Subtract business days from date - calculated column
Hello,
I had a calculated column on a library that took two dates and found the difference between them in business days, but I am not sure how to subtract business days from a date...for instance I get a start date from a form and I need to
subtract 10 business days from that date.
Can anyone help?
I've always resorted to Javascript/JQuery for that kind of function. I found an old fashioned loop worked the best for me - it supports going forward or backwards. I key it off of a change in a starting date, or sometimes a status change. My actual production
code takes into account another list where we remove holidays and non-work days.
newDate = getNextDate(newDate, -3);
$("input[title='Date Due']").val((newDate.getMonth() + 1) + "/" + newDate.getDate() + "/" + newDate.getFullYear());
function getNextDate(currentDate, offset) {
// offset is business days
var wkend = 0;
var index = Math.abs(offset); // need positive number for looping
var neg = true;
if(offset >= 0) { neg = false; }
var curDOW = currentDate.getDay();
var nextDate = new Date(currentDate);
for(var i=1; i <= index; i++) {
nextDate.setDate(nextDate.getDate() + (neg ? -1: 1));
var nextDOW = nextDate.getDay();
if(nextDOW == 0) {nextDate.setDate(nextDate.getDate() + (neg ? -2: 1));} // Sunday
if(nextDOW == 6) {nextDate.setDate(nextDate.getDate() + (neg ? -1: 2)); } // Sat
// alert("offset is " + offset + "start: " + currentDate + ", next date is " + nextDate);
return nextDate;
Robin
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AmolWe should probably add a function for this as I've certainly seen the requirement before.
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...too bad i don't get forum points for sharing this solution.
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Hi,
I have the following table with about 6000 rows of different year_month but I am compare and subtract v
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ABUJA JAN-2011 400000
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(q_name in VARCHAR2,
hmoCode in VARCHAR2)
return VARCHAR2
is
c1 number;
begin
select NHIS_CONTRIBUTION into c1 from CONTRIBUTION_MGT where upper(YEAR_MONTH)=upper(q_name) and upper(ORGANIZATION)=upper(hmoCode);
return c1;
exception when NO_DATA_FOUND THEN
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;I assume that the year_month column is a DATE. Date information always belongs in DATE columns.
As written, the two months do not need to be consecutive. If you always want consectuive months, this can be re-written so that you only have to enter the first month, not both of them.
I hope this answers your question.
If not, post a little sample data (CREATE TABLE and INSERT statements, relevant columns only) for all tables involved, and also post the results you want from that data.
Point out a few places where the query above is giving the wrong results, and explain, using specific examples, how you get those results from that data in those places. If you changed the query at all, post your code.
Always say which version of Oracle you're using.
See the forum FAQ {message:id=9360002} -
Group by month/year from DATE FORMAT column
Hi Chaps,
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This is what I have so far:
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SELECT
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<?php
$previousProject = '';
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Code:
// SHOW/HIDE CONTROL
<tr>
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Code:
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<?php $previousProject = $row_rsInvPending['projid']; } ?>
<tr class="proj1<?php echo $row_rsInvPending['projid'] ?>" style="display:none">
<td>column 1</td>
<td>column 2</td>
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<th>Job Title</th>
<th>Type</th>
<th>Language</th>
<th>Translator</th>
<th>Total</th>
<th>Full</th>
<th>Fuzzy</th>
<th>Proof</th>
<th>Full Price</th>
<th>Discount Price</th>
<th>Document Format</th>
<th>Pages</th>
<th>Typesetting Cost</th>
<th>EN Proofreading Cost</th>
<th>Total</th>
<th>Translator Charge</th>
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<?php
$previousProject = '';
if ($totalRows_rsInvPending > 0) {
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while ($row_rsInvPending = mysql_fetch_assoc($rsInvPending)) {
if ($previousProject != $row_rsInvPending['projid']) {
// for every Project, show the Project ID
?>
<tr>
<td colspan="18" class="highlight"><span class="blueBold"><a href="#" onclick="toggle2('proj1<?php echo $row_rsInvPending['projid'] ?>', this)"><img src="../../Images/plus.gif" border="0" /></a> <?php echo $row_rsInvPending['projid'] ?> - </a></span><span class="blueNOTBold"><em><?php echo $row_rsInvPending['projtitle'] ?></em></span></td>
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<?php $previousProject = $row_rsInvPending['projid']; } ?>
<tr class="proj1<?php echo $row_rsInvPending['projid'] ?>" style="display:none">
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<td><?php echo $row_rsInvPending['projtype']; ?></td>
<td><?php echo $row_rsInvPending['langtname']; ?></td>
<td><?php echo $row_rsInvPending['translator']; ?></td>
<td></td>
<td></td>
<td></td>
<td></td>
<td></td>
<td></td>
<td></td>
<td></td>
<td></td>
<td></td>
<td></td>
<td></td>
<td></td>
<td></td>
<td></td>
</tr>
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<?php } // Show if recordset not empty ?> -
Reg : subtracting weekoff day from the time difference
Hi ,
I have query where I am getting time difference first of all in terms of hours as:
DECLARE @L_TIME_DIFF INT
SET @L_TIME_DIFF= (DATEDIFF(hh, '05/09/2014', '05/12/2014'))
i.e. subtracting 12th may - 9th may and getting diff in hours ..in this case it will be 3*24 = 72 hours
All fine till here. Now next step I need to subtract from this 72 hours if falling any week off day. It can be Friday, it can be Saturday, or Sunday, or Half day Saturday, or Half Day Sunday, or Full 2 days Saturday/Sunday .
The above scenario is decided based on one table [CALENDER_WORKINGDAYS] which looks like below :
CAL_WORKING_DAY
CAL_AM
CAL_PM
CAL_SEQ_NO
Monday
1
1
1
Tuesday
1
1
2
Wednesday
1
1
3
Thursday
1
1
4
Friday
0
0
5
Saturday
1
1
6
Sunday
1
1
7
So in this case, if you look at friday (CAL_AM=0 and CAL_PM=0) this measn friday full day holiday for some country.
Hence I need to : 72 - 24 = 48 Hours should be my output.
In another scenario, let s say some country work half day Saturday and Sunday full day off.
So my output should be : 72 - 36 (because 24 hours for sunday off, and 12 hours for saturday half day off) =
36 should be my output
and so on..
Can you help to build such query based on the above table given ?
ThanksHi Jose Diz,
I am trying the below logic. Seems it should be working:
DECLARE
@L_START_TIME DATETIME,
@L_END_TIME DATETIME,
@L_TIME_DIFF INT, -- Time Difference considering week offs
@W_TIME_DIFF INT ,-- Whole Time Difference,
@PUBLICHOLS INT,
@WEEK_OFF INT
set @L_START_TIME = '5/9/2014';
set @L_END_TIME = '5/12/2014';
/* TO GET THE TIMDIFFERENCE */
SET @L_TIME_DIFF=(DATEDIFF(hh,@L_START_TIME,@L_END_TIME))
SET @W_TIME_DIFF = @L_TIME_DIFF
-- Subtracting day off in the week(i.e. sat/sun or in some case friday) if it is in between start and end
set @WEEK_OFF= 0;
SELECT @WEEK_OFF= @WEEK_OFF + 12 * (abs(CAL_AM -1) + abs(CAL_PM -1))
from CALENDER_WORKINGDAYS
IF DATENAME(WK,@L_END_TIME) > DATENAME(WK,@L_START_TIME)
BEGIN
SET @L_TIME_DIFF= @L_TIME_DIFF - @WEEK_OFF
END
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Hi all, I need to create a user exit to get a date value based on another variable. Basically, 100 days minus the first variable.
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l_s_range-opt = 'EQ'.
APPEND l_s_range TO E_T_range.
ENDLOOP.
EXIT.
ENDIF.
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