On 9 April 2010 19:35, Anna Powell-Smith <annapowellsm...@googlemail.com>wrote:

>
>
> On 9 April 2010 18:19, Gubatron <gubat...@gmail.com> wrote:
>
>> I suppose when you say FilePart, you mean this
>>
>> org.apache.commons.httpclient.methods.multipart.FilePart
>>
>> I'm thinking along these lines after looking at that API (I haven't
>> tested this)
>>
>> final File theFile = new File("yourFileLargerThan2Mb.ext");
>>
>> //Implement your own PartSource to feed your File
>> PartSource  partSource = new PartSource() {
>>   //this should return a buffered reader... right?
>>   public InputStream createInputStream() {
>>      return BufferedInputStream (new FileInputStream(theFile));
>>   }
>>
>>   //implement the other methods of the interface
>>   public String getFileName() {
>>     return theFile.getName();
>>   }
>>
>>   public long getLength() {
>>     return theFile.length();// although this might be how much is
>> left on the stream, not sure.
>>   }
>>
>> }
>>
>> FilePart part = FilePart(theFileName, partSource);
>>
>> then use your part on your multipart request.
>>
>
> Thanks. Constructing a FilePart from a File is actually as easy as this, I
> think - but do you think your method would provide some advantage?
>
>     File temp_file = new File(Environment.getExternalStorageDirectory(),
>  "myfile.gif");
>     FilePart vFile = new FilePart("fileupload", temp_file);
>
> As I say, this works for files of 50MB. The only reason I can't use it is
> that I can't construct a File object from an Android content Uri. I have to
> use an Android content Uri because that's what the video intent returns.
>

Ah. If I could transform my content Uri into FileInputStream (will have to
experiment) then I could use your method.

I can convert a content Uri to an InputStream already, so it'll just be
getting from an InputStream to a FileInputStream that might be the
challenge. I'll experiment.

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