Hi All,
I'm trying to make a image host for my members. Because I just moved hosts, I still have some time left on the "old" one. So, I was thinking I'll use it for the image host uploads. I got this code from php.net an I keep getting this error:
Warning: ftp_login() expects parameter 1 to be resource, boolean given in
/home/myusername/public_html/imghost/_upload.php on line
16
FTP connection has failed!
Attempted to connect to myftpserver.com for user
imghost@mysite.com
and the source:
PHP Code:
<?php
if(isset($_POST['start_upload']) && $_FILES['txt_file']['name'] != ""){
$local_file = $_FILES['txt_file']['tmp_name']; // Defines Name of Local File to be Uploaded
$destination_file = "/".basename($_FILES['txt_file']['name']); // Path for File Upload (relative to your login dir)
// Global Connection Settings
$ftp_server = "MyFTPServer.com"; // FTP Server Address (exlucde ftp://)
$ftp_user_name = "MyUsername"; // FTP Server Username
$ftp_user_pass = "MySecretPasswordThatIChaged"; // Password
// Connect to FTP Server
$conn_id = ftp_connect($ftp_server);
// Login to FTP Server
$login_result = ftp_login($conn_id, $ftp_user_name, $ftp_user_pass);
// Verify Log In Status
if ((!$conn_id) || (!$login_result)) {
echo "FTP connection has failed! <br />";
echo "Attempted to connect to $ftp_server for user $ftp_user_name";
exit;
} else {
echo "Connected to $ftp_server, for user $ftp_user_name <br />";
}
$upload = ftp_put($conn_id, $destination_file, $local_file, FTP_BINARY); // Upload the File
// Verify Upload Status
if (!$upload) {
echo "<h2>FTP upload of ".$_FILES['txt_file']['name']." has failed!</h2><br /><br />";
} else {
echo "Success!<br />" . $_FILES['txt_file']['name'] . " has been uploaded to " . $ftp_server . $destination_file . "!<br /><br />";
}
ftp_close($conn_id); // Close the FTP Connection
}
?>
<html>
<head>
<script type="text/javascript">
window.onload = function() {
document.getElementById("progress").style.visibility = "hidden";
document.getElementById("prog_text").style.visibility = "hidden";
}
function dispProgress() {
document.getElementById("progress").style.visibility = "visible";
document.getElementById("prog_text").style.visibility = "visible";
}
</script>
</head>
<body>
<form action="<?php echo $_SERVER['PHP_SELF'];?>" method="POST" enctype="multipart/form-data">
Please choose a file: <input name="txt_file" type="file" size="35" />
<input type="submit" name="start_upload" value="Upload File" onClick="dispProgress()" />
</form>
<img id="progress" src="http://images.mysite.com/progress.gif" />
<p id="prog_text" style="display:inline;"> Upload Started!</p>
</body>
<html>
Thanks for your help!
- Pc1203