Re: SQL Puzzle

Поиск
Список
Период
Сортировка
От Jeff Post
Тема Re: SQL Puzzle
Дата
Msg-id NGBBJHPOILLEGEKFHBEIEEMECPAA.postjeff@uwm.edu
обсуждение исходный текст
Ответ на SQL Puzzle  ("Ben-Nes Michael" <miki@canaan.co.il>)
Список pgsql-general
Fill an array with "Absent".
Do a select that will pull the correct entries.
Do a loop and overwrite the appropriate array elements with "Worked" based
on the select results.
Then display the array.

Hopefully this helps,
  Jeff Post

-----Original Message-----
From: pgsql-general-owner@postgresql.org
[mailto:pgsql-general-owner@postgresql.org]On Behalf Of Ben-Nes Michael
Sent: Sunday, June 30, 2002 7:33 AM
To: pgsql-general@postgresql.org
Subject: [GENERAL] SQL Puzzle


Hi All

I have a table where all worker logon and out every day:

table columns: record_id, worker, log_time, status, description

If a worker was absent there will be no record for the current day.

Now, I want to query this table to get a list of all days in month and in an
aditional field i would like to put worked/abscent ( depends if he was or
was not )

Is it possible in SQL ?

I thought about fuction with 'for' from 1 to 30 then subselect to check if
worker worked on a current day but its a complicated solution.

Does any body have idea for easier method ?

I can do it easily in php but i want to do as maximum as I can on the db
itself.





---------------------------(end of broadcast)---------------------------
TIP 2: you can get off all lists at once with the unregister command
    (send "unregister YourEmailAddressHere" to majordomo@postgresql.org)






В списке pgsql-general по дате отправления:

Предыдущее
От: "William N. Zanatta"
Дата:
Сообщение: Re: known bugs in sequences ?
Следующее
От: Gunther Schadow
Дата:
Сообщение: Re: recursing down a tree