php copy funtion

sucmanh

hi all
I still can't figure out how to copy an image (or file) from one location (a url, i.e http://st.com/image.png) to a url destination that I have the right to access (say http://mypage.com/image/)
anyone?
thank you.

rehfeld

have you tried reading the manual?

http://www.php.net/copy

cahva

You cant use copy function to copy from server to another. Use php's ftp functions to copy with ftp-protocol.

sucmanh

yes as ahcava put it, this is more like copy a file from one server to another.
I read the manual on copy() and ftp functions, yet, none of them address my question of copy from one server to another.
copy() documentation however say it's possible to have the destination and source be url, but no solid example is given.
ftp() function also provide ftp_nb_fget/get: but is to get a file from a fpt server that I already have access in, however, i only have access to the destination server, not the source. What I want is something just like IE, copy an image from a website to its cache for later use. I'm still stuck.
thanks.

cahva

Well this should work:

// Sourcefile(http)
$src='http://st.com/image.png';

//  destinationfile(absolute directory to server eg. /home/cahva/public_html/image/image.png)
$dest=$_SERVER['DOCUMENT_ROOT'].'/image/image.png';

copy($src,$dest);

Reason why I first suggested using ftp is that you cannot copy for example *.php files because webserver translates their output(you dont get the sourcefiles). But if you dont need those, use copy 🙂

rehfeld

<?php

$from = 'http://www.google.com/images/logo.gif';
$to = 'google_logo.gif';

copy($from, $to);

?>

sucmanh

I run this:
<?php
// Sourcefile(http)
<?php
$src='http://www.google.com/images/logo.gif';

$dest=$_SERVER['DOCUMENT_ROOT'].'/hehehe/logo.gif';

copy($src,$dest);

and get the error:
Warning: Unable to open 'http://www.google.com/images/logo.gif' for reading: Invalid argument in c:\documents and settings\sucmanh\phpdev\www\hehehe\copy.php on line 7

which exactly why I post this thread

rehfeld

what version of php?

sucmanh

4.2.3

quote from http://us3.php.net/manual/en/function.copy.php

"Note: As of PHP 4.3.0, both source and dest may be URLs if the "fopen wrappers" have been enabled. See fopen() for more details. If dest is a URL, the copy operation may fail if the wrapper does not support overwriting of existing files."

no solid example for this one...

rehfeld

youll need to use fopen() and fread() then because of your version

read the image into a string, the fwrite() it to a file