in [mailto:nj...@fb.com]
Sent: Wednesday, February 09, 2011 8:24 AM
To: user@hive.apache.org
Subject: Re: for each partition
You can use dynamic partitioning:
insert overwrite table item_view_aggregate partition
(date_hour) select iv.sid, count(*), date_hour from item_view iv where
(iv.date_hour=&#
ect iv.sid, count(*) from item_view iv where
>iv.date_hour='2011310116' group by iv.sid;
>
>..
>
>I have to repeat this for each partition. so I need something like a
>for loop, or do it with an external program.
>
>best regards,
>-c.b.
>
>On Wed, Feb 9, 20
='2011310116' group by iv.sid;
..
I have to repeat this for each partition. so I need something like a
for loop, or do it with an external program.
best regards,
-c.b.
On Wed, Feb 9, 2011 at 7:33 AM, Christopher, Pat
wrote:
> If you want to operate over all partitions in a table you
wrote:
Hello,
How can I do some process for each partition in some other table.
for example lets say table A has partitions 1,2,3
I want to be able to say
for each partition in A do {
select * from A where partition is ? into some othertable where partition is ?
}
Best Regards,
C.B.
Hello,
How can I do some process for each partition in some other table.
for example lets say table A has partitions 1,2,3
I want to be able to say
for each partition in A do {
select * from A where partition is ? into some othertable where partition is ?
}
Best Regards,
C.B.
