Manage the values of the actual day, month and year
Hello
I am trying to enter digital variables the same day (format: 1 / 10), month (format: 1 / 10) and year (format: 1900). I tried to use this sample:
"< mx:DateFormatter id ="DDformatter"formatString ="DD"/ > " "< mx:DateFormatter id ="MMformatter"formatString ="MM"/ >" "< mx:DateFormatter id ="YYYYformatter"formatString ="YYYY"/ >" In the Script: private var actualDate: Date = new Date(); [Bindable] private var actualDay:String = DDformatter(actualDate) .format. [Bindable] private var actualMonth:String = MMformatter(actualDate) .format. [Bindable] private var actualYear:String = YYYYformatter(actualDate) .format.
Part of getting an error when I run my application (am I missing some import?) I would like to have these data in numerical variables:
#1009 error - cannot access a property or method of a null object reference.
Pls find the solution of your problem... Let me know if you have any problem.
MainApplication.mxml
layout = "absolute" creationComplete = "onCreationComplete ()" > "
private var actualDate:Date = new Date(); [Bindable] private var actualDay:String; [Bindable] private var actualMonth:String; [Bindable] private var actualYear:String; private function onCreationComplete():void { actualDay = DDformatter.format (actualDate); actualMonth = MMformatter.format (actualDate); actualYear = YYYYformatter.format (actualDate); } ]]>
with respect,
Mayeul Singh Bartwal
Tags: Flex
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Thank you!I have something more good option here... Try to use this
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Hi all
I have this point of view
with (id, dt) t (select 1, to_date('02/06/2010','MM/DD/YYYY') of all the double union)
Select 2, to_date('11/29/2001','MM/DD/YYYY') of all the double union
Select 3, to_date('02/06/2011','MM/DD/YYYY') of all the double union
Select option 4, double to_date('10/10/2011','MM/DD/YYYY')
)
--
-end of test data
--
SELECT id, dt
floor (months_between(sysdate,dt)/12) years
, floor (mod (months_between (sysdate, dt), 12)) months
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, floor (sysdate-(add_months (dt, floor (months_between (sysdate, dt))) as dys
t
/
ID DT YEARS MONTHS DYS
---------- ----------- ---------- ---------- ----------
1-6 FEBRUARY 2010 5 8 22
2 29 NOVEMBER 2001 13 10 29
3-6 FEBRUARY 2011 4 8 22
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And subtract years, MONTHS and DYS as format
erase years, from 1 to 12 months and the days of the 1 to 30 as
91 days = 30 x 3 + 1 = > 1 day over 3 months
26 months + 3 months = 29 months = > 24 months + 5 months = 2 years + 5 months
and 26 years + 2 years = 28
desire to result: 28 years, 5 months and 1 day.
Is there a function that give me this result from the query above?
Kind regards
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Hello Gordan,.
I understand that you are the SUM of all the lines.
Here are two ideas:
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(a) SUM (TRUNC (sysdate) - dt)
I would give you the total number of days.
(b) adding the number of days to, for example, DATE ' 1900-01-01' would give another date...
(c) simply subtract the years 1900 and it takes several years,
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Maybe you want something else, for example: the total number of days divided by 365.25 for many years, the recall divided by 30 to get the number of months, the reminder being the number of days.(a) SUM (TRUNC (sysdate) - dt)
I would give you the total number of days.
(b) Division by 365.25 (le.25 is to take leap years into account, but you can choose for example "365")
(c) number of reminder of days (total - years * 365.25): divided by 30 is the number of months
(d) number reminder of days (total - years * 365.25 - months * 30): gives the number of days.
The following two options with your test data (by chance, this gives the same result in this case)
option 1:
with (id, dt) t (select 1, to_date('02/06/2010','MM/DD/YYYY') of all the double union)
Select 2, to_date('11/29/2001','MM/DD/YYYY') of all the double union
Select 3, to_date('02/06/2011','MM/DD/YYYY') of all the double union
Select option 4, double to_date('10/10/2011','MM/DD/YYYY')
)
target_date AS
(SELECT DATE ' 1900-01-01' + SUM (TRUNC (sysdate) - t.dt) FROM t trgt)
in_pieces AS
(SELECT TO_NUMBER (TO_CHAR (td.trgt, 'YYYY')) - 1900 y
, TO_NUMBER (TO_CHAR (td.trgt, 'MM')) - 1 m
, TO_NUMBER (TO_CHAR (td.trgt, 'DD')) - 1 d
Target_date TD
)
SELECT "result: ' |"
TRIM (CASE WHEN ip.y = 0 THEN NULL
WHEN ip.y = 1 THEN "1 year"
Of OTHER TO_CHAR (ip.y) | "years".
END |
CASE WHEN ip.m = 0 THEN NULL
WHEN ip.m = 1 THEN "1 month"
Of OTHER TO_CHAR (ip.m) | 'months '.
END |
CASE WHEN ip.d = 0 THEN NULL
WHEN ip.d = 1 THEN "1 day"
Of OTHER TO_CHAR (ip.d) | 'days '.
END
)
Of in_pieces ip
;
result: 28 years, 4 months and 28 daysoption 2:
with (id, dt) t (select 1, to_date('02/06/2010','MM/DD/YYYY') of all the double union)
Select 2, to_date('11/29/2001','MM/DD/YYYY') of all the double union
Select 3, to_date('02/06/2011','MM/DD/YYYY') of all the double union
Select option 4, double to_date('10/10/2011','MM/DD/YYYY')
)
nb_days AS
(SELECT SUM (TRUNC (sysdate) - t.dt) n t)
, y AS
(SELECT Nb.n, FLOOR (nb.n / 365.25) y nb_days n. b.)
ym AS
(SELECT y.n, y.y, FLOOR ((y.n-y.y * 365,25) / 30) FROM a)
ymd AS
(SELECT ym.y, ym.m, ym.n - ym.y * 365.25 - ym.m * ym FROM 30 d)
SELECT "result: ' |"
TRIM (CASE WHEN ymd.y = 0 THEN NULL
WHEN ymd.y = 1 THEN "1 year"
Of OTHER TO_CHAR (ymd.y) | "years".
END |
CASE WHEN ymd.m = 0 THEN NULL
WHEN ymd.m = 1 THEN "1 month"
Of OTHER TO_CHAR (ymd.m) | 'months '.
END |
CASE WHEN ymd.d = 0 THEN NULL
WHEN ymd.d = 1 THEN "1 day"
Of OTHER TO_CHAR (ymd.d) | 'days '.
END
)
WAN
;
result: 28 years, 4 months and 28 daysBest regards
Bruno Vroman.
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For example
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HTH
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Error definition day LOV based on the month and year
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...
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order by 2
That works well.
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Dear experts,
I have a year and the number of the day, I wanted to identify the date
ex: number of days = 203 year = 2009
Date: 22/07/2009
Please advice the best way possible to achieve.
Thank youDear user!
Please try this code.
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This figure first three are the day number and the last four digits are the year. The DDD format will give you the day and the month as JJ.MM. YYYY format join the year your date. To do this, try:
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Yours sincerely
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Published by: Florian W. the 22.07.2009 08:40
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