Re: updating rows which have a common value forconsecutive dates - Mailing list pgsql-general

Hi David,
Thanks for your reply.  I'm using 8.4.7, so window functions are
certainly an option, although I've admittedly never used them before.
I've spent the past few hours reading the dox, and I now have a
rudimentary understanding of window functions.  I tried to compose a
query based on your suggestion, but I think i'm running up against my
lack of experience.  This query seems to give me all failures but not
neccesarily when there are two in a row for a unique group (although
I'm still not 100% certain its actually returning only last_update
consecutive rows):
SELECT testname,os,arch,build_type,branch,current_status,last_update,rank()
OVER (PARTITION BY testname,os,arch,branch,build_type ORDER BY
last_update DESC) FROM mytable WHERE current_status='FAILED' AND
age(now(),last_update) <= INTERVAL '15 days'

However, if I'm understanding how this works, what I really care about
is when a rank=2 exists, as that's truly when something failed for two
consecutive last_update's.  I thought this might do it, but apparently
I'm doing it wrong:

SELECT testname,os,arch,build_type,branch,current_status,last_update,rank()
OVER (PARTITION BY testname,os,arch,branch,build_type ORDER BY
last_update DESC) FROM mytable WHERE current_status='FAILED' AND
age(now(),last_update) <= INTERVAL '15 days' AND rank()=2 ;
ERROR:  window function call requires an OVER clause
LINE 1: ... age(now(),last_update) <= INTERVAL '15 days' AND rank()=2 ;

How do I restrict the results to only show when rank=2 ?

thanks!


On Wed, Apr 13, 2011 at 9:57 AM, David Johnston <polobo@yahoo.com> wrote:
> If you have the ability to use Window functions you can group (as necessary), order by last_update, and then use
rank()to number each test run sequentially.  Then you can limit the results to  ( rank() <= 2 AND current_status =
'FAILED'). 
>
> David J.
>
> -----Original Message-----
> From: pgsql-general-owner@postgresql.org [mailto:pgsql-general-owner@postgresql.org] On Behalf Of Lonni J Friedman
> Sent: Wednesday, April 13, 2011 3:34 AM
> To: pgsql-general
> Subject: [GENERAL] updating rows which have a common value forconsecutive dates
>
> Greetings,
> I have a table full of automated test data, which continuously has new unique data inserted:
>
>    Column     |            Type             |
>  Modifiers
> ----------------+-----------------------------+-------------------------
> ----------------+-----------------------------+-------------------------
> ----------------+-----------------------------+-----------
>  id             | integer                     | not null default
> nextval('dbltests_id_seq'::regclass)
>  testname       | text                        | not null
>  last_update    | timestamp without time zone | not null default now()
>  current_status | text                        | not null
>  os             | text                        | not null
>  arch           | text                        | not null
>  build_type     | text                        | not null
>  branch         | text                        | not null
>
> The 'testname' column contains many different tests, and each unique 'testname' has numerous different associated
os,arch,build_type& branch values.  For example, testname='foo' will run on
os='Linux',arch='i686',build_type='debug',branch='t38',and also on
os='Windows7',arch='x86_64',build_type='release',branch='r40',etc, and there will be many other different testname's
withsimilar permutations of the os,arch,build_type & branch columns.  So for example, there will also be testname='bar'
ortestname='omega' or testname='sigma' for the other assorted os,arch,build_type & branch permutations. 
>
> The current_status column is either 'PASSED' or 'FAILED'.
>
> What I'm struggling with is how to put together a SQL query which shows me all instances of a given testname
permutation
> (os,arch,build_type,branch) which has a current_status value of 'FAILED' for two or more most recent consecutive
'last_update'values. 
>
> Suggestions welcome.
>
> thanks in advance!