Re: [libvirt] Creating a storage volume for raw format file

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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@xxxxxxx
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