POST['pass'] . '"><br>');
print('<input type=\'hidden\' name=host value="' . $_POST['host'] . '"><br>');
print('<input type=\'hidden\' name=root value="' . $_POST['root'] . '"><br>');
print('<input type=\'hidden\' name=run value="1">');
print('<input type=\'hidden\' name=out value="1">');
print('Page Title <input type=\'text\' name=title value="' . $title . '">');
print('Any files in the library folder you would like to use in ANY page <textarea name="lib" cols=5 rows=8>' . $lib . '</textarea>');
print('The top background image, set to "" for none <input type=\'topimage\' name=title value="' . $topimage . '">');
print('The top background color <input type=\'topcolor\' name=title value="' . $topcolor . '">');
print('The side background image, set to "" for none <input type=\'sideimage\' name=title value="' . $sideimage . '">');
print('The side background color <input type=\'sidecolor\' name=title value="' . $sidecolor . '">');
print('The middle background image, set to "" for none <input type=\'midimage\' name=title value="' . $midimage . '">');
print('The middle background color <input type=\'midcolor\' name=title value="' . $midcolor . '">');
print('The links for your side bar<br>(note: if theres a link without an image below it will not show) <textarea name="links" cols=5 rows=8>' . $links . '</textarea>');
print('The images for the links for your side bar <textarea name="linkimage" cols=5 rows=8>' . $linkimage . '</textarea>');
print('<input type=\'submit\'>');
}else{
echo "Couldn't connect as $ftp_user\n please refresh and do not click retry";
}
// close the connection
ftp_close($conn_id);
?>

and i get this error

QUOTE

Parse error: parse error, unexpected $ in /home/waterfou/public_html/Editor.php on line 56

does any 1 see the problem

Added quote around the error message. You must use quotes, even when copying from your own site!

Spectre

Oct 29 2005, 06:31 AM

The error I received:

QUOTE

Parse error: parse error, unexpected $end in test.php on line 54

This is because you aren't closing all your if() statements. You need to include a closing squigly bracket ('}') to inform the script that you wish to end the statement there (or depending on your style of coding, there are other ways to acheive this).

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