[libvirt-users] [libvirt] Creating a storage volume for raw format file
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) Thanks -- 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@hpe.com Hewlett-Packard Limited registered Office: Cain Road, Bracknell, Berks RG12 1HN Registered No: 690597 England. The contents of this message and any attachments to it are confidential and may be legally privileged. If you have received this message in error, you should delete it from your system immediately and advise the sender. To any recipient of this message within HP, unless otherwise stated you should consider this message and attachments as "HP CONFIDENTIAL".
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
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@hpe.com Hewlett-Packard Limited registered Office: Cain Road, Bracknell, Berks RG12 1HN Registered No: 690597 England. The contents of this message and any attachments to it are confidential and may be legally privileged. If you have received this message in error, you should delete it from your system immediately and advise the sender. To any recipient of this message within HP, unless otherwise stated you should consider this message and attachments as "HP CONFIDENTIAL".
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?
On 04/05/16 14:30, Cole Robinson wrote:
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?
Cole,
Thanks, that is what I will use! -- 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@hpe.com Hewlett-Packard Limited registered Office: Cain Road, Bracknell, Berks RG12 1HN Registered No: 690597 England. The contents of this message and any attachments to it are confidential and may be legally privileged. If you have received this message in error, you should delete it from your system immediately and advise the sender. To any recipient of this message within HP, unless otherwise stated you should consider this message and attachments as "HP CONFIDENTIAL".
participants (2)
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Cole Robinson -
Paul Carlton