[libvirt-users] [libvirt] Creating a storage volume for raw format file

[Cole Robinson]

On 05/04/2016 09:25 AM, Paul Carlton wrote:
> 
> 
> On 04/05/16 14:07, Cole Robinson wrote:
>> On 05/04/2016 06:54 AM, Paul Carlton wrote:
>>> Hi
>>>
>>> I'm trying to create a volume in an existing storage pool using python by
>>> calling createXML() on the pool object.
>>> If the file that is the subject of the new volume exists you get and error,
>>> also if it doesn't exist
>>>
>>> I worked out that specifying
>>>
>>> <source>
>>>    <path>path to copy of file</path>
>>> </source>
>>>
>>>   works, seemingly by copying the 'source' file to the intended location of
>>> the
>>> volume file
>>>
>>> However I am concerned about the overhead of copying the file, is there a way
>>> to tell it to move the file (i.e. rename)
>> Unfortunately there isn't any simple API to 'rename' a storage volume
>>
>> - Cole
>>
> Actually I don't want to rename a storage pool, in my inept attempt a brevity
> I failed to explain myself clearly.
> 
> I want to create a raw format file in a directory that is referenced by
> directory storage pool and register it as a storage volume
> 
> If I place the file in the directory then try to create a storage volume to
> reference it the operation fails due to the file already existing.
> however if I try to create the storage pool then copy the file to the
> directory it fails to create the storage pool because the file does not exist.
> I figured out that if I create the file in another directory then create the
> storage volume as follows...
> 

Not sure if I fully follow, but if you manually create the file in the
directory, outside of libvirt's APIs, just do a pool.refresh() and libvirt
will notice the new volume and it will show up in listVolumes. createXML is
only for telling libvirt to create an entirely new disk image (likely by
invoking qemu-img), not to teach it about an existing image... that's what
pool.refresh() is essentially for

- Cole

>     def create_volume(self, vol_name, src_path):
>         def create_vol_xml(name, src):
>             xml = """<volume type='file'>
>                       <name>{name}</name>
>                       <allocation unit='bytes'>{allocation}</allocation>
>                       <capacity unit='bytes'>{capacity}</capacity>
>                       <source>
>                         <path>{src_path}</path>
>                       </source>
>                       <target>
>                         <permissions>
>                           <mode>{mode}</mode>
>                           <owner>{uid}</owner>
>                           <group>{gid}</group>
>                         </permissions>
>                       </target>
>                      </volume>"""
>             stat = os.stat(src)
>             qmeu_info = images.qemu_img_info(src)
>             return xml.format(**{'name': name,
>                                  'capacity': qmeu_info.virtual_size,
>                                  'allocation': qmeu_info.disk_size,
>                                  'src_path': src,
>                                  'format': qmeu_info.file_format,
>                                  'mode': oct(stat.st_mode)[-3:],
>                                  'uid': stat.st_uid,
>                                  'gid': stat.st_gid})
> 
>         vol_xml = create_vol_xml(vol_name, src_path)
>         self._vols[vol_name] = self._pool.createXML(vol_xml, 0)
> 
> Where 'src_path' is the path to a file in another directory it seems to copy
> src_path to my storage pool directory as 'vol_name' and
> creates a storage volume that references it.
> 
> This is fine except, my concern is what if the file is very large the copy
> operation could take a long time and we'd use twice the
> disk space we should do.
> 
> Since I could not find any documentation on the <source> elements for a
> directory pool's volume xml I was wondering if there
> was a setting that would make it move/rename src_file to the vol_name file in
> my pool directory rather than copy it?
> 
>