Dynamic calculation of the number of days between two dates in a table
Hello
I'm working on request where I dynamically calculate the number of days between two dates in a table.
The calculation must be dynamic, i.e., when I recover the Start_date and End_date and move to the field following (call_duration) in the same row, the difference must be calculated dynamically in this area and make sure the field read-only.
APEX version: 5.0
Hi BO123,
BO123 wrote:
Hello
I'm working on request where I dynamically calculate the number of days between two dates in a table.
The calculation must be dynamic, i.e., when I recover the Start_date and End_date and move to the field following (call_duration) in the same row, the difference must be calculated dynamically in this area and make sure the field read-only.
APEX version: 5.0
one of the way to do this by calling ajax on change of end_date.
See the sample code given below to fetch the resulting duration and making the field read only after calculation
Step 1: Change your page
under CSS-> Inline, put the code below
.row_item_disabled { cursor: default; opacity: 0.5; filter: alpha(opacity=50); pointer-events: none; }
Step 2: Create on demand Ajax process I say CALC_DURATION
Please check Procces Ajax, see line 6.7 How to assign a value to the variable sent by ajax call
Declare p_start_date date; p_end_date date; p_duration number; Begin p_start_date := to_date(apex_application.g_x01); p_end_date := to_date(apex_application.g_x02); --do your calculation and assign the output to the variable p_duration select p_end_date - p_start_date into p_duration from dual; -- return calculated duration sys.htp.p(p_duration); End;
Step 3: Create the javascript function
Change your page-> the function and the declaration of the Global Variable-> put the javascript function
You must extract the rowid in the first place, for which you want to set the time, see line 2
assuming f06, f07 and f08 is the id of the start date, and end date columns respectively, and duration
See no line no 8 how set the value returned by the process of ajax at the duration column
Replace your column to the respective column identifiers in the code below
function f_calulate_duration(pThis) { var row_id = pThis.id.substr(4); var start_date = $('#f06_'+row_id).val(); apex.server.process ( "CALC_DURATION", { x01: start_date,x02: $(pThis).val() }, { success: function( pData ) { // set duration to duration column $('#f08_'+row_id).val(pData); // disable duration column $("#f08_" + row_id).attr("readonly", true).addClass('row_item_disabled'); } }); }
Step 4: choose the end date call the javascript function
Go to report attributes-> edit your Date column end-> column-> Attrbiutes element attributes-> put the code below
onchange="javascript:f_calulate_duration(this);"
hope this helps you,
Kind regards
Jitendra
Tags: Database
Similar Questions
-
JavaScript anomaly on the number of days between two dates
Use ApEx 4.0, I found an anomaly in a javascript code that calculates the number of days between two dates, the current_date and the past_date. If the past and present is the or before March 10, 2013, and the current_date lies between 10 March 2013 and November 3, 2013, the number of days will be from 1 day to less than the actual number. Between November 3, 2013 and on 4 November 2013, the increments of number by 2, then the count will be accurate from this date forward.
Here are some examples:
March 10, 2013 = 69 days of December 31, 2012
March 11, 2013 = 69 days of December 31, 2012
March 12, 2013 = 70 days of December 31, 2012
November 3, 2013 = 306 days in December 31, 2012
November 4, 2013 = 308 days in December 31, 2012
11 March should be 70 and 12 March should be 71. November 3 is 307 and 4 November corrects the number of fake, which began March 11.
Change the past_date to March 10, 2013 produces the following:
March 10, 2013 = 0 days of March 10, 2013
March 11, 2013 = 0 days of March 10, 2013
March 12, 2013 = 1 days of March 10, 2013
But change the past_date to 11 March 2013, product of the correct numbers:
March 11, 2013 = 0 days of March 11, 2013
March 12, 2013 = 1 days of March 11, 2013
March 13, 2013 = 2 days of March 11, 2013
I would certainly all help to determine the cause of this anomaly. Here's the javascript code:
var w1 = ($v ("P48_PAST_DATE"));
W1 = (w1.toString ());
vmon var = (w1.substr (3.3));
vyr var = (w1.substr (7));
var r = (vyr.length);
If (r == 2)
vyr. = (parseFloat (vyr) + 2000);
vday var = (w1.substr (0.2));
var y = (vmon.concat ("", vday, ",", vyr));
y = Date.parse (y);
var w2 = ($v ("P48_CURRENT_DATE"));
var vmon2 = (w2.substr (3.3));
var vyr2 = (w2.substr (7));
var vday2 = (w2.substr (0.2));
var x = (vmon2.concat ("", vday2, ",", vyr2));
x = Date.parse (x);
var numdays = (x - y);
numdays = (Math.floor(numdays / 86400000));
$s ("P48_NUMBEROFDAYS", numdays);Did you google for something like "javascript number of days between two dates. I think you will find the explanation to this observation:
This method does not work correctly if there is an advanced economies jump between the two dates.
There are examples available to calculate the difference between two dates.
-
Calculate the number of days between two dates
Hello
Can someone help please change my formcalc script to calculate the number of working days between two date fields. My script currently calculates the total number of days between two dates, including the weekends which must be excluded from the total.
If
(HasValue (Start_Date1) & HasValue (End_Date1)) then
$
= Date2Num (End_Date1, "YYYY-MM-DD" "en_IE") - Date2Num (Start_Date1, "YYYY-MM-DD" "en_IE") + 1
on the other
""
endif
Any help will be most appreciated.
Thank you.
Check...
(1) you said that you put the script on the event «days1» calculate My sample imitates the variable names used in the original message, "Start_Date1" and "End_Date1". If the names of variables for the start and end dates are different, you will need to modify the script to account for these names.
(2) the Date2Num functions in the calculation of the "totalDays" use the date format "YYYY-MM-DD". If your date habits differ from "YYYY-MM-DD" FormCalc will complain.
Steve
-
Dynamically calculate the number of days between two dates and amounts of split
Hello
I have searched for a solution for this, but had no success.
I need to show the amounts broken down by days.
I have a table that has an amount column and start and end dates.
I need to write a query so that the amounts will be broken evenly based on the number of days between the start date and end date.
For example, for this line.
insert into my_test values (' 1, '' 3-mar-2010, ' 7 - mar - 2010 ", 1000);
the query returns this (split $1,000 over 5 days)
ID Date amount
1 ' 3-mar-2010' 200,00
1 ' 4-mar-2010' 200,00
1 ' 5-mar-2010' 200,00
1 ' 6-mar-2010' 200,00
1 ' 7-mar-2010' 200,00
create table my_test)
ID number (10),
start_date date,
End_date date,
amount number (10.2)
);
Select * from my_test
insert into my_test values (' 1, '' 3-mar-2010, ' 7 - mar - 2010 ", 1000);
insert into my_test values (2, 10-mar-2010 ", 19-mar-2010", 2000);
insert into my_test values (3, 20-mar-2010 ',' 21-mar-2010, 5000);
Thanks in advance.Hello
One way is to join a Meter of Table , a table, or (more often) a set of results includes a line for eery number 1, 2, 3,... until the maximum number of times you need to divide a line.
For example:WITH cntr AS ( SELECT LEVEL - 1 AS n FROM ( SELECT MAX (end_date - start_date) AS max_day_cnt FROM my_test ) CONNECT BY LEVEL <= 1 + max_day_cnt ) SELECT t.id , t.start_date + c.n AS dt , t.amount / (t.end_date + 1 - t.start_date) AS amt FROM my_test t JOIN cntr c ON c.n <= t.end_date - t.start_date ORDER BY id , dt ;
This assumes that all dates have the same number of hours, minutes, and seconds, as is the case in your sample data.
If this isn't the case, then use TRUNC (start_date) and TRUNC (end_date) instead of start_date and end_date or post some sample data and results if some lines do not represent a whole number of days. -
Calculation of the difference in days between two Dates
Hi all
I have two text fields with dates in them (P2_CURRENT_TIME_POINT_TEST_START_DATE, P2_ORGINAL_TEST_START_DATE). I would like to assign another text field (P3_TIME_FROM_INITIAL_TEST), the difference between the two days. I use a calculation page and tried suptracting them right or a date function diff, I found online, but nothing seems to work. Could someone lend me a hand on this one?
Thank you.Ben,
Sorry, please try the following:
:P3_TIME_FROM_INITIAL_TEST := TO_DATE(:P2_CURRENT_TIME_POINT_TEST_START_DATE, 'DD-MON-YYYY') - TO_DATE(:P2_ORGINAL_TEST_START_DATE, 'DD-MON-YYYY');
Remember to substitute the correct format string MON-DD-YYYY.
Kind regards
Danhttp://danielmcghan.us
http://sourceforge.NET/projects/tapigenYou can reward this answer by marking as being useful or correct ;-)
-
to calculate the number of days between 2 dates
Hi all
I need to calculate the number of days between 2 dates.
Y at - it no oracle built on that basis.
Thanks in advance
Kind regards
Nickuser646786 wrote:
HelloThanks a lot its really useful, but I must make a correction, I just want weekdays rather working days
As already said, there are many examples on the forum of the calculation of the working days or the days of the week or whatever you want between certain dates of restrictions. You will need to have a go at adapting what has already been given before and then when you get stuck, come here and your postcode, and someone will take a look for you.
-
How to find the number of days between 2 date elements in the XSLT file
Hello
I need to calculate the number of days between 2 date elements (type xs: date). Can you please direct me as to how I can do the same thing.
I work in 11g and using XSLT 1.0. I tried several options but yet to get a solution for this. I think that this can be done also using XSLT 2.0, but who has not worked for me.
Can someone please help with this problem, thanks in advance!
Thank you
AnjuHello
Have you seen this message?
Re: Get the Date difference between 2 values of date in daysYou can do in the XSLT file since the dates are in ISO 8601 format.
http://www.w3.org/TR/NOTE-datetimeHere is a sample XSLT...
The XSLT above will result in * 4 * for the next entry...
2012-01-11T00:00:00.000-05:00 2012-01-15T00:00:00.000-05:00 You can test this example here...
http://xslttest.appspot.com/I hope this helps...
See you soon,.
VladIt is considered good etiquette to the answerers rewards with points (as "useful" - 5 pts - or 'correct' - 10pts)
https://forums.Oracle.com/forums/Ann.jspa?annID=893 -
Calculate the difference in days between two Dates
Hello
I'm trying to understand how to calculate the difference in days between two dates using JavaScript in LiveCycle. (Minimum = JavaScript knowledge)
Where 'Start_Date' and 'Current_Date' are the names of the two dates in the palette of the hierarchy. (the two Date/time field)
* Current date is using the object > value > execution property > current Date/time
I need a text or number field showing the difference in days. (Difference_in_Days)
I noticed the following code is pretty standard among other responses:
var
Start_date = new Date (Start_Date);
var
Current_Date = new Date (Current_Date);
var
nAgeMilliseconds = Current_Date.getTime) - Start_date.getTime ();
var
nMilliSecondsPerYear = 365 * 24 * 60 * 60 * 1000 ;
I know there is lack of code and code above are may not be not correct.
Please notify.
OK, that's because of the way that javascript and works of the calculate event. The field will be filled with whatever the script resolves at the end of execution. Technically, your script does not have a value because the last thing you do is an assignment to a variable. Change the last line as follows:
Math.ABS ((firstDate.getTime)
((- secondDate.getTime (()) / (oneDay));
(eliminate the variable assignment) and get rid of the app.alert. Your script will "return" (have) regardless of the value of calculation from the East and which will be stored in the field.
-
find the difference in days between two dates
Hello world
I'm trying to find out the difference in days between two dates and the execution of the query that I'm passing
SELECT TO_char(sysdate, 'dd/mm/yyyy') - TO_char('15/11/2011', 'dd/mm/yyyy') DAYS FROM DUAL
the error I get is
ORA-01722: invalid number
01722 00000 - "invalid number."
* Cause:
* Action:
Could someone please help.
Thanks in advanceuser10636796 wrote:
Hello worldThanks a lot for all the replies. I am trying to apply it in a statement to my table like this
SELECT trunc (sysdate) - TO_char (date_last_recommended, ' dd/mm/yyyy') DAYS OF recommendation;
SELECT trunc (sysdate) - TRUNC (date_last_recommended) DAYS OF recommendation;
-
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 13:12An 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
-
Calculation of the number of rows between two rows
Hello
For a query I want to connect two tables with an outer join, and the external joined table instead of display NULL values display the value of the last row containg data. I try it using the LAG function, but I need the number of lines from which to extract the data to be dynamic, as the gap between the current line (NULL) and the last line displays data.
For example:
SELECT NVL (fadd.deprn_reserve,
LAG (fadd.deprn_reserve, "XXX") OVER (ORDER BY fadp.period_counter)
),
fadp.period_counter
OF fa_deprn_detail fadd, fa_deprn_periods naturally
WHERE fadd.period_counter (+) = fadp.period_counter
, where 'fadd' contains amounts and accounting periods 'naturally '. For example, there would be a balance in NOV - 08 with the next transaction is made in MAY - 09. Periods DEC-08 Apr-09 all shall display the balance of NOV - 08. The "XXX" would be the difference of the position of each line to the ' NOV-08' containing the amount to use.»
Is there a way to calculate the difference in positions of line like this, or this approach is not feasible?
ConcerningHello
I think LAST_VALUE is better for this task.
As Salim said, it is always useful to post a small example of data and the results desired from these data. Your version of Oracle can be important.
Using the table scott.emp, here is an example of using LAST_VALUE:
SELECT ename , comm , LAST_VALUE (comm IGNORE NULLS) OVER ( ORDER BY ename ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING ) AS prev_comm FROM scott.emp ORDER BY ename ;
Output:
ENAME COMM PREV_COMM ---------- ---------- ---------- ADAMS ALLEN 300 BLAKE 300 CLARK 300 FORD 300 JAMES 300 JONES 300 KING 300 MARTIN 1400 300 MILLER 1400 SCOTT 1400 SMITH 1400 TURNER 0 1400 WARD 500 0
-
Dates of the fortnight of days between two dates
Hello
I want to show every fortnight dates based on the date, to date and the day of week (sunday, monday...)
Input parameters
Date: 31/05/2011
To date: 30/06/2011
day of week: Thursday
Output
06/02/2011
16/06/2011
30/06/2011
Input parameters
Date: 25/06/2011
To date: 30/06/2011
day of week: Monday
Output
27/06/2011
can someone help me?
Published by: user10594152 on May 30, 2011 22:10
Published by: user10594152 on May 30, 2011 22:19Hello
Here's one way:
WITH parameters AS ( SELECT DATE '2011-05-31' AS start_date , DATE '2011-06-30' AS end_date , 'Thursday' AS day_o_week FROM dual ) , got_dt AS ( SELECT end_date , NEXT_DAY ( start_date - 1 , day_o_week ) + (14 * (LEVEL - 1)) AS dt FROM parameters CONNECT BY LEVEL <= CEIL ( (end_date - start_date) / 14 ) ) SELECT dt FROM got_dt WHERE dt <= end_date ;
This ID on NLS_DATE_LANGUAGE load. If you do not what it is, you can change the sub-quewry like this:
WITH parameters AS ( SELECT DATE '2011-06-25' AS start_date , DATE '2011-06-30' AS end_date , TO_CHAR ( DATE '2011-05-30' -- or any Monday , 'Day' ) AS day_o_week FROM dual ) ...
Published by: Frank Kulash, 30 May 2011 13:13
Adding a warning about NLS_DATE_LANGUAGE -
Behavior of days between two dates
Hello!
If anyone can help? ... with differences in Date? :
All I want is to calculate the number of days between two dates:
InitDate has been entered as 01/10/2010 planning, to recover is 20101001
HireDate was entered into the 1980-10-01 planning, to recover is 19801001
I tried
NumDays = @DATEDIFF (InitDate, HireDate, DP_DAY);
but give me only 0 and -1.
When I try the regular difference:
NumDays = InitDate - HireDate;
He returned to 30,000 (which in the example is 20101001-19801001) - which should not.
I use Essbase/Planning 11.1.1.2, I tried also to break dates in years and months using INT and operations, but when I want to get the calendar days he is ugly. The two dates live in the same block.
Someone at - it an idea?
Thank you very much!Hello
Here's a piece of code in my trunk of code which may help you. This assumes you have string CDF pack (https://www.samplecode.oracle.com/tracker/tracking/linkid/prpl1004/remcurreport/true/template/ViewIssue.vm/id/S441/nbrresults/9)var y1, y2, m1, m2, d1, d2, difx1, difx2;
FIX (SOLUTION YOUR MEMBERS HERE)
"BLOCK MEMBER")Y1=@round ("Date of 1" / 10000,0);
Y2=@round ("Date 2" / 10000,0);
M1=@round (("Date 1"-y1*10000)/100, 0);
M2=@round (("Date 2"-y2*10000)/100, 0);
D1 = ' Date of 1 "-(y1*10000+m1*100);"»
D2 = "Date 2"-(y2*10000+m2*100); "»difx1=@TODATEEX ("mm/dd/yyyy",
@JconcatStrings (@LIST (@JgetStringFromDouble (m2,@_false,@_false), ' / ', @JgetStringFromDouble (d2,@_false,@_false), ' / ', @JgetStringFromDouble (y2,@_false,@_false)))
)
) ;
difx2 = @TODATEEX ("mm/dd/yyyy", @JconcatStrings (@LIST)
@JgetStringFromDouble (m1,@_false,@_false), ' / '.
@JgetStringFromDouble (d1,@_false,@_false), ' / '.
@JgetStringFromDouble (y1,@_false,@_false))
)
);diff=@DateDiff (difx2, difx1, DP_DAY);
)
ENDFIXSee you soon,.
Alp -
County of days between two dates without weekend
Hello
I need a solution in the query or another thread, which returns the number of days between two dates without considering the weekend (Saturday and Sunday), I have a column of type Date, and return need in the form of a column HH hours and days in another column.
Concerning
JonasHi and welcome to the forum.
Don't forget you can do a search on this forum.
Your question has been asked before.Some other tips:
http://asktom.Oracle.com/pls/asktom/asktom.download_file?p_file=6551242712657900129
http://asktom.Oracle.com/pls/asktom/f?p=100:11:0:P11_QUESTION_ID:185012348071 -
Find the number of days between the dates...
Hi friends,
I am new to this development of blackberry applications so I do not know how to find the number of days between two days. Y at - there any API is avilable otherwise we have to write our own code. In fact, I tried GregorianCalendar but I get error that cannot find the symbol but I already imported net.rim.device.api.util and package java.util also... Please give an idea how to solve this problem.
with respect,
s.Kumaran.
Use DateField class instances to represent dates on the screen.
And your code will be as follows:
long date1 = date1DateField.getDate ();
long date2 = date2DateField.getDate ();
Maybe you are looking for
-
Control of search engine seems invalid or corrupt
In parameters, the function of search engine to customize does not appear to offer any option of withdrawal or the command option. Actually when I search, the auto-usages Yahoo browser. What could happen?
-
Hi all I'm new here and I need your help! My AXC600 hangs whenever I turn it on after 5 min and gives the blue windows so known and recognized 8.1 blue screen with an error in registry (but not the error number). I got the impression that it occurs a
-
How do I get another copy of Windows XP, if the disk has been lost?
Lost my cd of Windows XP. I would like to format my computer, real slow. I have the product key sticker on the side of my computer. Can I get a copy of the op and if I can use the same key? * original title - OP Windows XP have since lost the cd to r
-
Cannot connect Windows Marketplace error connection to the service.
Original title: Cannot connect to Windows Marketplace When I try to connect, I get the following message: Error connecting to the service. There was a problem connecting to the Games for Windows - LIVE service. Click Retry to try connecting, or click
-
Windows recovery error, does not start in windows 7
Description of the problem: "Windows error recovery" appears during the implementation with 2 options- (1) starting repair (recommended) (2) start Windows normally I can't access my desktop. If I choose 'Launch Startup Repair', it goes to the screen