Re: Finding duplicates only.

In response to Greenhorn :
> Hi,
>
> Can someone please help me with this duplicate query.
>
> I'm trying to:
>
> 1. Return duplicates only. (without including the first valid record), and

I will try to help you. Assuming this table:

test=*# select * from greenhorn order by id;
id | inspection_time
----+----------------------------
1 | 2010-03-11 07:14:14.290259
1 | 2010-03-11 07:14:14.290259
2 | 2010-03-11 07:14:14.290259
3 | 2010-03-11 07:15:14.290259
4 | 2010-03-11 07:16:14.290259
5 | 2010-03-11 07:24:14.290259
6 | 2010-03-11 07:34:14.290259
(7 rows)

The record with id=1 is twice.

> 2. Return as duplicate if the difference between a.inspection_time
> and b.inspection time is under 5 minutes.

Assuming you have a 8.4-version:

with the table above, and time-difference < 2 minutes, rows 2, 3 and 4:

test=*# select * from (
select id,
inspection_time,
lag(inspection_time) over (order by inspection_time RANGE UNBOUNDED PRECEDING)
from greenhorn
group by 1,2
) foo
where inspection_time-lag < '2minutes'::interval;

id | inspection_time | lag
----+----------------------------+----------------------------
2 | 2010-03-11 07:14:14.290259 | 2010-03-11 07:14:14.290259
3 | 2010-03-11 07:15:14.290259 | 2010-03-11 07:14:14.290259
4 | 2010-03-11 07:16:14.290259 | 2010-03-11 07:15:14.290259
(3 rows)

>
> Here's the query string I'm using to retrieve the duplicates but it is
> returning every duplicate records.
>
> select a.rego,
> a.inspection_date,
> a.inspection_time,

Why do you have 2 fields, one for date and one for time? Use one timestamp-field instead.

Regards, hope that helps, Andreas
--
Andreas Kretschmer
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