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Posted by Rik Wasmus on 10/29/07 19:30
On Mon, 29 Oct 2007 20:07:29 +0100, Fons <fons@fons.invalid> wrote:
> My script acts on a file (approx. 100k) offered by the user. Actually,=
it
> acts on a variable that should have the contents of that file in it. I=
> could try to let users upload that file and then read that file into a=
> variable, but before I figure out how to do that: is this unnecessary
> overhead? Is it possible to do it without creating a file on the serve=
r
> filesystem and do it completely within the PHP framework? What method
> would you recommend?
Just let it do a normal fileupload to a/the tmp directory and read it in=
to =
a variable. Anything else will be a waste of time. The code is pretty =
simple:
if(isset($_FILES['your_html_fileinput_name'])){
if($_FILES['your_html_fileinput_name']['error']!=3DUPLOAD_ERR_OK){
echo 'upload error #'.$_FILES['your_html_fileinput_name']['error'];
} else {
$filecontents =3D =
file_get_contents($_FILES['your_html_fileinput_name']['tmp_name']);
}
} else {
echo 'no upload took place';
}
-- =
Rik Wasmus
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