No, I'm not sending it with other files. I just want to send this one,
a bytearray of a jpg file.  How do I do step 2 on your list?

If I make a UrlRequest, and set it's "data" property to my ByteArray,
 then what will the variable name be in php's $_FILES array?  

$_FILES['what??'] ?

It's not $_FILES['Filedata'], that's blank.  That's what it is for a
FileReference.upload(), but not for a URLRequest/URLLoader.






--- In [email protected], "Tim Scollick" <[EMAIL PROTECTED]>
wrote:
>
> I might be wrong but I think that if you wanted it in the files
array, you
> would have to:
> 1.  Get the ByteArray.
> 2.  Send it to the server to make the jpg.
> 3.  Load the jpg into Flash.
> 4.  Send it again in the files array.
> 
> I think it would be easier and more effiicient to send the ByteArray
to make
> the jpeg and then send the other files independently (I'm assuming that
> you're sending it with other files).
> 
> On 6/6/06, rigidcode <[EMAIL PROTECTED]> wrote:
> >
> >
> > If I make a ByteArray of jpeg data, how do I send it to the server in
> > such a way that PHP will have it in it's $_FILES array? If I just set
> > the URLRequest.data property to the ByteArray, then it won't have a
> > filename, or a variable name, to refer to in the php.
> >
> > http://us3.php.net/manual/en/features.file-upload.php
> >
> >  
> >
>







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