Find maximun days of a month

Hi. Someone can tell me if exist one method to find the maximun days of a month?
I think that i must use Calendar Class.
Thanks.
Daniele.

I've made a mistake :
Use Cal.getActualMaximum(Cal.DAY_OF_MONTH);
instead of
Cal.getMaximum(Cal.DAY_OF_MONTH);
Jean.
U're right, use Calendar class like that :
Calendar Cal = Calendar.getInstance();
Cal.set(Cal.MONTH, Cal.JUNE);
int maxDay = Cal.getMaximum(Cal.DAY_OF_MONTH);
I didn't try it, but I think this should work.
Regards
Jean

Similar Messages

Function for finding first day of the month !!

Hi,
I know we have function to find the last day(DD) of the month. Do we have any functions for finding the first day of the month ??? if not is there any way i can find the first date(DD) of the month .
Bcoz i m trying to incorporate the logic for finding the first day of the month partition.
Thank you!!!

Shahid Ali Tcs wrote:
There are many solution given by member in relation with your question, and all are correct .
I want to make u know something else.
Have you ever think why oracle has given function for last_day but no function for first.....
I u think this question u will get the answer of your "first day finding " question.
Because first day is alwasy 01 of every, while last day can be 30,31,29,28.....
Got my words ........
One more solution from my side,,,,
SQL> select '01' || to_char(sysdate,'-MON-YY') from dual;
'01'||TO_
01-SEP-09Which is a completely poor way of doing it.
Using TRUNC or LAST_DAY(..) + 1 will return a DATE datatype result.
Your method is converting the DATE into a VARCHAR2, which then prevents further processing/date based calculations unless it is explicitly converted back to a DATE again.
The reason Oracle hasn't provided a FIRST_DAY is because the TRUNC function already caters for it as this works with DATE's as well as NUMBER's, not because the first day is always 1.

Elapse days - calculation from month and year

Duplicate thread ...
coding required for converting month into days
I have Month and year field in my DSO --Year( 2009), Month(4)
I want below logic to calculate -
MTD qty % = (qty * elapse days) / no of days in month
so from the above, I want to know,
1. How can we get the no of days from the above 2 objects (year and month).
2. Elapse days are the days that are over from current date to Ist day of that month.
eg: current date - 04.03.2009 , 1st day of the month - 04.01.2009,
elapse days = 2. (you need to consider all days in the month, not only working days)
so my questions is how can we get, # of days n a month from above 2 fields and elapse days based on the above condition.
also want to know, where can I have the logic in transformations or query level.
please provide your suggestions.
Thanks,
Pra
Edited by: Arun Varadarajan on Apr 5, 2009 10:44 PM
Edited by: Arun Varadarajan on Apr 5, 2009 10:45 PM

Hello,
I think the sample program below does what you ask: it finds the days in the month (bit of an overkill to use function modules for that) and then computes the QTD.
Note that the internal string representation of a data variable is always YYYYMMDD, regardless of the "externalized" form (e.g. yyyy/mm/dd or dd.mm.yyyy), so this code will work regardless of your custom date format.
Regards,
Mark
REPORT zqty_to_date.
PARAMETERS:
p_date     TYPE dats,
p_qty      TYPE i.
DATA:
days      TYPE i,
n_year    TYPE numc4,
n_month   TYPE numc2,
n_day     TYPE numc2,
qtd(6)    TYPE p DECIMALS 1.
START-OF-SELECTION.
n_year = p_date+0(4).
n_month = p_date+4(2).
n_day = p_date+6(2).
PERFORM days_in_month USING n_year n_month CHANGING days.
qtd = ( p_qty * ( n_day - 1 ) ) / days.
WRITE: / 'Days in month:', days,
         / 'Qty to date :', qtd.
*&      Form days_in_month
*       text
FORM days_in_month USING year month CHANGING days.
DATA: ymod4   TYPE i,
        ymod100 TYPE i,
        ymod400 TYPE i.
CASE month.
    WHEN 4 OR 6 OR 9 OR 11.
      days = 30.
    WHEN 2.
      ymod4 = year MOD 4.
      ymod100 = year MOD 100.
      ymod400 = year MOD 400.
      IF ( ymod4 = 0 AND ymod100 > 0 ) OR ( ymod100 = 0 AND ymod400 = 0 ).
        days = 29.
      ELSE.
        days = 28.
      ENDIF.
    WHEN OTHERS.
      days = 31.
ENDCASE.
ENDFORM.                    "days_in_month

How to find first and last day of previous month?.

Based on current month, I want to find start and end day of
previous month.
For example,
For august, Start date is July 1st and End date July 31st.
How can i get first of previous month as start date and last
day of the previous month as end date?.
Same way,
i want to find start date of current month and end date of
next month.
Example:
For august,
i want to get start date, august 1st and end date : september
30.
How can i do this from current date or now().
I am looking for best and easy way to find start and end
dates..
Thanks

<cfset today = now()>
<cfset firstOfThisMonth = createDate(year(today),
month(today), 1)>
<cfset lastOfNextMonth = dateAdd("d", -1, dateAdd("m", 2,
firstOfThisMonth))>
<cfoutput>
today = #today#<br>
firstOfThisMonth = #firstOfThisMonth#<br>
lastOfNextMonth = #lastOfNextMonth#<br>
</cfoutput>
Edit - To find the start and end day of previous month, get
the first of THIS month. Use Subtract 1 month ("m") to get the
start date. Subtract 1 day ("d") to get the end date.

Finding no. of days in a month in formula variable

Hi all
I wanted to find out no. of days in a month in formula variable customer exit at bex. I did the following code, which returning X value. Can anybody helps here.
WHEN 'ZFV_NO_DAYS_MONTH'.
     IF I_STEP = 2.
       CLEAR L_S_RANGE.
       READ TABLE I_T_VAR_RANGE INTO XTAB WITH KEY VNAM = 'zdate1'.
       V_DATE = XTAB-LOW.
       P_DATE = V_DATE+6(2).
       CALL FUNCTION 'HR_E_NUM_OF_DAYS_OF_MONTH'
         EXPORTING
           P_FECHA        = V_DATE
         IMPORTING
           NUMBER_OF_DAYS = V_DATE1.
       V_RES = V_DATE1 - P_DATE.
       L_S_RANGE-LOW = V_RES.
       APPEND L_S_RANGE TO E_T_RANGE.
     ENDIF.
Regrds
swetha

Hi Swetha
in your code itself just change something as mentioned below
VDATE1 type as integer.
WHEN 'ZFV_NO_DAYS_MONTH'.
     IF I_STEP = 2.
       CLEAR L_S_RANGE.
       READ TABLE I_T_VAR_RANGE INTO XTAB WITH KEY VNAM = 'zdate1'.
       V_DATE = XTAB-LOW.
    *Hide No need***   P_DATE = V_DATE+6(2).
       CALL FUNCTION 'HR_E_NUM_OF_DAYS_OF_MONTH'
         EXPORTING
           P_FECHA        = V_DATE
         IMPORTING
           NUMBER_OF_DAYS = V_DATE1.
   *hide**    V_RES = V_DATE1 - P_DATE.
   *hide**     L_S_RANGE-LOW = V_RES.
*Add the below new line
L_S_RANGE-LOW = V_DATE1.
       APPEND L_S_RANGE TO E_T_RANGE.
     ENDIF.
Regards
Sureshkumar C

ICal launches, but no calendars (day, week or month) will open. All menus in the top bar seem normal, but none of them create any action. I cannot find any troubleshooting or instructional articles explaining how to restore or reload iCal. Anyone had this

iCal launches, but no calendars (day, week or month) will open. All menus in the top bar seem normal, but none of them create any action. I cannot find any troubleshooting or instructional articles explaining how to restore or reload iCal. Anyone had this happen? Is there a fix?

I tried using Time Machine to restore iCal, tried doing disk repairs... nothing changes the outcome. No windows will open in the application.

Find number of days in current month

Hi Folks ,
I needed some help in knowing the number of days in any month ... I need to develop an application which will be triggering a mail with an excel attachment at the end of every month ...after populating the excel sheet from an Oracle database ..
Kindly help...
Thx..

Thx a lot for your immediate reply .. I tried using
the calender class. But for my application .. i need
to automate this process by sending a mail every last
day of the month... but i am finding it a bit
difficult to really automate setting the current
month to the Timer class
Timer timer;
     public Reminder ( int seconds ) {
          timer = new Timer ( ) ;
timer.schedule( new RemindTask ( ), seconds*1000
000 ) ;
     }The Calendar class is still the best solution IMHO.

How do I find the start day of each month

Hi all,
Hope someone can give a a hand here I've to do a calender project for college and I'm not allowed to use any built in Java classes so fun all the way so far :-)
I've got one question that is bugging me at the moment. and that is how to find the start day for each month in a year and store them in an array.
I'm starting my week at Monday (0) and stopping at Sunday (6) and I do have a formula for discovering the first day in Jan of the given year its
((y-1900)*365 + (y-1901)/4)%7
I was just wondering how I could amend this to find the first day in Feb, March etc.
Any suggestions greatly accepted.
Thanks in advance.

Irish-Student wrote:
newark maybe I didnt make clear exactly what I want to do.
I have each day of the week starting at 0 for Monday up to 6 for Sunday I want to map what the start day of a month as a number between 0 and 6.
The formula I have discovering the first day of Jan of a given year will give you 1 i.e. Tuesday
what I need to do is find out what number between 0 and 6 the start day of each other month of a given year is.
For example the 1st of Feb is a Friday which is 4 on the weekday map.
Your suggestion of 31%7 won't work for this, sorry more than likely my fault for not being clearer in what my requirements were.
Cheers for the help so far and yep your right not being able to use the built in classes is a pain :-)
Edited by: Irish-Student on 03-Mar-2008 20:01This doesn't really change much of my first suggestion. If you can already calculate the first day of the month of January for any given year, then there's no problem figuring out the other months. Say I want to know the first day April 1, 1978. You already know how to find January 1, 1978. So find that. Then find the number of days between January 1 and April 1 (remember to account for leap years). Once you have the number of days, you can figure out how many weeks and days that is, and thus what day of the week it is.

How to find out average files i am getting every day in a month?

How to find out average files I am getting every day in a month?

Do you mean average count of files received?
Please Mark This As Answer if it solved your issue
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What do you do in a query in order to find the 1st day of last month ?

i use a
select trunc(trunc(sysdate,'mon')-1,'mon') from dual;
what about you guys ?

Hi,
Of all the solutions, I like yours
add_months(trunc(sysdate,'mm'),-1)or doing the operations in the opposite order, like Sentinel did
trunc(add_months(sysdate,-1),'mm')best, mainly because:
(1) it's easy to read and understand
(2) it's easy to modify (in case you want the first day from three months ago, or six months in the future)
By the way, after you post something on this forum, if you see that you've made a mistake, you can correct it. Click on the second icon to the left of "Reply". (What is that: a hand holding a pencil?) You don't need to post a second message.

How to find the last 4th business day of the month (not include weekend)

Hi,
I need help how to code the ABAP program to locate the last 4th business day of the month (not include Satuday, Sunday and holidays).
For example for the last 4th business day in 2008 are:
1/28/2008,
2/26/2008
3/26/2008
4/25/2008
5/27/2008
6/25/2008
7/28/2008
8/26/2008
9/25/2008
10/28/2008
11/21/2008
12/26/2008
Thank you so much!
Helen

Hello Helen
I would use a two-step approach:
(1) Set date to the first of day of month (e.g. 20080201 = 01.02.2008)
(2) Subtract four working days from this date
If I remember correctly there is a function group BKK available which contains fm's for adding workdays to a given date (search for 'BKKWORKDAY). To subtract a number of working days use a negative number.
See also: [BKK_GET_MONTH_LASTDAY|reg :BKK_GET_MONTH_LASTDAY;
Regards
Uwe

Query to find the fifteenth day of a month

i want to know the fifteenth day of a month in oracle
can anyone send the query

Hi,
Try this:
"afiedt.buf" 2 lines, 57 characters
1* SELECT TO_CHAR(TO_DATE('15-AUG-2006'),'Day') FROM DUAL
SQL>/
TO_CHAR(T
Tuesday
1 row selected.
SQL>Regards
Avinash

View Outlook Calendar on iPhone In Day, Week, Or Month Views--NOT As A List

I apologize in advance if this has been asked already but couldn't find the answer after an extensive search.
My question is, after syncing my Outlook calendar to the iPhone, how can I view my appointments other than as a list in the iPhone? I would like to view my appointments in either day, week, or month views as opposed to only knowing what my next two or three appointments are. To me it just seems like a simple setting I'm overlooking somewhere. I appreciate any help you can afford
Thanks in advance,
-V5

*iPhone calendar views are a joke!*
My 90's Palm Vx had far more usable views via Pimlico Software. The gray scale display with poor resolution was able to show one or even two weeks with appointment texts.
What does iPhone have? List view is the best there is. One sees all appointments with only just scrolling. Day view is terrible. The whole day should be visible without scrolling for it to be at least somewhat efficient. Now one has to first select a day with arrows and after that scroll thru the day and after that select another day and scroll again. No way. Month view is very poorly designed! The grid takes way too much space and who decided to put the appointment list in the bottom of the screen when one's index finger blocks it if sliding thru days in the grid!!! Oh man!!! Unlock the landscape view and please make the grid a little bit smaller and please place the appointment list on the left or at least on the right side. And make its font way smaller! No point in seeing two appoinments without scrolling.
iPhone's wonderful screen would be able to show one or two weeks with written appointments in a list like manner especially landscape. I mean seven columns titled Mo-Su and all appointments listed in that day's column.
Hopefully SDK will let someone else do it if Apple won't.
And we have to get to copy appointments and paste them to other days also. I am spending 30 min writing the same info one at a time to our monthly plan. Everyone else is already having a cup of coffee.

Count days in a month for a date range

i am trying to find no. of days between 2 Date Ranges for a list of Ids. i used the logic in the below link:
count days of the month
My query is giving duplicates since, I have list of Ids.
Doctor_ID     Patient_ID     ARRIVE_DT_TM     DISCH_DT_TM
755722     42972229     10/18/2012 7:50     3/14/2013 20:45
763305     42972232     1/7/2013 20:27     3/15/2013 19:15
25391509     42972298     2/4/2013 22:45     3/8/2013 22:03
746779     42972331     1/4/2013 23:00     3/26/2013 21:50
763305     42972338     3/4/2013 22:19     3/6/2013 19:35
763305     42972411     11/4/2013 22:32     3/29/2013 17:30
I am looking for query to give me for Patient_ID = 42972229
MONTH     COUNT_DAYS
201210     14
201211     30
201212     31
201301     31
201302     28
201303     14
I am running the following code and it loops through the months and gives duplicates when I remove where Patient_id IN (42972229)
select
Doctor_ID
, Patient_ID
, AR_DTTM
, DSC_DTTM
, TO_CHAR(ADD_MONTHS(TRUNC(date1, 'MONTH'), LEVEL - 1), 'YYYY MM') MONTHS_BET
, (LEAST(date2, ADD_MONTHS(TRUNC(date1, 'MONTH') - 1, LEVEL)) - GREATEST(date1, ADD_MONTHS(TRUNC(date1, 'MONTH'), LEVEL - 1)))+ 1 AS DAYSCOUNT
from (select
Doctor_ID
, Patient_ID
, ARRIVE_DT_TM AR_DTTM
, DISCH_DT_TM DSC_DTTM
,TRUNC(ARRIVE_DT_TM,'DDD') AS date1
,TRUNC(DISCH_DT_TM,'DDD') AS date2
from temp where Patient_id IN (42972229)
CONNECT BY LEVEL <= MONTHS_BETWEEN(TRUNC(date2, 'MONTH'), TRUNC(date1, 'MONTH')) + 1
Please help!

Hi,
ASTRA_007 wrote:
Results I would like to see are:
Doctor_ID     Patient_ID     ARRIVE_DT_TM     DISCH_DT_TM     Month     CountofDays
755722     42972229     10/18/2012 7:50     3/14/2013 20:45     2012 10     14
755722     42972229     10/18/2012 7:50     3/14/2013 20:45     2012 11     30
755722     42972229     10/18/2012 7:50     3/14/2013 20:45     2012 12     31
755722     42972229     10/18/2012 7:50     3/14/2013 20:45     2013 01     31
755722     42972229     10/18/2012 7:50     3/14/2013 20:45     2013 02     28
755722     42972229     10/18/2012 7:50     3/14/2013 20:45     2013 03     14
763305     42972232     1/7/2013 20:27     3/15/2013 19:15     2013 01     25
763305     42972232     1/7/2013 20:27     3/15/2013 19:15     2013 02     28
763305     42972232     1/7/2013 20:27     3/15/2013 19:15     2013 03     15
and so on...So each row represents a patient-month, and you want to display several columns from the temp table on each output row. In that case, include all those columns in both the SELECT and GROUP BY clauses, like this:
WITH     universe     AS
     SELECT     *
     FROM     temp
--     WHERE     patient_id     IN (42972229)
,     date_range     AS
     SELECT     TRUNC (MIN (arrive_dt_tm))     AS first_date
     ,     TRUNC (MAX (disch_dt_tm))     AS last_date
     FROM     universe
,     all_dates     AS
     SELECT     first_date + LEVEL - 1     AS a_date
     FROM     date_range
     CONNECT BY     LEVEL     <= (last_date + 1) - first_date
SELECT    u.doctor_id
,       u.patient_id
,       u.arrive_dt_tm
,       u.disch_dt_tm
,       TO_CHAR ( TRUNC (a.a_date, 'MONTH')
              , 'YYYY MM'
            )          AS month
,       COUNT (*)          AS count_days
FROM       all_dates a
JOIN       universe   u ON a.a_date BETWEEN TRUNC (u.arrive_dt_tm)
                                     AND      u.disch_dt_tm
GROUP BY u.doctor_id
,       u.patient_id
,       u.arrive_dt_tm
,       u.disch_dt_tm
,        TRUNC (a.a_date, 'MONTH')
ORDER BY u.patient_id
,       TRUNC (a.a_date, 'MONTH')
;Output from your sample data (with no filtering):
`DOCTOR_ID PATIENT_ID ARRIVE_DT_TM     DISCH_DT_TM      MONTH   COUNT_DAYS
    755722   42972229 10/18/2012 7:50 3/14/2013 20:45 2012 10         14
    755722   42972229 10/18/2012 7:50 3/14/2013 20:45 2012 11         30
    755722   42972229 10/18/2012 7:50 3/14/2013 20:45 2012 12         31
    755722   42972229 10/18/2012 7:50 3/14/2013 20:45 2013 01         31
    755722   42972229 10/18/2012 7:50 3/14/2013 20:45 2013 02         28
    755722   42972229 10/18/2012 7:50 3/14/2013 20:45 2013 03         14
    763305   42972232 1/7/2013 20:27   3/15/2013 19:15 2013 01         25
    763305   42972232 1/7/2013 20:27   3/15/2013 19:15 2013 02         28
    763305   42972232 1/7/2013 20:27   3/15/2013 19:15 2013 03         15
25391509   42972298 2/4/2013 22:45   3/8/2013 22:3    2013 02         25
25391509   42972298 2/4/2013 22:45   3/8/2013 22:3    2013 03          8
    746779   42972331 1/4/2013 23:0    3/26/2013 21:50 2013 01         28
    746779   42972331 1/4/2013 23:0    3/26/2013 21:50 2013 02         28
    746779   42972331 1/4/2013 23:0    3/26/2013 21:50 2013 03         26
    763305   42972338 3/4/2013 22:19   3/6/2013 19:35   2013 03          3
In the end the objective is to count the no. of days in each month between the arrival and discharge dates by Physician and for his/her patients.Then is the output above really what you want? Say you're interested in physician 763305. That physician had 18-patient days in March, 2013, but the output doesn't make it clear.
I ran your query, it works great but I have a long list of patients for whom I have to run these counts.the query above includes all patient_ids.
That's a separate problem, to be solved in the first sub-query, universe. The rest of the query will be unchanged.
How will you know which patients to include? If you can derive the list from temp itself, just use a WHERE clause in universe. If you need to look at other tables, join them in universe, or use them in sub-queries in universe, or both.
For exmple, if you decide that the list of patient_ids has no pattern, and that you'll need to store their ids in a separate table (perhaps a global temporary table), then universe might be:
WITH     universe     AS
     SELECT     t.*     -- or list columns needed
     FROM     temp                        t
     JOIN     patient_ids_to_include p
                  ON p.patient_id = t.patient_id
) ...The rest of the query can be the same as above.
If a same patient is admitted again then Patient_ID will be different no matter when readmitted.Are you saying that patient_id identifies a visit, not a patient, and that the same person is assigned a different patient_id every time that person is admitted?
For
INSERT INTO temp (doctor_id, patient_id, arrive_dt_tm, disch_dt_tm)
VALUES ( 755722
, 42972229
, TO_DATE ('03/14/2013 23:00', 'MM/DD/YYYY HH24:MI')
, TO_DATE ('04/01/2013 12:00', 'MM/DD/YYYY HH24:MI')
First the Patient ID will be different from the earlier admission. Second the results will show like:
Doctor_ID     Patient_ID     ARRIVE_DT_TM     DISCH_DT_TM     Month     CountofDays
755722     42972229     3/14/2013 23:00     4/1/2013 12:00     2012 03     14
755722     42972229     3/14/2013 23:00     4/1/2013 12:00     2012 04     1Are you saying that temp.patient_id is unique, and so the situation is impossible?
Edited by: Frank Kulash on May 7, 2013 10:23 AM

Calculating given occurance of a given day in every month

Hi,
I'm using oracle 10.2.0.1.0 and forms 6i.
I have inputs like frequency, day and start date.
Suppose
Start Date : 12-03-2011
Day : Saturday
Frequency : 2
What i want to do is, i want to calculate 2nd occurance of saturday(2nd Saturday), in every month, starting from the given date, till the end of the year.
The given start date, is anyway the 2nd occurance of saturday in March.
So now in april -- its is 09-04-2011
May -- 14-05-2011
June -- 11-06-2011
Is there a way to calculate like this?
Please help
Thanks

Hi,
Another way is to use the NEXT_DAY function, like this:
WITH     parameters     AS
     SELECT     DATE '2011-03-12'     AS start_date
     ,     'SATURDAY'          AS week_day
     ,     2               AS week_num     -- This isn't really a frequency
     FROM     dual
,     all_dates     AS
     SELECT start_date
     ,     NEXT_DAY ( ADD_MONTHS ( TRUNC (start_date, 'MONTH')
                          , LEVEL - 1
                          ) + (7 * week_num)
                         - 8
                 , week_day
                )     AS a_date
     FROM     parameters
     CONNECT BY     LEVEL     <= 12
SELECT       a_date
FROM       all_dates
WHERE       a_date          >= start_date
AND       TRUNC (a_date, 'YEAR') = TRUNC (start_date, 'YEAR')
ORDER BY a_date
;If you're doing this in Oracle Forms, then you don't need a sub-query like parameters. You can pass the parameters in PL/SQL variables, and use dual as the base table in all_dates (or use a PL/SQL loop).
The calculation of a_date works like this:
First, we find the beginning of the month.
Then, we add wekk_num weeks to that. For example, if week_num=2, that will give us the beginning of the 3rd week.
We're interested in the week before that, so we subtract 7, and we also subtract another 1 because NEXT_DAY always returns a date after (not the same as) its 1st argument.
Unfortunately, NEXT_DAY is always dependent on NLS_LANGUAGE. If you specify 'Saturday', but that day is called 'Lördag' in your NLS language, you'll get an error. If that's a possibility, then you can change the first sub-query like this:
WITH     parameters     AS
     SELECT     DATE '2011-03-12'     AS start_date
     ,     TO_CHAR ( DATE '2011-04-02'     -- or any Saturday
               , 'DAY'
               )          AS week_day
     ,     2               AS week_num
     FROM     dual
) ...Instead of April 2, 2011, you can you any Saturday you want; just be sure it's a Saturday.

Find maximun days of a month

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