File transfer with Apache2::Upload and XMLHttpRequest

--Apple-Mail-1--789621525
Content-Transfer-Encoding: quoted-printable
Content-Type: text/plain;
charset=us-ascii

I tried to switch from the 'common' way to send a file to the server, =
which is in my case:

a) define a form with an file-input-element:

<form method=3D"post" action=3D"gallery?action=3Dupload_image" =
enctype=3D"multipart/form-data">
<input type=3D"file" name=3D"file"/>
<input type=3D"submit" value=3D"Submit" />
</form>

b) define mod_perl code, that saves the file-data inside a handler, =
something like:

my $apr =3D Apache2::Request->new($r);
my $upload =3D $apr->upload('file');
....

That works fine as it is easy to work with $upload and save the data.
But I really can not figure out how to handle the XHR-Approach on the =
server-side because now, of cause, the 'file' param is missing.

The JS client side code is (where 'file' is a HTML5 File-Object):

var xhr =3D new XMLHttpRequest();
xhr.open("POST", url + "?filename=3D" + encodeURIComponent(file.name), =
true);
xhr.setRequestHeader("Content-Type", "multipart/form-data");
xhr.send(file);

Any idea how now the server side part has to look like now?

Thanks, Michael=

--Apple-Mail-1--789621525
Content-Transfer-Encoding: quoted-printable
Content-Type: text/html;
charset=us-ascii

<html><head></head><body style=3D"word-wrap: break-word; =
-webkit-nbsp-mode: space; -webkit-line-break: after-white-space; ">
 I tried to switch from the 'common' way to send a =
file to the server, which is in my case: a) define a form with an =
file-input-element: <span=
class=3D"Apple-style-span" style=3D"font-family: verdana, geneva; =
font-size: 12px; "><form method=3D"post" =
action=3D"gallery?action=3Dupload_image" =
enctype=3D"multipart/form-data"> <input =
type=3D"file" name=3D"file"/> <input =
type=3D"submit" value=3D"Submit" /> </form> b) define mod_perl code, that saves the file-data =
inside a handler, something like: my $apr =3D =
Apache2::Request->new($r); my $upload =3D =
$apr->upload('file'); .... That works fine as it is easy to =
work with $upload and save the data. But I really =
can not figure out how to handle the XHR-Approach on the server-side =
because now, of cause, the 'file' param is missing. The JS client side code is (where =
'file' is a HTML5 File-Object): var xhr =3D new XMLHttpRequest(); xhr.open("POST", url + "?filename=3D" + =
encodeURIComponent(<a =
href=3D"http://file.name/">file.name</a>), true); xhr.setRequestHeader("Content-Type", =
"multipart/form-data"); xhr.send(file); Any idea how now the server side =
part has to look like now? Thanks, Michael</body></html>=

--Apple-Mail-1--789621525--

File transfer with Apache2::Upload and XMLHttpRequest

Re: File transfer with Apache2::Upload and XMLHttpRequest