[code]Warning: ftp_put(): error opening [/code]

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Grigory
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[code]Warning: ftp_put(): error opening [/code]

Postby Grigory » Fri Dec 06, 2002 11:45 am

I get this error

Code: Select all

Warning: ftp_put(): error opening (address to the image file on my computer)
when using this line of code...

Code: Select all

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


I've checked the drive on my computer and it is set to share and read-only so that should be fine. Why else would it not be able to open it?[/code]

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WiZARD
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Re: [code]Warning: ftp_put(): error opening [/code]

Postby WiZARD » Sat Dec 07, 2002 12:56 am

Grigory wrote:I get this error

Code: Select all

Warning: ftp_put(): error opening (address to the image file on my computer)
when using this line of code...

Code: Select all

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


I've checked the drive on my computer and it is set to share and read-only so that should be fine. Why else would it not be able to open it?[/code]

Hi!
ftp_put -- Uploads a file to the FTP server It's from manual... :!:
If at your HDD folder shared as read-only PHP actually give you message about error.....
It's means what folder only for reading not for putting files :wink:

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Postby Grigory » Sat Dec 07, 2002 1:21 am

Yes I know. It's reading from the file from that drive which is fine and uploading it to my webspace. The error is coming from reading the file from my local machine. I don't think you need "full" read and write access just to read it from there to upload somewhere else.

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Postby WiZARD » Sat Dec 07, 2002 2:16 am

Grigory wrote:Yes I know. It's reading from the file from that drive which is fine and uploading it to my webspace. The error is coming from reading the file from my local machine. I don't think you need "full" read and write access just to read it from there to upload somewhere else.

You somewere have a mistake send me code....

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Postby Grigory » Sat Dec 07, 2002 11:02 am

Ok, here is the code I'm using...

Code: Select all

if (stristr($type,".jpg") || stristr($type,".jpeg")) {
               $ftp_server = "ftp.myserver.com";
               $conn_id = ftp_connect($ftp_server);
               $ftp_user_name = "myusername";
               $ftp_user_pass = "mypassword";
               $login_result = ftp_login($conn_id, $ftp_user_name, $ftp_user_pass);
               // check connection
               if ((!$conn_id) || (!$login_result)) {
                    echo "FTP connection has failed!";
                    echo "Attempted to connect to $ftp_server for user $ftp_user_name";
                    die;
                   }
                  else {
                    echo "Connected to $ftp_server, for user $ftp_user_name";
                   }
               // upload the file
               $destination_file = "/html/catalog/Images/$prodname.jpg";
               $upload = ftp_put($conn_id, $destination_file, $image_no_slashes, FTP_BINARY);
               // check upload status
               if (!$upload)    {
                    echo "FTP upload has failed!";
                   }
                  else {
                    echo "Uploaded $source_file to $ftp_server as $destination_file";
                   }
               ftp_close($conn_id);


And the path to the where the file is to upload to the destination is stored in the variable $image_no_slashes

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Postby WiZARD » Tue Dec 17, 2002 5:48 am

Hi try cgange one string:

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$destination_file = "/html/catalog/Images/$prodname.jpg";

at the

Code: Select all

$destination_file = "/html/catalog/Images/" . $prodname.jpg;


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