zfs discuss - Oct 2012 - removing upgrade notice from 'zpool status -x'

Hi,

I have a machine whose zpools are at version 28, and I would like to
keep them at that version for portability between OSes. I understand
that ''zpool status'' asks me to upgrade, but so does
''zpool status -x''
(the man page says it should only report errors or unavailability).
This is a problem because I have a script that assumes "zpool status
-x" only returns errors requiring user intervention.

Is there a way to either:
A) suppress the upgrade notice from ''zpool status -x'' ?
B) use a different command to get information about actual errors
w/out encountering the upgrade notice ?

I''m using OpenIndiana 151a6 on x86.

Jan

On Oct 4, 2012, at 8:58 AM, Jan Owoc <jsowoc at gmail.com> wrote:
> Hi,
> 
> I have a machine whose zpools are at version 28, and I would like to
> keep them at that version for portability between OSes. I understand
> that ''zpool status'' asks me to upgrade, but so does
''zpool status -x''
> (the man page says it should only report errors or unavailability).
> This is a problem because I have a script that assumes "zpool status
> -x" only returns errors requiring user intervention.
The return code for zpool is ambiguous. Do not rely upon it to determine
if the pool is healthy. You should check the health property instead.
> Is there a way to either:
> A) suppress the upgrade notice from ''zpool status -x'' ?
Pedantic answer, it is open source ;-)
> B) use a different command to get information about actual errors
> w/out encountering the upgrade notice ?
> 
> I''m using OpenIndiana 151a6 on x86.

 -- richard

--

Richard.Elling at RichardElling.com
+1-760-896-4422









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On Thu, Oct 4, 2012 at 9:14 AM, Richard Elling <richard.elling at
gmail.com> wrote:
> On Oct 4, 2012, at 8:58 AM, Jan Owoc <jsowoc at gmail.com> wrote:
> The return code for zpool is ambiguous. Do not rely upon it to determine
> if the pool is healthy. You should check the health property instead.
Huh.  Learn something new everyday.  You just simplified my "pool
health check" script immensely.  Thank you!

pstatus=$( zpool get health storage | grep "health" | awk ''{
print $3 }'' )
if [ "${pstatus}" != "ONLINE" ]; then

Much simpler than the nested ifs and grep pipelines I was using before.

Not sure why I didn''t see "health" in the list of pool
properties all
the times I''ve read the zpool man page.

-- 
Freddie Cash
fjwcash at gmail.com

2012-10-04 20:36, Freddie Cash ?????:
> On Thu, Oct 4, 2012 at 9:14 AM, Richard Elling <richard.elling at
gmail.com> wrote:
>> On Oct 4, 2012, at 8:58 AM, Jan Owoc <jsowoc at gmail.com> wrote:
>> The return code for zpool is ambiguous. Do not rely upon it to
determine
>> if the pool is healthy. You should check the health property instead.
>
> Huh.  Learn something new everyday.  You just simplified my "pool
> health check" script immensely.  Thank you!
>
> pstatus=$( zpool get health storage | grep "health" | awk
''{ print $3 }'' )
> if [ "${pstatus}" != "ONLINE" ]; then
Simplify that too with "zpool list":

# zpool list -H -o health rpool
ONLINE

;)
//Jim

On Thu, Oct 4, 2012 at 9:45 AM, Jim Klimov <jimklimov at cos.ru>
wrote:
> 2012-10-04 20:36, Freddie Cash ?????:
>>
>> On Thu, Oct 4, 2012 at 9:14 AM, Richard Elling <richard.elling at
gmail.com>
>> wrote:
>>>
>>> On Oct 4, 2012, at 8:58 AM, Jan Owoc <jsowoc at gmail.com>
wrote:
>>> The return code for zpool is ambiguous. Do not rely upon it to
determine
>>> if the pool is healthy. You should check the health property
instead.
>>
>>
>> Huh.  Learn something new everyday.  You just simplified my "pool
>> health check" script immensely.  Thank you!
>>
>> pstatus=$( zpool get health storage | grep "health" | awk
''{ print $3 }'' )
>> if [ "${pstatus}" != "ONLINE" ]; then
>
>
> Simplify that too with "zpool list":
>
> # zpool list -H -o health rpool
> ONLINE
Thanks!  Was trying to figure out how to remove the heading as I use
-H a lot with zfs commands, but it didn''t work for "zpool
get" and I
didn''t bother reading to figure it out.  :)


-- 
Freddie Cash
fjwcash at gmail.com