ldmwndletsm
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Can anyone explain or show an example of how to use the 'show aggregate' command ? I was playing around with 'show bfo' and 'show invo', but this one is a bit perplexing.
The IBM documentation (Show commands for the server or storage agent) states that it's run as:
'show aggregate aggrID-high aggrID-low'
where each of these is the high and low order 32-bit words of the 64-bit aggregate ID. So am I supposed to somehow convert the aggregated ID into binary and then separate the two parts ?
As an example, I just ran 'show bfo 7205804175' where this specific object_id was returned from a "select * from contents where volume_name='volume_name'" (and/or a "select * from contents where node_name='node_name' and filespace_name='/filespace' and hl_name='/dir1/' and ll_name='filename'")
It reports thus:
Bitfile Object: 7205804175
Active
**Sub-bitfile 7205804175 is stored in the following aggregate(s)
Super-bitfile: 7205804172, Offset: 16797, Length 758, Deduped: F, CopySdType: F
If I run 'show bfo 7205804172' then of course I can see the volumes, and 'show invo 7205804175' and 'show invo 7205804172' also work.
Next, I then ran 'show aggr 7205804172' and that reports:
AggrId: 7205804172, AggrSize: 22791, LogSize: 22791, RptSize: 22791, NumFiles: 8, NumChunks: 0, Analyzed: False, Type: 0.
Aliases: None.
No idea if I did this correctly, but it appears that the aggregate id is 7205804172.
1. Is it in fact the case that the aggregate ID is the Super-bitfile value ?
2. I'm still unclear on why IBM specifies high and low as separate arguments and whether I ran it correctly. The documentation seems to suggest that I would pass it two arguments (maybe one would be the binary representation for the high order bits and the other the low order bits ?). What would be the low in this case ? If you can just use the AggrId (Super-bitfile) then why don't they just say that ?
The IBM documentation (Show commands for the server or storage agent) states that it's run as:
'show aggregate aggrID-high aggrID-low'
where each of these is the high and low order 32-bit words of the 64-bit aggregate ID. So am I supposed to somehow convert the aggregated ID into binary and then separate the two parts ?
As an example, I just ran 'show bfo 7205804175' where this specific object_id was returned from a "select * from contents where volume_name='volume_name'" (and/or a "select * from contents where node_name='node_name' and filespace_name='/filespace' and hl_name='/dir1/' and ll_name='filename'")
It reports thus:
Bitfile Object: 7205804175
Active
**Sub-bitfile 7205804175 is stored in the following aggregate(s)
Super-bitfile: 7205804172, Offset: 16797, Length 758, Deduped: F, CopySdType: F
If I run 'show bfo 7205804172' then of course I can see the volumes, and 'show invo 7205804175' and 'show invo 7205804172' also work.
Next, I then ran 'show aggr 7205804172' and that reports:
AggrId: 7205804172, AggrSize: 22791, LogSize: 22791, RptSize: 22791, NumFiles: 8, NumChunks: 0, Analyzed: False, Type: 0.
Aliases: None.
No idea if I did this correctly, but it appears that the aggregate id is 7205804172.
1. Is it in fact the case that the aggregate ID is the Super-bitfile value ?
2. I'm still unclear on why IBM specifies high and low as separate arguments and whether I ran it correctly. The documentation seems to suggest that I would pass it two arguments (maybe one would be the binary representation for the high order bits and the other the low order bits ?). What would be the low in this case ? If you can just use the AggrId (Super-bitfile) then why don't they just say that ?
