[mythtv-users] Two new physical drives appear as subdirectories of the same drive: why?

[Scott Theisen]

Since I have been looking at that part of the MythTV code, I'll add my 
two cents:

MythTV gets its filesystem data by (from my WIP PR 
https://github.com/MythTV/mythtv/pull/485 ):
```
bool FileSystemInfo::refresh()
{
     struct statfs statbuf {};

     // there are cases where statfs will return 0 (good), but f_blocks and
     // others are invalid and set to 0 (such as when an automounted 
directory
     // is not mounted but still visible because --ghost was used),
     // so check to make sure we can have a total size > 0
     if ((statfs(getPath().toLocal8Bit().constData(), &statbuf) == 0) &&
         (statbuf.f_blocks > 0) && (statbuf.f_bsize > 0)
        )
     {
         m_total = statbuf.f_blocks * statbuf.f_bsize;
         //free  = statbuf.f_bavail * statbuf.f_bsize;
         m_used  = m_total - statbuf.f_bavail * statbuf.f_bsize;
         m_blksize = statbuf.f_bsize;

         // TODO keep as B not KiB
         m_total >>= 10;
         m_used  >>= 10;

#ifdef Q_OS_DARWIN
         char *fstypename = statbuf.f_fstypename;
         m_local = !(
                     (strcmp(fstypename, "nfs")   == 0) ||  // NFS|FTP
                     (strcmp(fstypename, "afpfs") == 0) ||  // AppleShare
                     (strcmp(fstypename, "smbfs") == 0)     // SMB
                    );
#elif defined(__linux__)
         long fstype = statbuf.f_type;
         m_local = !(
                     (fstype == 0x6969)  ||   // NFS
                     (fstype == 0x517B)  ||   // SMB
                     (fstype == 0xFF534D42L)  // CIFS
                    );
#else
         m_local = true; // for equivalent behavior
#endif
         return true;
     }
     return false;
}
```

This path can include network mounted filesystems, which would be 
available to more than one backend.  Consolidate() uses a default fuzz 
value of ~14 MiB.

On 3/6/22 21:27, Stephen Worthington wrote:
> On Sun, 6 Mar 2022 23:03:07 +0000, you wrote:
>
>> On 06/03/2022 22:03, UB40D via mythtv-users wrote:
>>> On Sun, 6 Mar 2022 at 00:33, Gary Buhrmaster <gary.buhrmaster at gmail.com>
>>> wrote:
>>>
>>>> On Sat, Mar 5, 2022 at 11:56 PM UB40D via mythtv-users
>>>> <mythtv-users at mythtv.org> wrote:
>>>>
>>>>> Thanks for this explanation but I find it mindboggling... Why does Myth
>>>> need "heuristics" when perfectly reliable first-hand information is
>>>> available from the computer and OS about which drive is which?
>>>>
>>>> Because some people choose to use different
>>>> directories on the same drive, and therefore
>>>> the amount of space available on the drive is
>>>> not always a simple thing to calculate.  Only
>>>> when the available space is different does
>>>> MythTV "guess" that they must be different
>>>> drives (and calculates space as if they are
>>>> different).
>>>>
>>> I am not sure I understand the problem the programmers experienced when
>>> they wrote the code that way. It seems to me that there is no need for
>>> heuristics (= guesswork) when unequivocal information is readily available.
>>>
>>> for every directory d listed in mythtv setup as containing recordings:
>>>      df -h $d will return something like
>>> Filesystem      Size  Used Avail Use% Mounted on
>>> /dev/sdd1        17T   16T  294G  99% /mnt/wd18000a1
>>>
>>> so it is trivial to figure out both
>>> a) how much free space is left on the drive that hosts that directory
>>> b) whether that directory is on the same filesystem or drive of any other
>>> directory (to avoid double-counting)
>>>
>>> I may be missing something crucial but I still fail to see why any
>>> guesswork might be needed.
>>>
>>> BTW I did copy some files from the "recording" drive to one of those new
>>> "read-only" drives and indeed they now appear as separate drives, and the
>>> reported total amount of free space is now what I expected, so thanks to
>>> David and you for that suggestion.
>>>

In order to use the partitions, you must identify which partition a 
folder is on.  How to do this will be different depending on the OS 
(especially Windows vs. UNIX-like).  Using the UUID of the partition may 
be a better solution, but that is not the purpose of my PR and I will 
not investigate it.

>> Your arrangement implies that one partition = one storage directory. In the past this was not always
>> the case. Users may have created more than one storage directory on a single partition, perhaps to
>> keep different recordings separate or for other reasons. When that is true it is not possible to
>> discover the free space in each storage directory. Indeed, there may have been other non-myth
>> directories on those devices as well, adding to the fun.
>>
>> In these days, when storage is stupid cheap, such configurations are not usually necessary.
> But they do still exist.  I have two directories in different storage
> groups on each of my recording drives.  One is for my recordings, and
> the other is for recordings I did for my mother from my pay satellite
> service.  Her MythTV box has access to the directories containing her
> recordings, and I can transfer the database entries for them to her
> box so that they show up there are normal recordings.  All of that is
> now redundant as her MythTV box now has access to SAT>IP tuners for my
> satellite service and she records the satellite programmes that way.
> But there are still quite a few old recordings left in her
> directories.

Not only do they still exist, but it is (was?) the default for Ubuntu to 
have different folders for recordings and live TV.