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Editorial Team
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Asked: 2026-06-13T02:39:40+00:00

Let’s say I have following code: $file=http://example.com/path/to/file.zip; $in = fopen($file, r); var_dump($in); // result

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Let’s say I have following code:

$file="http://example.com/path/to/file.zip";
$in = fopen($file, "r");
var_dump($in);             // result - resource(8) of type (stream)

$put = ftp_fput($ftp->getConnectionId(), $file, $in, FTP_BINARY);
var_dump($put);           // result - bool(false)

And I am getting following error:

Warning: ftp_fput(): http://example.com/path/to/file.zip: No such file or directory in /var/www/inc/application.php

Connection is established correctly, because I can change directory and read files, but problem appears when I want to upload file as above. What am I doing wrong? Thanks in advance for any help.

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1 Answer

  1. Editorial Team

    The second parameter of ftp_fput allows you to define the remote file path & name.

    You are attempting to use a URL instead of a path for your second parameter.

    It should be like so:

    $put = ftp_fput($ftp->getConnectionId(), '/path/to/file.zip', $in, FTP_BINARY);

    Source: http://au.php.net/ftp_fput

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