uploading files to ftp via php

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October 19th, 2003, 01:30 PM

kaen

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uploading files to ftp via php

I have a problem. This is my code:

PHP Code:


		
			
<?

//Get FTP Pass, User, Server, and Category

 $ftp = "SELECT ftp_server, ftp_user, ftp_pass, ftp_cat, address FROM 



layout_servers where      ftp_cat = '$_SESSION[cat_id]'"; 

 $ftp_res = mysql_query($ftp) or die (mysql_error());



   while ($row5 = mysql_fetch_assoc($ftp_res)) 

    { 

        $ftp_server = $row5['ftp_server']; 

        $address = $row5['address']; 

        $ftp_user = $row5['ftp_user']; 

        $ftp_pass = $row5['ftp_pass']; 

        $ftp_cat = $row5['ftp_cat']; 

    }



$conn_id = ftp_connect($ftp_server); 



// login with username and password

$login_result = ftp_login($conn_id, $ftp_user, $ftp_pass); 





// check connection

if ((!$conn_id) || (!$login_result)) { 

        echo "FTP connection has failed!<br>";

        echo "Attempted to connect to $ftp_server for user $ftp_user"; 

        exit(); 

    } else {



ftp_chdir ($conn_id, $_GET[type]);

ftp_mkdir ($conn_id, $_GET[dirname]);



$dir = "temp/$_GET[dirname]/";

$files = array();

$open = opendir($dir);

while ($file = readdir($open)) { if ($file != "." && $file != "..") { $files[] = 



$file; } }

closedir($open);

sort($files);

reset($files);







for ( $i=0; $i < count($files); $i++ ) { 



//Upload

$source_file="temp/$_GET[dirname]/$files[$i]";

$destination_file="$_GET[dirname]/$files[$i]";



$upload = ftp_put($conn_id, $destination_file, $source_file, FTP_BINARY); 



unlink($source_file);

}



if($upload) {



rmdir("temp/$_GET[dirname]");



echo "Files moved to $ftp_server";

echo "<img src=$address/$_GET[dirname]/$zip>";

        

} else {

echo "Error uploading. Contact the webmistress immediately.";

}    





}



// close the FTP stream

ftp_close($conn_id); 





//++++++++++++++++++++++++++++++++++++++

}

?>

However, once I upload the two images in the array- when I try to display them with this PHP script:

PHP Code:


		
			
<?

$thumb = "$server/$cat/$item_id/$item_image";

?>

It says that it's broken. And if I go to the file address itself, it says that the image cannot be accessed. However, if I refresh, it works and works for the rest of the time.

Can you help me?

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~ Kaen

October 19th, 2003, 02:34 PM

jabba_29

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Hi, bit confussed

Where do these vars come from?

PHP Code:


		
			
 $thumb = "$server/$cat/$item_id/$item_image";

Are they ftp links or http?

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October 19th, 2003, 04:12 PM

kaen

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Join Date: Apr 2003

Posts: 63

Time spent in forums: < 1 sec
Reputation Power: 6

well, on a seperate page, i pulled the information of the image out of my database- I just didn't show it to you. sorry ^^

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