Set,
Thank you. This definitely does something...to be honest I'm not certain
exactly what... I'm definitely going to need to study this for a while.
I sincerely appreciate the effort - though I think I am going to
reconsider my table structure.
--
Daniel
On 9/28/2017 1:42:18 PM, "setysvar [email protected]
[firebird-support]" <[email protected]> wrote:
Sorry again, Daniel, I had to get to a place where I could test things
before replying. My problem was that what I'd done before was to try to
find identical sets, not trying to find unique sets, and that made me
mess up the logic. I hope this query will get you what you want (and
this time I've tested before replying):
with tmp(NODE, ROUTE_UPDATED) as
(select distinct r.NODE, rh.ROUTE_UPDATED
from routes r
left join route_history rh on r.NODE = rh.NODE),
new_routes(NODE) as
(select distinct t.NODE
from tmp t
where exists(select *
from routes r
left join route_history rh on r.NODE = rh.NODE
and r.ROUTE_INDEX =
rh.ROUTE_INDEX
and r.LINK_NODE = rh.LINK_NODE
and r.QUALITY is not distinct
from rh.QUALITY
and t.ROUTE_UPDATED =
rh.ROUTE_UPDATED
where t.NODE = r.NODE
and rh.NODE is null)
or exists(select *
from route_history rh
left join routes r on r.NODE = rh.NODE
and r.ROUTE_INDEX = rh.ROUTE_INDEX
and r.LINK_NODE = rh.LINK_NODE
and r.QUALITY is not distinct from
rh.QUALITY
where t.NODE = rh.NODE
and t.ROUTE_UPDATED = rh.ROUTE_UPDATED
and r.NODE is null))
select r.NODE, r.ROUTE_INDEX, r.LINK_NODE, r.QUALITY
from new_routes nr
join routes r on nr.NODE = r.NODE
HTH,
Set
Den 28.09.2017 11:00, skrev 'Daniel Miller' [email protected]
[firebird-support]:
This appears no different than a simple:
select * from routes
So...not there yet. I think what I'm fighting is the lack of pivot or
crosstab functionality - and I may have no choice (though it offends
me deeply) but to setup my archive table for this comparison. Either
via the hated structure of column names ROUTE1, ROUTE2, ROUTE8...or
the only slightly less offensive VARCHAR concatenation.
A concatenated LIST seems like it would be a elegant solution - and
while operating over the whole table is time-consuming a single node
isn't too bad. And if I can get the history filtered and old
duplicate records deleted it would be fine. But the inconsistency is
killing me.
There MUST be a Firebird-friendly, normalized, compact storage
structure that will work. I'm just not seeing it.
--
Daniel
On 9/28/2017 1:19:18 AM, "Svein Erling Tysvær [email protected]
[firebird-support]" <[email protected]> wrote:
Sorry again, Daniel, fourth query required (I forgot that things may
vary depending on time, this is something I've never done before):
with tmp(NODE, ROUTE_UPDATED) as
(select distinct rh.NODE, rh.ROUTE_UPDATED
from route_history rh
join routes r on rh.NODE = r.NODE), //*This join is only for
performance reasons in case ROUTE_HISTORY contains a lot more records
than ROUTES and may be removed*/
new_routes(NODE) as
(select distinct t.NODE
from tmp t
where not exists(select *
from routes r
left join route_history rh on r.NODE = rh.NODE
and r.ROUTE_INDEX =
rh.ROUTE_INDEX
and r.LINK_NODE =
rh.LINK_NODE
and r.QUALITY is not
distinct from rh.QUALITY
and t.ROUTE_UPDATED =
rh.ROUTE_UPDATED
where t.NODE = r.NODE
and rh.NODE is null)
and not exists(select *
from route_history rh
left join routes r on r.NODE = rh.NODE
and r.ROUTE_INDEX =
rh.ROUTE_INDEX
and r.LINK_NODE =
rh.LINK_NODE
and r.QUALITY is not
distinct from rh.QUALITY
where t.NODE = rh.NODE
and t.ROUTE_UPDATED = rh.ROUTE_UPDATED
and r.NODE is null))
select r.NODE, r.ROUTE_INDEX, r.LINK_NODE, r.QUALITY
from new_routes nr
join routes r on nr.NODE = r.NODE
Hope I finally got it right,
Set
2017-09-28 10:01 GMT+02:00 Svein Erling Tysvær <[email protected]>:
Sorry, I forgot that should probably use 'IS NOT DISTINCT' and not
'=' for QUALITY.
Note that this checks only checks if there has ever been anything
identical in ROUTE_HISTORY, it doesn't restrain itself to the latest
entry only (i.e. the query I've written is slightly similar to the
second query I wrote, it has to be extended if you want something
slightly similar to the first query). Also, my third query doesn't
consider duplicate rows (since the fields seem to be the primary key
in ROUTE, it isn't theoretically possible in your example).
Set
2017-09-28 9:52 GMT+02:00 Svein Erling Tysvær <[email protected]>:
OK, that's completely different, but I've done something similar
once before. It is not possible to directly compare sets for
equality, but it is possible to check that set A doesn't contain
anything that isn't in set B and that set B doesn't contain
anything that isn't in set A. I assume NODE is the one common
denominator:
with tmp (NODE) as
(select distinct NODE from routes),
new_routes(NODE) as
(select t.NODE
from tmp t
where not exists(select *
from routes r
left join route_history rh on r.NODE = rh.NODE
and r.ROUTE_INDEX =
rh.ROUTE_INDEX
and r.LINK_NODE =
rh.LINK_NODE
and r.QUALITY =
rh.QUALITY
where t.NODE = r.NODE
and rh.NODE is null)
and not exists(select *
from route_history rh
left join routes r on r.NODE = rh.NODE
and r.ROUTE_INDEX =
rh.ROUTE_INDEX
and r.LINK_NODE =
rh.LINK_NODE
and r.QUALITY =
rh.QUALITY
where t.NODE = rh.NODE
and r.NODE is null))
select r.NODE, r.ROUTE_INDEX, r.LINK_NODE, r.QUALITY
from new_routes nr
join routes r on nr.NODE = r.NODE
Does this get you the result you want?
Set
2017-09-28 9:19 GMT+02:00 'Daniel Miller' [email protected]
[firebird-support] <[email protected]
<mailto:[email protected]>>:
Thanks - but that still doesn't work. I get a huge return set
with a ton of duplicates. I think something's being lost in
translation. A simple match of one row isn't sufficient - I need
to match the "set".
With the following six records:
5557 1 1116 0
5557 2 1115 0
5557 3 5104 2
5557 4 5073 2
5557 5 5222 2
5557 6 5110 2
That is the list of routes a given node has at a specific point in
time. So it's perfectly possible at another time to have fewer,
or more, or different routes. And I need to consider ALL the
routes in play for a given timestamp to be a single "set". So if
today I have the above list of routes, and yesterday route index 4
was looking at a different node - that would constitute a
different and unique set even though the other 5 records match.
5557 1 1116 0
5557 2 1115 0
5557 3 5104 2
5557 4 9999 2
5557 5 5222 2
5557 6 5110 2
So it's quite possible I will have lots of duplicates for an
individual row in ROUTES when compared in ROUTE_HISTORY - what I'm
trying to filter is the pattern of rows. Only if ALL the rows of
a current entries in ROUTES exist in ROUTE_HISTORY should it be
considered a duplicate condition.
--
Daniel
On 9/27/2017 11:53:48 PM, "Svein Erling Tysvær [email protected]
[firebird-support]" <[email protected]
<mailto:[email protected]>> wrote:
Sure it is possible to write such a query:
select r.NODE, r.ROUTE_INDEX, r.LINK_NODE, r.QUALITY
from routes r
left join route_history rh1 on r.node = rh1.node
and r.route_index = rh1.route_index
and r.link_node = rh1.link_node
and r.quality is not distinct from
rh1.quality
left join route_history rh2 on rh1.node = rh2.node
and rh1.route_index = rh2.route_index
and rh1.route_updated <
rh2.route_updated
where rh1.node is null
and rh2.node is null
I use two LEFT JOINS because I assume you want things to be
returned if things are changed back, e.g. if you have
5557 1 1116 0
5557 1 1116 0
then you only want one row in route_history, but if you have
5557 1 1116 0
5557 1 1116 1
5557 1 1116 0
then you don't want the last record to be considered a duplicate,
but know when it changed back. If ROUTE_INDEX is just a running
number, you consider
5557 1 1116 0
5557 2 1116 0
to be duplicates, and there are no gaps between the ROUTE_INDEX
for each NODE, then the query can be simplified:
select r.NODE, r.ROUTE_INDEX, r.LINK_NODE, r.QUALITY
from routes r
left join route_history rh1 on r.node = rh1.node
and r.route_index+1 = rh1.route_index
and r.link_node = rh1.link_node
and r.quality is not distinct from
rh1.quality
where rh1.node is null
Note that I use IS NOT DISTINCT so that NULLs are considered
equal. If you prefer, the left joins can be replaced by (nested)
NOT EXISTS.
HTH,
Set
2017-09-28 7:22 GMT+02:00 'Daniel Miller' [email protected]
[firebird-support] <[email protected]
<mailto:[email protected]>>:
On 9/27/2017 9:20:54 PM, "liviuslivius
[email protected] [firebird-support]"
<[email protected]
<mailto:[email protected]>> wrote:
group by + having
is your friend.
E.g.
Select field1, field2, count(*)
From tablex
Group by field1, field2
Having count(*)>1
But in your scenario i do not know if this is the solution.
Question is if record was changed and in next update "restored"
and once again changed is this duplicate or no?
If yes above query is ok if no then procedure is your friend.
And for the future change your audit trigger and check before
if there was any change
Thank you, but I this isn't quite what I need. I need to match
against all the records of a set (up to 8 rows per set) - not
just individual rows.
--
Daniel
