Re: SQL Puzzle
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)
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