Yes, it is.
the return value of fopen is a resource of type file.
var_dump($data); will show that
if you want to read the data use : fread($data,$buf);
but I think its better with file() function. The example from the docs show a better
method:
<?php
$data = implode("", "bigfile.txt");
$gzdata = gzencode($data, 9);
$fp = fopen("bigfile.txt.gz", "w");
fwrite($fp, $gzdata);
fclose($fp);
?>
Regards,
Andrey Hristov
IcyGEN Corporation
http://www.icygen.com
BALANCED SOLUTIONS
----- Original Message -----
From: "Keith Whyman" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Tuesday, November 27, 2001 4:23 PM
Subject: [PHP-DB] gzip image files
> I create a png dynamically - store the path in mysql.
>
> And want to be able to offer the user a zipped version to download
> Unfortunately haven't been able to get it to work up until now !
> I'm sure it's something obvious but...........
>
>
> Open the png
> $data = fopen('kunden/'.$name.'.png', "rb");
> encode it
> $gzdata = gzencode($data);
> where to write
> $fp = fopen('kunden/zip/'.$name.'.gz', "w");
> write it
> fwrite($fp, $gzdata);
> fclose($fp);
>
> what happens is the name gets included in the gz but written to it is only
> kunden/example.png
>
> Can anyone help ??
>
> thanks in advance
> regards
> keith
>
>
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