I see what you mean; the RFID tags I am using are 13.56 Mhz, and the
Nexus S can detect them and .

If my understanding is correct, the way it happens is that when an
13.56 RFID or NFC tag is detected, a new intent is thrown with some
kind of special added data and we "resolve" this intent to get the
information out of it.
I am doing the following:

1) call the resolveintent(Intent intent) function
2) Check if the action in the intent is "TAG_DISCOVERED" (this is
always the case with my tags, maybe because they're not a type android
is familiar with)
3) Then, I want to extract this UID: what should i call?
intent.get.... ?
          Note: I tried "  Parcelable[] rawMsgs =
intent.getParcelableArrayExtra(NfcAdapter.EXTRA_NDEF_MESSAGES);"  as
is        described on the android website, but it returned null,
probably because the tag type is unknown



Thank you very much

Pierre

On May 18, 2:34 am, nemik <ne...@nemik.net> wrote:
> I don't know about a "bit vector" but most NFC tags (whether type A,
> B, Mifare 1k/Ultralight/etc) all have a "UID". This is a unique
> identifier for the tag and what you're looking for.
>
> But the tags you use will have to be of the 13.56 MHz hi-frequency
> variety. You won't be able to read most older HID Prox cards for
> example because they operate on the 125 kHz frequency.
>
> On May 17, 5:19 pm, Pierre Yaacoub <pierreyaac...@gmail.com> wrote:
>
>
>
>
>
>
>
> > Hello,
>
> > I am trying to build an application that uses the new Android
> > capability of detectingRFID(NFC) tags.
> > The tags that I have are not NFC tags, that is, they are not of a
> > special type (like URI ...).
> > For my purposes, I do not care about parsing the tag data in a certain
> > way (as would be the case in some NFC tags like URI or SmartText). I
> > just need to access the bit vector that the tag sends, and use it as a
> > unique identifier.
>
> > Is it possible to do this? That is, if the tag is anRFIDtag that is
> > being detected by the phone, but isn't of any special type, can I
> > access its bit vector?
>
> > Thank you very much
>
> > Pierre

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