Number of days in a month

I have created a form with 2 date objects:

"Date1" is a date object in which the user will enter the first day of a month gave "MM/DD/YYYY";

I need "Date2" to calculate the last day of the month of 'Date1' ' MM/DD/YYYY ".

Obviously, the date displayed in Date2 will vary, because months contain a different number of days, so I can't just say '+ 30 '. Can someone help me with a JavaScript for the FomCalc code to achieve this?

Thank you!!!

Gene-O

Maybe you're view model is different from the trend in the script.

Here is an example of work form:

https://Acrobat.com/#d=rqvjWvn5WIBrWD * GLDRgfA

Tags: Adobe LiveCycle

Similar Questions

How to get the maximum number of days in a month?

How to get the maximum number of days in a month?

I use a dateField. where I can get the month. Now, I want to get the maximum number of days in a month.

How can I get it?

Please help me. and thanks in advance... []

NET. RIM. Device.API.util
Class DateTimeUtilities

getNumberOfDaysInMonth (int month, int year)
Returns the number of days in the specified month

It took me like 10 seconds to find it in the API, I guess you spent more time writing the post...

Get the number of days in a month based on the month and year of fields

I have a column in my form which lists the days in a month. I want to configure a hidden field that calculates the total number of days in a month, based on the month and year of the field inputs. The number of days will determine what appears on the column. For example, if I put 4 months, and 2016 in the field of the year, I get 30 in the hidden field. Thus, on the column 'Day', I'll have numbers 1-30. Or if I put 2 months and 2016 in the field of the year, I get the 29 in the hidden field. If the numbers 1-29 appears in the column 'day '.
Found this on some forum javascript code:
//Month is 1 based function daysInMonth(month,year) { return new Date(year, month, 0).getDate(); } //July daysInMonth(7,2009); //31 //February daysInMonth(2,2009); //28 daysInMonth(2,2008); //29
I do not know how to convert this code in JavaScript to adobe and don't really know how to use it. All I know how to do is to configure the field values for the field month and year as variables. I am a novice programmer and would appreciate it really all the help I can get. Thank you in advance!

The code seems to be JavaScript and runs as needed by using the JavaScript console.

I would like to consider making more general code, so if you have a date string that includes at least the month and year we could just call the function and get the number of days for that month.

The following script will calculate the number of days in a month, by using at least the month and year values can display the result on the JavaScript console and all of the value field for the field that has this code as the custom calculation Script.

function daysInMonth (oDate) {}
return new Date (oDate.getFullYear (), oDate.getMonth () + 1, 0) .getDate ();
}

nMonth var = this.getField("Month").valueAsString; get the value of month;
nYear var = this.getField("Year").valueAsString; get the value of the year;

Event.Value = "";

If (nMonth! = "" & nYear!) = "") {}
var MyDate = util.scand ("' / mm/yyyy ', nMonth +" / "+ nYear); convert to date object;
var nDaysInMonth = daysInMonth (MyDate); get the number of days;

Console.Open (); Open the JavaScript console;

Console.clear(); clear the console;

Console.println ("Days in" + nMonth + ":" + nDaysInMonth); show days in month;

Event.Value = nDaysInMonth; Set the value of the field;

}

Need to obtain equal records the number of days of the month.

Dear all
My table and its data is given:
Table name: Transaction_Master
Trn_Num Trn_Dte
1 January 2, 2014
2 January 3, 2014
3 February 15, 2014
4 December 29, 2013
5 December 30, 2013
and so on
Now, I wish that my query must return equal records the number of days for each month of Trn_Dte. Be careful, I don't have Trn_Num column in my result. Its cities here only for explanation.
It should, for example, records 31 to January 31 to December and 28 February.
I tried following query but it passes loop and my db session gets hang.
With T as
(
Select Distinct (Trn_Dte, 'Month') To_Char Trn_Mon, To_Number (To_Char (Last_Day (Trn_Dte), 'DD')) Tot_Day
Of Transaction_Master
Where Trn_Dte between parameter_1 and PARAMETRE_2
)
Select Distinct Trn_Mon, level
T
Connect by level < = Tot_Day
Order By Trn_Mon, level
Could you give me a solution for this case?

Another solution for your exact needs. Try this and let me know in case of any problems. Because I changed the connection by the clause double itself.

SELECT DISTINCT TRUNC (Trn_Dte, 'MY') + (lvl - 1)

OF Transaction_Master aa (SELECT LEVEL lvl

OF the double

CONNECT BY LEVEL<= 31)="">

WHERE Trn_Dte<= to_date('15-feb-2014','dd-mon-yyyy')="" ---="" add="">

AND TO_CHAR(Trn_Dte,'MON') = TO_CHAR (TRUNC(Trn_Dte,'MON') + (lvl-1), 'MY');

Calculate fees that pay based on the number of days in a month

Hello
I'm writing the BR to fresh wages, based on the number of days in the month.
Fix(Jan:DEC)
"Salary" = ("Annual salary"-> "BegBalance" / 12) * "workforce";
The existing calculation was ' wages divided by months no .of (12) ", now instead of number months wages that need to be calculated ("BegBalance"/ No..) OF DAYS IN THE MONTH) * 12.
Can someone pls guide me, how to begin writing the BR for 'No. DAYS OF THE MONTH.

You can do an admin enter number of days in the month. I remember CL reminding me the rhyme thirty days has September - Wikipedia, the free encyclopedia

Jokes aside, you can use SQL to get the number of days. You will get different options for SQL and Oracle.

If you want to continue to do this in BR then

Fix(Jan:DEC)

IF (@ISMBR (SEP) OR @ISMBR (Apr) OR @ISMBR (Jun) OR @ISMBR (Nov))

("Salary" = ("annual salary"-> "BegBalance" / 30) * 12) * "workforce";

ELSEIF (@ISMBR (Jan) OR @ISMBR (Mar) OR @ISMBR (May) OR @ISMBR (July) OR @ISMBR (Aug))

("Salary" = ("annual salary"-> "BegBalance" / 31) * 12) * "workforce";

ELSE (@ISMBR (Feb))

IF (@Remainder($CurrYr/4) == 0)

("Salary" = ("annual salary"-> "BegBalance" / 29) * 12) * "workforce";

ON THE OTHER

("Salary" = ("annual salary"-> "BegBalance" / 28) * 12) * "workforce";

ENDIF

ENDIF

ENDFIX

Concerning

Celvin

http://www.orahyplabs.com

Post edited by: CelvinKattookaran

Function to return the number of days in a month?
In one of my rules, I need to allocate an amount of benefits based on the number of days remaining in the calendar month.

OPM does have a function that can provide me with the number of days in a calendar month?

Thank you very much
Isamu
It is not a direct function to do this, however, it can be done with a bit of messing around. The following rules you will get the number of days of the month of the date of the date

Basically, you get the first day of the month, add a month to the first day of the following month and then get the number of days between them, which will give you the number of days of the month of the date
```
the month = ExtractMonth(the date)
the year = ExtractYear(the date)
the start of the month =MakeDate(the year, the month, 1)
the start of the next month = AddMonths(the start of the month, 1)
The number of days in the month = DayDifference(the start of the month, the start of the next month)
```
These rules can be a little more compact, but it is basically how to do it.

Trn_Num	Trn_Dte
1	January 2, 2014
2	January 3, 2014
3	February 15, 2014
4	December 29, 2013
5	December 30, 2013
and so on

How to get the number of days in a month of belonging to a range of dates

Hi, I'm going crazy around a problem, I have 2 dates and one month I wanto to retrieve the number of days belonging to the months that fall within the range.

for example:

month of January 2011. begin_date = January 11, 2011, May 30, 2011 end_date result is 21
month of January 2011. begin_date = December 11, 2010, result of end_date, January 10, 2011 10
month January 2011 .begin_date = 2 February 2011, end_date may 25, 2011 result 0
month of January 2011. begin_date = January 3, 2011, January 5, 2011 result DATE END 3

and so on...

I appreciate any suggestion
Thank you

Andrea

Something like that... ?

SQL> with t as
  2  (select  to_date('11/01/11','dd/mm/yy') from_dt,
  3           to_date('30/05/11','dd/mm/yy') to_dt,
  4           'Jan-11' mnth from dual)
  5  select least(last_day(to_date(mnth,'Mon-yy')),to_dt) -
  6         greatest(to_date(mnth,'Mon-yy'),from_dt) cnt
  7  from t  ;

       CNT
----------
        20

If 0 is also expected...

SQL> with t as
  2  (select  to_date('11/01/11','dd/mm/yy') from_dt,
  3           to_date('30/05/11','dd/mm/yy') to_dt,
  4           'Jan-11' mnth from dual union all
  5           select  to_date('11/01/11','dd/mm/yy') from_dt,
  6           to_date('30/05/11','dd/mm/yy') to_dt,
  7           'Jan-12' mnth from dual
  8           )
  9  select from_dt,to_dt,mnth,
 10         greatest(
 11              least(last_day(to_date(mnth,'Mon-yy')),to_dt)
 12              -
 13              greatest(to_date(mnth,'Mon-yy'),from_dt)
 14                 ,0) cnt
 15  from t;

FROM_DT   TO_DT     MNTH          CNT
--------- --------- ------ ----------
11-JAN-11 30-MAY-11 Jan-11         20
11-JAN-11 30-MAY-11 Jan-12          0

Published by: JAC on February 9, 2012 19:22

number of days in the month

Dear members,

I have available dates,
Let's say date1 = January 1, 2011"
and date2 = April 15, 2011"

My goal is to calculate the sum of the no. of days between this range.

i.e.

month = 01 days = 22
month = 02 days-28
month = 03 days = 31
month = 04 days = 15

so total days = 22 + 28 + 31 + 15 = 96 between individuals because of the date range.
My goal is to have the figure 96 days that is the sum of the days of the month.

Kind regards

Teefu
Developer

Published by: teefu on July 21, 2011 10:09

Hello

If all you want is the total number of days, then you can simply subtract the earlier DATE of the last one, like this:

SELECT     TO_DATE ( '15-04-2011'
          , 'DD-MM=YYYY'
          ) + 1 - TO_DATE ( '10-01-2011'     -- Jan. 10; '01-10-' would be Oct. 1
                    , 'DD-MM-YYYY'
                    )     AS days_diff
FROM     dual;

If you want to see a breakdown by month, here's one way:

WITH     parameters     AS
(
     SELECT     TO_DATE ( '10-01-2011'
               , 'DD-MM-YYYY'
               )     AS start_dt
     ,     TO_DATE ( '15-04-2011'
               , 'DD-MM-YYYY'
               )     AS end_dt
     FROM     dual
)
SELECT       TO_CHAR ( start_dt + LEVEL - 1
            , 'FMMonth YYYY'
            )     AS month
,       COUNT (*)     AS days
FROM       parameters
CONNECT BY     LEVEL <= 1 + end_dt - start_dt
GROUP BY  TO_CHAR ( start_dt + LEVEL - 1
            , 'FMMonth YYYY'
            )
ORDER BY  MIN (LEVEL)
;

Output:

MONTH                      DAYS
-------------------- ----------
January 2011                 22
February 2011                28
March 2011                   31
April 2011                   15

Published by: Frank Kulash, July 21, 2011 01:10

N ° of days in a month in Hyperion Essbase/Planning-For EN reports

Hi all
I know with your helpHow to reach no. Hyperion/Essbase days planning service hierarchy to create EN reports based on the number of days in a month

Example: I need to create a new relationship with 3 columns and 1 row EN

POV - year of the period, the scenario.
Lines - Dimension entity

Column1 - PTD Avg (account, Version, measure, Products_Type, products)
Column2 - amount (counts, Version, measure, Products_Type, products)

Column3 (formula column) - Eval ([Column2] /number of days per month* 36500 / [Column1] )

Thanks in advance for the help!

You can create a member of formula to account for the number of days in the month. It should look something like below. You must also include check year no.. of days in February (to assign 29 for leap years).

IF (@ISMBR ("JAN") OR)

...

... etc.

)

'No days' = 31;

ElseIf (@ISMBR ("FEB"))

'No days' = 28;

on the other

'No days' = 30;

endif;

Kind regards

Sunil

Query or logic to find several days through each month.
Dear friends,

I need your help to solve this problem.

We have a now where we have to pay compensation to the employees based on their number of working days.

Say, for example, if an employee has worked since March 3, 2012-April 5, 2012.

We have a fixed value for a month 300 dirhams. But the number of days on s March 31 and number of days in the month of April are 30.

So by compensation daily for March day would be 300/31 and April would be 300/30.

We are looking for a logical opr query that calcluates first right number of days in each month (per month) and then calculate as below

Number of working days in the month of March is 31-3 + 1 = 29

Compensation of A1 = (300 * 29) / 31

Number of working days in the month of April is 5 (must also find logical I guess)
Allowance for A2 = (300 * 5) / 30

Then A1 + A2.

(E) would be the total quota when the number of months in all.

Please guys any help would be greatly appricated.

R
Just a follow-up ;-)

Note that you can almost get what you want with standard MONTHS_BETWEEN function [url http://docs.oracle.com/cd/E11882_01/server.112/e26088/functions102.htm#i78039].
But notice this quote from the documentation:

>
If date1 and date2 are the same days of the month or the last two days of the month, the result is always an integer. Otherwise the Oracle database calculates the fractional part of the result based on a month of 31 days and consider the time difference components date1 and date2.
>

It will always use one even if 31 days a month has fewer days.

Then you could also implement the logic of my previous answer as your own function months_between implementation:
```
create or replace function my_months_between (
   p_todate    date
 , p_fromdate  date
)
   return number
is
begin
   return ((extract(day from last_day(p_fromdate))-extract(day from p_fromdate)+1) / extract(day from last_day(p_fromdate)))
        + (months_between(trunc(p_todate-1,'MM'),trunc(p_fromdate,'MM'))-1)
        + (extract(day from p_todate-1) / extract(day from last_day(p_todate-1)));
end my_months_between;
/
```
I changed it to make it "compatible" with the standard function that it calculates up to but not including the parameter to date (which is correct when you are considering dates with time included, but I guess in your case, you have just date values.)

Then, you can compare the results of the standard months_between with the new my_months_between:
(Note that I have call the functions with the d2 + 1 to give the desired result of inclusion and to this day in the calculation).
```
SQL> with testdata as (
  2     select date '2012-03-03' as d1, date '2012-04-05' as d2 from dual union all
  3     select date '2012-02-03' as d1, date '2012-05-02' as d2 from dual union all
  4     select date '2012-03-03' as d1, date '2012-03-10' as d2 from dual union all
  5     select date '2012-03-01' as d1, date '2012-07-31' as d2 from dual union all
  6     select date '2012-02-01' as d1, date '2012-03-01' as d2 from dual union all
  7     select date '2012-01-01' as d1, date '2012-03-25' as d2 from dual
  8  )
  9  --
 10  -- end of test data --
 11  --
 12  select months_between(d2+1, d1) std_months
 13       , 300 * months_between(d2+1, d1) std_allowance
 14       , my_months_between(d2+1, d1) my_months
 15       , 300 * my_months_between(d2+1, d1) my_allowance
 16    from testdata
 17  /

STD_MONTHS STD_ALLOWANCE  MY_MONTHS MY_ALLOWANCE
---------- ------------- ---------- ------------
1.09677419    329.032258 1.10215054   330.645161
         3           900 2.99555061   898.665184
.258064516    77.4193548 .258064516   77.4193548
         5          1500          5         1500
1.03225806    309.677419 1.03225806   309.677419
2.80645161    841.935484 2.80645161   841.935484

6 rows selected.
```
For a large part of the data, both give the same result, especially when the two dates are in month of 31 days.
To work from ' 2012-03-03' to '' 2012-04-05 standard service gives too little because it uses always 31 days, while my_months_between uses 30 days of April.

"But note working from '' 2012-02-03 to 2012 - 05 - 02. He jumped 2 days in February and has worked 2 days in May - the standard function concludes he works precisely 3 months.
But my_months_between computes fraction 27/29 months in February, then March and April like 2 months, and then the fraction 2/31 months in May - which gives a total of 2.99555061 instead of 3 months.

Just something to be aware of ;-)

Number of days exclude sat'day and Sunday
Hello

I am able to get the number of days in the month starting today, but not able to exclude weekends. Please notify.

My query

SELECT TRUNC (SYSDATE) - TRUNC (SYSDATE, 'MM') OF DOUBLE

I want to request something very simple as show below. Please, do not use login and keyword level because they are not supported by OBIEE.

SELECT TRUNC (SYSDATE) - TRUNC (SYSDATE, 'MM') OF DOUBLE
where wday not in ('Sam', 'Sunday');

Thank you
Jay
What is OBIEE?
You want to say it does not support "connection by ' and 'level '? These keywords are for the SQL engine and therefore has nothing to do with your frontend.

The correct solution would be something like:
```
SQL>   select count(*)
  2    from (select sysdate + l trans_date
  3          from (select level - 1 l from dual connect by level <= ((sysdate+7) - sysdate)))
  4          where to_char(trans_date, 'dy') not in ('sat', 'sun');

  COUNT(*)
----------
         5

SQL> 
```

Count of the number of days in a calendar month according to the year

Hi all

On my table, column 1 is the list of months over several years.

Column 2 is the activity number od each month.

Column 3 I want an average rate of activity per day during the month, which should be: (C2 / number of days according to C1) %

Problem: I can't find a fomula give me the number of days, particularly in February when the year is leap.

Any help is appreciated.

Thank you all,

Lopez

PS: by the way, with this new version of Apple Comunity I seem to have lost my previous questions about the numbers.

Hey Lopez,

EOMONTH(A1,0) (end of month) will GET help you.

Column B is Date & time format to display only the month and year (actually the 1st of the month, but the day is not displayed).

Formula in C2 (fill down)

= EOMONTH(B2,0)

The end of the month 0 months later.

Formula in D2 (fill down)

= DAY (C2)

February 2015 (non-leap years) shows 28

February 2016 (leap year) shows 29

Kind regards

Ian.

Number of days remaining in the month

Hi guys,.
I'm looking to create a function that returns the number of working days (Monday to Friday) between sysdate and the last day of the month.
If someone has done something similar.
I tried the following:
< code >
CURSOR days_c
IS
SELECT THE LEVEL
, to_char (LAST_DAY (ADD_MONTHS (SYSDATE-1)) + LEVEL,'Dy DD-Mon-YYYY "") AS DOW - day of the week
OF the double
WHERE ROWNUM < = EXTRACT (DAY FROM LAST_DAY (SYSDATE))
CONNECT BY LEVEL = ROWNUM;
v_cnt NUMBER: = 0;
BEGIN
FOR days_r IN days_c
LOOP
To_char (days_r.DOW) NOT AS "Sat %" If GOLD to_char (days_r.DOW) NOT LIKE '% of Sun '.
- AND to_date (days_r.dow) > sysdate
THEN
v_cnt: = v_cnt + 1;
END IF;
END LOOP;
RETURN v_cnt;
< code >
At the present time, IF logic around the days_r.DOW does not work as I am also bad to only return the greater these days but less dates at the end of the month.
Thanks in advance
Chris

Try this

SQL > with input_data
2 as
(3)
4. Select sysdate current_day
5, last_day (sysdate) last_day
6 double
7)
8. select count (*)
9 of)
10. Select current_day + (level - 1) d
input_data 11
12 connect
13 by level<= last_day="" -="" current_day="" +="">
14 )
15 where to_char (d, 'fmday') not in ('Saturday', 'Sunday');

COUNT (*)
----------
20

How to calculate the number of days/weeks/months between 2 dates?
Hello

I would like to know how to calculate the number of days/weeks/months between 2 dates in OBIEE 11 g, for example, I have 26/05/2013 and 19/05/2013, then I want to get 7 days.

Thank you!
Jamie
Hi Jamie,

Through this links...

http://www.bravesoft.com/blog/?p=682
http://twobiee.blogspot.in/2012/01/working-with-date-differences.html

Mark as correct it allows u...
Thank you...

Number of days between two Dates in the Correct month

I am trying to write a SELECT statement that returns the number of days spent on a project.

Here's my problem. I have a table with several columns of date (start_date, end_date, etc.) as well as several other columns. Here is an example of the table.

PROJECT... START_DATE... END_DATE
123 ................. 1ST JANUARY 13... 15 JANUARY 13
456 ................. 25 JANUARY 13... FEBRUARY 5, 13
789 ................. 30 JANUARY 13... FEBRUARY 5, 13
999 ................. 1ST FEBRUARY 13... FEBRUARY 8, 13

I'm counting the number of days spent on projects in each month. For example, with the above data, I would come back...

MONTHS... PROJECT_DAYS
Jan ............. 24
Feb ............. 18

I tried to use a CASE statement, but I'm having a hard time to understand. Any help would be greatly appreciated!

Published by: 987079 on February 8, 2013 13:12

An option would be something like

  1  with project as (
  2    select 123 project_id, date '2013-01-01' start_date, date '2013-01-15' end_date from dual union all
  3    select 456, date '2013-01-25', date '2013-02-05' from dual union all
  4    select 789, date '2013-01-30', date '2013-02-05' from dual union all
  5    select 999, date '2013-02-01', date '2013-02-08' from dual
  6  ),
  7  all_days as (
  8    select start_date + level - 1 dt
  9      from (select min(start_date) start_date,
 10                   max(end_date) end_date
 11              from project)
 12   connect by level <= end_date - start_date + 1
 13  )
 14  select trunc(dt,'MM'),
 15         count(*)
 16    from all_days ad
 17         join project p on (ad.dt between p.start_date and p.end_date)
 18   group by trunc(dt,'MM')
 19*  order by trunc(dt,'MM')
SQL> /

TRUNC(DT,'MM')        COUNT(*)
------------------- ----------
2013-01-01 00:00:00         24
2013-02-01 00:00:00         18

Justin

Number of days in a month

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