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FTP upload script errors

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  #1  
Old October 9th, 2002, 04:02 PM
Mobius Man's Avatar
Mobius Man Mobius Man is offline
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FTP upload script errors
I've been working on making a script to upload image files to the server through a PHP script. I have 2 files: images.html (which is the upload form), and image_manager.php (which uploads the image file thru FTP). I've been going through the code for an hour, and still can't figure out my errors. It has something to do with my $connect variable.

Here's the errors I'm getting:

Warning: ftp_login() expects parameter 1 to be resource, string given in /usr/home/hostultra/hostultra.com/html/hosted/sites/Mobius_Man/image_manager.php on line 18

Warning: ftp_mkdir() expects parameter 1 to be resource, string given in /usr/home/hostultra/hostultra.com/html/hosted/sites/Mobius_Man/image_manager.php on line 24

Warning: ftp_chdir() expects parameter 1 to be resource, string given in /usr/home/hostultra/hostultra.com/html/hosted/sites/Mobius_Man/image_manager.php on line 25
uploaded!
Warning: ftp_close() expects parameter 1 to be resource, string given in /usr/home/hostultra/hostultra.com/html/hosted/sites/Mobius_Man/image_manager.php on line 50


And here's my script:
PHP Code:
<?php

/***********************************************
                    Image FTP Uploading Script            
                            (image_manager.php)        
***********************************************/


//enter your FTP server's name, as well as your username and password into these variables
$ftp_server="ftp.hostultra.com";
$ftp_user="Mobius_Man";
$ftp_pass="********";

//variable that connects to the FTP server
$connect="ftp_connect(ftp.hostultra.com, 21, 120)";

//logins into FTP server account
ftp_login($connect, $ftp_user, $ftp_pass);

//checks for image directory
if (! @ftp_chdir($ftp_server, "/news/images"))
{
    //if directory doesn't exist, it is now created
    ftp_mkdir($ftp_server, "/news/images");
    ftp_chdir($ftp_server, "/news/images");    
}

else
{
//else statement
}

//destination file variable
$des_file="ftp_pwd($connect)";

$upload="ftp_put($connect, $des_file, $source_file, FTP_BINARY)";

//checks to see if the file was successfully uploaded
if    (!$upload)
{
    echo "Upload of $source_file was unsuccessful!";
}

else
{
    echo "$source_file uploaded!";
}

//closes FTP server connection
ftp_close($connect);

?>
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  #2  
Old October 9th, 2002, 04:31 PM
Tzicha's Avatar
Tzicha Tzicha is offline
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Read: http://www.php.net/manual/en/ref.ftp.php

You need to send a connection id(or connection resource) to the ftp functions.
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  #3  
Old October 9th, 2002, 05:01 PM
hedge hedge is offline
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Why are you putting the function call in quotes.

ie.
//variable that connects to the FTP server
$connect="ftp_connect(ftp.hostultra.com, 21, 120)";


s/b $connect = ftp_connect('ftp.hostultra.com',21,120);

... all the way down
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  #4  
Old October 9th, 2002, 09:10 PM
Mobius Man's Avatar
Mobius Man Mobius Man is offline
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Well, I changed that stuff, however it still isn't working. But, now I only get these errors:

Warning: ftp_mkdir() expects parameter 1 to be resource, string given in /usr/home/hostultra/hostultra.com/html/hosted/sites/Mobius_Man/image_manager.php on line 24

Warning: ftp_chdir() expects parameter 1 to be resource, string given in /usr/home/hostultra/hostultra.com/html/hosted/sites/Mobius_Man/image_manager.php on line 25

Warning: ftp_put(): error opening in /usr/home/hostultra/hostultra.com/html/hosted/sites/Mobius_Man/image_manager.php on line 36
Upload of was unsuccessful!

Here's the form code:
Code:
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">

<html>
<head>
	<title>Image Upload</title>
</head>
<script language="JavaScript">

		function insertImage(imgPath)
		{
			parent.opener.iView.focus();
			parent.window.close();
			parent.opener.iView.document.execCommand('insertimage', false, imgPath);
		}

	</script>
<body>
<form action="image_manager.php" method="post" enctype="multipart/form-data">
<input type="hidden" name="doUpload" value="true">
		<div class="text">Upload File:</div>
		<input type="file" name="source_file"> <input type="submit" value="Upload" onClick="this.value='Please Wait...'">
	
</form>


</body>
</html>


PHP Code:
<?php

/***********************************************
                    Image FTP Uploading Script            
                            (image_manager.php)        
***********************************************/


//enter your FTP server's name, as well as your username and password into these variables
$ftp_server="ftp.hostultra.com";
$ftp_user="Mobius_Man";
$ftp_pass="TummyKitte";

//variable that connects to the FTP server
$connect = ftp_connect('ftp.hostultra.com',21,120);

//logins into FTP server account
ftp_login($connect, $ftp_user, $ftp_pass);

//checks for image directory
if (! @ftp_chdir($ftp_server, "/news/images"))
{
    //if directory doesn't exist, it is now created
    ftp_mkdir('$ftp_server', "/news/images");
    ftp_chdir('$ftp_server', "/news/images");    
}

else
{
//else statement
}

//destination file variable
$des_file = ftp_pwd($connect);

$upload = ftp_put($connect, $des_file, $source_file, FTP_BINARY);

//checks to see if the file was successfully uploaded
if    (!$upload)
{
    echo "Upload of $source_file was unsuccessful!";
}

else
{
    echo "$source_file uploaded!";
}

//closes FTP server connection
ftp_close($connect);

?>



Oh the joys of trying to learn PHP...
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  #5  
Old October 10th, 2002, 11:42 AM
hedge hedge is offline
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Quote:
Originally posted by Mobius Man
Well, I changed that stuff, however it still isn't working. But, now I only get these errors:

Warning: ftp_mkdir() expects parameter 1 to be resource, string given in /usr/home/hostultra/hostultra.com/html/hosted/sites/Mobius_Man/image_manager.php on line 24

Warning: ftp_chdir() expects parameter 1 to be resource, string given in /usr/home/hostultra/hostultra.com/html/hosted/sites/Mobius_Man/image_manager.php on line 25

Warning: ftp_put(): error opening in /usr/home/hostultra/hostultra.com/html/hosted/sites/Mobius_Man/image_manager.php on line 36
Upload of was unsuccessful!

Here's the form code:
Code:
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">

<html>
<head>
	<title>Image Upload</title>
</head>
<script language="JavaScript">

		function insertImage(imgPath)
		{
			parent.opener.iView.focus();
			parent.window.close();
			parent.opener.iView.document.execCommand('insertimage', false, imgPath);
		}

	</script>
<body>
<form action="image_manager.php" method="post" enctype="multipart/form-data">
<input type="hidden" name="doUpload" value="true">
		<div class="text">Upload File:</div>
		<input type="file" name="source_file"> <input type="submit" value="Upload" onClick="this.value='Please Wait...'">
	
</form>


</body>
</html>


PHP Code:
<?php

/***********************************************
                    Image FTP Uploading Script            
                            (image_manager.php)        
***********************************************/


//enter your FTP server's name, as well as your username and password into these variables
$ftp_server="ftp.hostultra.com";
$ftp_user="Mobius_Man";
$ftp_pass="TummyKitte";

//variable that connects to the FTP server
$connect = ftp_connect('ftp.hostultra.com',21,120);

//logins into FTP server account
ftp_login($connect, $ftp_user, $ftp_pass);

//checks for image directory
if (! @ftp_chdir($ftp_server, "/news/images"))
{
    //if directory doesn't exist, it is now created
    ftp_mkdir('$ftp_server', "/news/images");
    ftp_chdir('$ftp_server', "/news/images");    
}

else
{
//else statement
}

//destination file variable
$des_file = ftp_pwd($connect);

$upload = ftp_put($connect, $des_file, $source_file, FTP_BINARY);

//checks to see if the file was successfully uploaded
if    (!$upload)
{
    echo "Upload of $source_file was unsuccessful!";
}

else
{
    echo "$source_file uploaded!";
}

//closes FTP server connection
ftp_close($connect);

?>



Oh the joys of trying to learn PHP...


Well it's the same problem.. when you put variables in quotes they have became literals, so instead of passing your connectionid you are passing a string '$connectionid' to the function... which it hates.
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  #6  
Old July 15th, 2003, 12:01 PM
alexcd alexcd is offline
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I am getting upload one image in php and check if upload is successful, however, when the script finish for close the connection ftp, I receive this error: Fatal error: Call to undefined function: ftp_close() in c:\apache\htdocs\test.php on line 11

I am using ftp_close($conn_id); for close the $conn_id connection. Can you tell me what is happening?
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