Try:
select users."User-Name", ss."Date", ss."Time" from users left outer
join ( select distinct on ("User-Name") "User-Name", "Date", "Time" order by
1, 2 DESC, 3 DESC) as sson ( users."User-Name" = ss."User-Name")
order by 1;
It help if yo uhave an index on logs( "User-Name", "Date" desc, "Time"
desc).
JL
PS Can't you convert "Date" || "Time" to timestamp?
"Sill-II, Stephen" wrote:
>
> I'm trying to come up with an efficient way to do the following task, but I
> haven't found the correct join syntax to do it.
>
> I have table "users" for keeping a list of users I need to check logins for.
> It contains the following fields:
>
> id,Full-Name,User-Name
>
> I have table called "logs" that contains the actual radius log data. The
> three fields I am concerned with are:
>
> User-Name,Date,Time
>
> I have gotten thus far almost what I want with the following query.
>
> SELECT users."User-Name",max(logs."Date") as login_date,max(logs."Time") as
> login_time FROM logs where logs."User-Name"=users."User-Name" GROUP BY
> users."User-Name";
>
> This gives me the User-Name, date, and time of people WHO HAVE LOGGED IN.
> I'm looking to have a query that returns all of the users in the first
> table, including those who have not logged in. I have an external perl
> script that generates a nice html report for the manager. I have a script
> that parses the raw .csv files, but I'm trying to move it entirely to
> postgres, including if possible stored-procedures in plperl. I already have
> a perl script that imports the raw log files into the logs table.
>
> Am I on the right track?
>
> Thanks,
>
> Stephen Sill II
>
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