Delete scheduled jobs from previous days

Post Author: supermeerkat
CA Forum: Administration
I look after a Business Objects server that runs about 400 reports per day.
Over the weekend the CMC stopped, nothing was reported in the logs, no attemps at restarting occured. But this isn't the problem.
When I restarted the CMC it started up with no problems, but in the
Instance Manager, it has started running the scheduled reports from the
day it stopped (Sunday the 4th May). It looks like it'll work through
all of the scheduled reports from Sunday, then Monday, before it gets
to today (Tuesday).
Is there any way I stop it from running the scheduled reports from
Sunday and Monday, and just have it run Tuesdays reports? Any help
would be gratefully recieved.

Nope. You can't get that, if house keeping jobs are scheduled.
Regards,
Subhash

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Hi,
1.
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<<removed_by_moderator>>
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Edited by: Vijay Babu Dudla on Jan 2, 2009 4:18 AM

hi,
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Get records from previous day to today

hi
i have 1 main table, table A, which has all the data .
now i have table b which has previous day data and table c which has current data. i need to compare table a(each column)
to each column in table b and c
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table B and table C doesnt have any date which say which date it got changed

>> I have 1 main table, table A, which has all the data .<<
What is a “main table”? I never used or heard that term. I am old, so I remember “main” or “master” tape files, however.
No table should have all the data in a schema that has more than one relationship in it. Where is the DDL?
>> now I have table b which has previous day data and table c which has current data <<
NO! This design error is called “Attribute splitting” and it usually comes from programmers like you who are using SQL to write 1950's magnetic tape files. Would you have a “Male_Personnel” and a “Female_Personnel” table or a “Personnel” table? Those
two absurd tables were split on sex_code, just like you want to split this unknown data on a “<something>_date” attribute.
>> I need to compare table a(each column) to each column in table b and c and need to find out rows that has been changed from yesterday to today and send that data. <<
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CREATE TABLE Foobar_History
(foo_id CHAR(9) NOT NULL,
start_date DATE NOT NULL,
end_date DATE, --null means current
CHECK (start_date <= end_date),
foo_status INTEGER NOT NULL,
PRIMARY KEY (foo_id, start_date));
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SELECT *
FROM Foobar
WHERE @in_cal_date
BETWEEN start_date
AND COALESCE (end_date, CURRENT_TIMESTAMP);
There are more tricks in the DDL to prevent gaps, etc
CREATE TABLE Events
(event_id CHAR(10) NOT NULL,
previous_event_end_date DATE NOT NULL
CONSTRAINT Chained_Dates
REFERENCES Events (event_end_date),
event_start_date DATE NOT NULL,
event_end_date DATE UNIQUE, -- null means event in progress
PRIMARY KEY (event_id, event_start_date),
CONSTRAINT Event_Order_Valid
CHECK (event_start_date <= event_end_date),
CONSTRAINT Chained_Dates
CHECK (DATEADD(DAY, 1, previous_event_end_date) = event_start_date)
-- CHECK (previous_event_end_date + INTERVAL '01' DAYS) = event_start_date)
-- disable the Chained_Dates constraint
ALTER TABLE Events NOCHECK CONSTRAINT Chained_Dates;
GO
-- insert a starter row
INSERT INTO Events(event_id, previous_event_end_date, event_start_date, event_end_date)
VALUES ('Foo Fest', '2010-01-01', '2010-01-02', '2010-01-05');
GO
-- enable the constraint in the table
ALTER TABLE Events CHECK CONSTRAINT Chained_Dates;
GO
-- this works
INSERT INTO Events(event_id, previous_event_end_date, event_start_date, event_end_date)
VALUES ('Glob Week', '2010-01-05', '2010-01-06', '2010-01-10');
-- this fails
INSERT INTO Events(event_id, previous_event_end_date, event_start_date, event_end_date)
VALUES ('Snoob', '2010-01-09', '2010-01-11', '2010-01-15');
--CELKO-- Books in Celko Series for Morgan-Kaufmann Publishing: Analytics and OLAP in SQL / Data and Databases: Concepts in Practice Data / Measurements and Standards in SQL SQL for Smarties / SQL Programming Style / SQL Puzzles and Answers / Thinking
in Sets / Trees and Hierarchies in SQL

Delete scheduled jobs from previous days

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