HELP: MCI Control for VB 5 TimeFormat (CDAudio) 
Author Message
 HELP: MCI Control for VB 5 TimeFormat (CDAudio)

Quote:

>Any help would be greatly appreciated. I am just writing a simple CD
>Player and then I want it to access the CDDB servers to look up CD
>information which needs the track offsets in MM:SS:FF format.

I am working on the same thing (sort of).  I have found a Microsoft article
in their KB (Q94012) but it is for VB 2.0.  It then tells you how to extract
the information from the 4-byte integer returned.

HOWEVER, it does not seem to work under VB 5 Enterprise.  When I try what
they say, I still do not get the correct information.  I use the TMSF format
and on a CD where the first song is 4m:19s long, it says track is 0, M is 2,
S is 0 and F is 0 (at least that is what I remember).  Since it says each
part is returned as a byte (byte1 is least and byte4 is most significant), I
assumed that their formulas (byte1 = mmcontrol.position and &HFF&) return
actual values and you do not need to do any other transfomration.

If you found out anything new, please let me know.




Sat, 17 Jun 2000 03:00:00 GMT  
 HELP: MCI Control for VB 5 TimeFormat (CDAudio)

Quote:


>>Any help would be greatly appreciated. I am just writing a simple CD
>>Player and then I want it to access the CDDB servers to look up CD
>>information which needs the track offsets in MM:SS:FF format.
>I am working on the same thing (sort of).  I have found a Microsoft article
>in their KB (Q94012) but it is for VB 2.0.  It then tells you how to extract
>the information from the 4-byte integer returned.

>HOWEVER, it does not seem to work under VB 5 Enterprise.  When I try what
>they say, I still do not get the correct information.  I use the TMSF format
>and on a CD where the first song is 4m:19s long, it says track is 0, M is 2,
>S is 0 and F is 0 (at least that is what I remember).  Since it says each
>part is returned as a byte (byte1 is least and byte4 is most significant), I
>assumed that their formulas (byte1 = mmcontrol.position and &HFF&) return
>actual values and you do not need to do any other transfomration.

>If you found out anything new, please let me know.



First off, the cddb data base does not use the same ID scheme as MS.

Your 0:02:00 is the starting position of the first track.

--
Wolfgang John    - AM-WOLJO Shareware -

 - Phone Dialer and Logger - CD Player and Library - Win3.x / Win95 / NT
- CD Player imports data from GIANT internet library -
- FREE programmable 6 cities/countries world clock -

Please remove "nospam" from my signature if replying by email.



Mon, 19 Jun 2000 03:00:00 GMT  
 HELP: MCI Control for VB 5 TimeFormat (CDAudio)

Quote:
>First off, the cddb data base does not use the same ID scheme as MS.

Yes, I know that, he and I were talking about getting the offsets of each
track from the values returned via the MCI control.

Quote:
>Your 0:02:00 is the starting position of the first track.

But at the time I was just trying to get the length of each track, not its
offset.


Tue, 20 Jun 2000 03:00:00 GMT  
 
 [ 4 post ] 

 Relevant Pages 

1. MCI Help with CDAudio

2. VB 4.0 and CDAudio using MCI ????

3. MCI TimeFormat ??????!!!

4. MCI and the timeformat property

5. MCI Multimedia Timeformat confusion

6. MCI CDAUDIO Question/Problem

7. MCI CDAUDIO

8. MCI CDAUDIO FF OR RR?

9. How to play CDaudio(MCI) on a redbook CDROM

10. mci and cdaudio

11. MCI CDAUDIO

12. I need Help for "TIMEFORMAT" at MCI Control

 

 
Powered by phpBB® Forum Software