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

[Paul Carlton]

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...

     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?


-- 
Paul Carlton
Software Engineer
Cloud Services
Hewlett Packard
BUK03:T242
Longdown Avenue
Stoke Gifford
Bristol BS34 8QZ

Mobile:    +44 (0)7768 994283
Email:    mailto:paul.carlton2 at hpe.com
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