first day of the year and Monday next year

Hello all;

Given a year, how to determine the first day of this year and the next two Mondays for this year...

This gives a shot:

SELECT TRUNC(TO_DATE('2010','YYYY'),'YEAR')                        AS first_day_of_year
     , NEXT_DAY(TRUNC(TO_DATE('2010','YYYY'),'YEAR')-1,'MONDAY')   AS first_monday_of_year
     , NEXT_DAY(TRUNC(TO_DATE('2010','YYYY'),'YEAR')-1,'MONDAY')+7 AS second_monday_of_year
FROM   dual
;

Tags: Database

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Find the first day of the third quarter of next year

Hello
I'm writing a sql to find "the first day of the third quarter of the year (from the current year) next»
I am able to get the first and the last day of the current quarter:
Select trunc (sysdate, 'Q'), trunc (add_months (sysdate, + 3), 'Q')-1 double
But, how to find the first day of the third quarter of the year next (from the current year) based on any date entry (i.e. sysdate)
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If my entry date = 07/04/2005 should be released 07/01/2015
If my entry date = 20/02/2013 should be released 07/01/2015
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Hello

The 1st day of the 3rd quarter of next year is:

ADD_MONTHS (TRUNC (SYSDATE, 'YEAR')

18 = 12 + 6

)

TRUNC (SYSDATE, 'YEAR') is the 1st day of the 1st quarter of this year

The first day of the new year is 12 months after that, and the first day of quarter 3 is another 6 months after that.

user634169 wrote:

... But, how to find the first day of the third quarter of the year next (from the current year) based on any date entry (i.e. sysdate)

for example if my entry date = 12/01/2001 should be released 07/01/2015

If my entry date = 07/04/2005 should be released 07/01/2015

If my entry date = 20/02/2013 should be released 07/01/2015

... etc.

Thanks in advance

You did some errors above?

If the effective date in 2001, why is the output in 2015, not 2002?

If the output is based only on the date of the day (currently in 2014), why even mention a start date?

first day of the month and time

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{code}
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{code}

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of the double

resp.

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First day with the mobile and shatter proof - broken

Hello

I bought Xperia double Z3, mobile is great and every thing,

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How to extract the first day of the year quarter
Hello

How can I write a query to extract the first day in a neighborhood of declaration

for example: -.

Q1 starts from April 1 to June 30, second quarter starts from July 1 to September 30, third quarter starts from 1 October to 31 December and the fourth quarter begins from January 1 to March 31.

How can I recover then the first day of these quarters, regardless of the year. The year can be any year.
Hello

don't know exactly what you want, but if you want the first day of a quarter for any given date and then
```
select trunc(sysdate,'Q') from dual;
```
Sysdate obviously could be a valid date.

André

1st day of the year then the Monday group

Hello expert;

I have the following tables below

create table table_one
(
 v_id varchar2(60),
 close_date date
);

insert into table_one
  (v_id, close_date)
values
  ('A', to_date('1/1/2010 4:47:09 PM', 'MM/DD/YYYY HH:MI:SS:AM'));   --- First day of the year

insert into table_one
  (v_id, close_date)
values
  ('B', to_date('1/4/2010 2:47:09 PM', 'MM/DD/YYYY HH:MI:SS:AM'));   -- monday
  
  insert into table_one
  (v_id, close_date)
values
  ('C', to_date('1/11/2010 1:47:19 PM', 'MM/DD/YYYY HH:MI:SS:AM'));  -- monday
  
 insert into table_one
  (v_id, close_date)
values
  ('D', to_date('5/6/2010 5:47:09 PM', 'MM/DD/YYYY HH:MI:SS:AM'));  --- not a monday
  
 insert into table_one
  (v_id, close_date)
values
  ('E', to_date('5/3/2010 4:47:09 PM', 'MM/DD/YYYY HH:MI:SS:AM'));  ---monday
  
 insert into table_one
  (v_id, close_date)
values
  ('F', to_date('2/9/2010 5:17:09 PM', 'MM/DD/YYYY HH:MI:SS:AM'));  -- not monday

I need to modify my query below and I can't think of a good way to go about it

select t.v_id, to_char(t.close_date, 'YYYY-MM-DD') from table_one t
where t.close_date between to_date('2010/1/1', 'YYYY-MM-DD') and to_date('2010/12/31', 'YYYY-MM-DD')
group by t.v_id, to_char(t.close_date, 'YYYY-MM-DD');

what I'm trying to do is in fact I want to group by only Mondays

Edited by: user13328581 on Nov 17, 2010 8:04 AM

Edited by: user13328581 on Nov 17, 2010 8:04 AM

Edited by: user13328581 on Nov 17, 2010 8:29 AM

select t.v_id, to_char(t.close_date, 'YYYY-MM-DD')
  from table_one t
where t.close_date between to_date('2010/1/1', 'YYYY-MM-DD') and to_date('2010/12/31', 'YYYY-MM-DD')
and   to_char(trunc (t.close_date), 'dy' ) = 'mon'

V_ID     TO_CHAR(T.CLOSE_DATE,'YYYY-MM-DD')
B     2010-01-04
C     2010-01-11
E     2010-05-03

If you also want the first day of the year that you mentioned.

select t.v_id, to_char(t.close_date, 'YYYY-MM-DD')
  from table_one t
where t.close_date between to_date('2010/1/1', 'YYYY-MM-DD') and to_date('2010/12/31', 'YYYY-MM-DD')
and   to_char(trunc (t.close_date), 'dy' ) = 'mon'
or  extract( month from t.close_date) + extract( day from t.close_date) = 2

V_ID     TO_CHAR(T.CLOSE_DATE,'YYYY-MM-DD')
A     2010-01-01
B     2010-01-04
C     2010-01-11
E     2010-05-03

Published by: pollywog on November 17, 2010 11:38

First day of the month
```
How to find the first day of the month if the month format is like TO_CHAR(SYSDATE,'YYYYMM')

Sanjay
```
Published by: user12957777 on May 23, 2012 21:12
There is no 'day' in this date format.

What you trying to do? What do you expect the result to look like?

Are you wanting to set a date for the first day of the month and year of your example?
```
SELECT TO_DATE(TO_CHAR(SYSDATE,'YYYYMM') || '01', 'YYYYMMDD') FROM DUAL;
```

How to get the last day of the year
Hi all

Thanks in advance...

How do I get the last day of the year that I'm passing date at run time.

I can manage to get the first day of the year by
SELECT TRUNC(SYSDATE,'YEAR') AS FDAY_YEAR
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Concerning
Sachin
```
  1*  select ADD_MONTHS(trunc(sysdate,'yyyy'),12)-1 dd from dual
SQL> /

DD
-----------
31-DEC-2010
```

How to find the first day of the month following two years one year from now?

Hello
I need to find the first day of the month following two years ago, based on the current date.
For example: If today is the 31/08/2015 so I need to find the date range between 31/08/2015 and on 09/01/2013. How can I write an SQL which allows to automate this calculation instead of hard-coding the values. Please do help me out with this so that I can build the SQL that can produce the desired result.
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find the start of the month - 2 years in advance, early - 2 years + 1 months in advance and go back 1

Select trunc (add_months(sysdate,24), 'MM'), trunc (add_months(sysdate,25), 'MM') - 1 double;

First day of the month between 08:00 and 18:30

DB version: 10.2.0.4

Due to the nature of our the first working day of each month (like today 1 MS), the DB will be very busy.
We execute a proc purge planned Windows Scheduler every 5 minutes every day.

But the first day of every month (like today 1 MS), if it's a day of the week (Monday to Friday) we do not want this proc running between 08:00 and for dinner from 18:30.

So, I thought including the logic in the procedure itself.

How can I implement this?

Edited by: user872043, Sep 1 2010 08:48

Here is an example of using DBMS_SCHEDULER which should meet your needs:

Create schedules

BEGIN
        DBMS_SCHEDULER.CREATE_SCHEDULE
        ( schedule_name   => 'every_5_schedule'
        , repeat_interval => 'FREQ=MINUTELY; BYMONTHDAY=2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31; INTERVAL=5; BYSECOND=0'
        , comments        => 'Schedule to run every 5 minutes'
        );

        DBMS_SCHEDULER.CREATE_SCHEDULE
        ( schedule_name   => 'busy_hourly_schedule'
        , repeat_interval => 'FREQ=MINUTELY; BYMONTHDAY=1; BYHOUR=0,1,2,3,4,5,6,7,19,20,21,22,23; INTERVAL=5;'
        , comments        => 'Schedule to run every 5 minutes except certain hours on the first of the month'
        );

        DBMS_SCHEDULER.CREATE_SCHEDULE
        ( schedule_name   => 'busy_6pm_schedule'
        , repeat_interval => 'FREQ=MINUTELY; BYMONTHDAY=1; BYHOUR=18; BYMINUTE=30,35,40,45,50,55'
        , comments        => 'Schedule to run every 5 minutes from 6:30 - 7'
        );
END;
/

There are three lists.

every_5_schedule - runs every 5 minutes for all the days that are not on the first day of the month
busy_hourly_schedule - runs every 5 minutes for the acceptable hours on the first day of the month
busy_6pm_schedule - retrieves the half hour from 6:30 - 7 on the first day of the month.

Then, create your business:

Job creation

BEGIN
        DBMS_SCHEDULER.CREATE_JOB
        ( job_name        => 'YOUR JOB'
        , job_type        => 'PLSQL_BLOCK'
        , job_action      =>'BEGIN YOURPROC; END;'
        , repeat_interval => 'EVERY_5_SCHEDULE, BUSY_HOURLY_SCHEDULE, BUSY_6PM_SCHEDULE'
        , enabled         => TRUE
        , comments        => 'Some comments'
        );
END;
/

HTH!

Deriving from the first day of the week ww in the year yyyy

I need to be able to display the first day of a particular week in addition to the "week of the year" and the year. for example:

select to_char("ServiceDate",'yyyy-ww') as "Year/Week", sum("Points") as "Points"
from my_table
group by to_char("ServiceDate",'yyyy-ww')
order by to_char("ServiceDate",'yyyy-ww')

give me the following:

Year/Week     Points
2008-33            1                      
2008-35            1                      
2008-42            4                      
2008-43            2                      
2009-01            1                      
...

and so on. What I would like to add to the dataset is the same day, the first day of the week for this week/year. For example, the first line would look like this:

Year/Week     1st day of week     Points
2008-33          08-AUG-2008*         1

* Or whatever the first day of the week 33 of 2008 is (assuming that "first day of the week" is a Sunday, in my case).

My brain is on "dim" right now; any help is greatly appreciated.
Thanks in advance,
Carl

select  to_char(trunc("ServiceDate",'ww'),'yyyy-ww') as "Year/Week",
        to_char(trunc("ServiceDate",'ww'),'DD-MON-YYYY') "Week Start Date",
        sum("Points") as "Points"
from my_table
group by trunc("ServiceDate",'ww')
order by trunc("ServiceDate",'ww')
/

SY.

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The only solution I can imagine at this stage is to calculate the month and the day of the day of the year, which would imply a choice of the month table and a check of the leap year.

Smart solutions, I'm missing?

DaveT

Dave,

I found a Julian Georgian so far.

Why my laptop wireless computer connect to the internet first thing in the morning and late at night, but not during the day.

I wonder if you can help me please?

I have problems with my satellite internet access on my laptop wireless. He beautifully connects to the router and the signal is excellent. However, I'm unable to connect to the internet. I can use the laptop first thing in the morning because it connects right away, but I would say that 95% of the day I can't use it at all. Sometimes it will work later in the evening. I learned this could be due to the intensity of the signal and that some suppliers to fall during the day. My desk top connects perfectly (if the weather permits it) but is not on the wireless network. I had no problem with my laptop when I used my previous service provider. It becomes more frustrating that the inability to use the internet visited my lap top unnecessary. I can't use it on my neighbors wireless network no problem at all but they we internet different provider. Can you please help because I don't know why it works first thing in the morning and sometimes late at night, but not during the day? No matter where the lap top is the first thing in the morning, it will work even if it is forty feet away. I can sit during the day with my phone right by my router and it still not connect to internet even though the router signal strength is excellent and desktop is on the internet. Any help would be great, because it leads me into the wall! I am using WIndows 7 on my lap top from acer.

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Why Windows 8 'Calendar' app has no option to change the first day of the week?

Why can't I change the first day of the week in 'Calendar' app?

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Windows calendar - like some other ' standard' Windows 8 applications - is pretty basic and was obviously designed to speed up tasks 'on the fly'. As soon as it comes to advanced settings for these applications, you will need to upgrade to the "long" version

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Hope that answers & help

See you soon

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each first day of the month
Hi friends here is the querry I wrote to display every first day of the month, but there is something wrong in it... Help me...
create or replace procedure minmax is
cursor minmax_cur is
Select min (hiredate), max in emp (hiredate);
number of v_minmonths;
number of v_maxmonths;
number of v_minyears;
number of v_maxyears;
x date of publication;
date of y;

Start
Open minmax_cur;
extraction minmax_cur
x, y;
v_minmonths: = extraction (x months);
v_minyears: = extraction (year x);
While (x, y) loop
because me in 1... v_minmonths loop
If (v_minmonths < 12) then
v_minmonths: = v_minmonths + 1;
x: = x + 28.
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v_minyears);
on the other
v_minmonths: = 0;
v_minyears: = v_minyears + 1;
x: = x + 28.

end if;
end loop;
end loop;
minMax end;
and the results are

01/01/1981
01/02/1981
01/03/1981
..................
.........---
01/04/1987
01/05/1987
01/06/1987
01/07/1987
01/08/1987
01/09/1987
01/10/1987
01/11/1987 the problem is max (hiredate) is 5/23/1987...it must show up to 01/05/1987...
I get a few more months...

first day of the year and Monday next year

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