Extract Time from date and Time and Need XLMOD Funtion to find the Difference between Two Time.
X6 = "1/5/15 5:16 AM" & NOW ....................difference by Only Time
not date
X6 date and Time will be changing, Its not Constant
Dim myDateTime As DateTime = X6
Dim myDate As String = myDateTime.ToString("dd/MM/yy")
Dim myTime As String = myDateTime.ToString("hh:mm tt")
Dim myDateTime1 As DateTime = Now
Dim myDate1 As String = myDateTime1.ToString("dd/MM/yy")
Dim myTime1 As String = myDateTime1.ToString("hh:mm tt")
Need to use this function to find the Difference between Two Time. due to 12:00 AM isuue
Function XLMod(a, b)
' This replicates the Excel MOD function
XLMod = a - b * Int(a / b)
End Function
Output Required
dim dd = XLMod(myTime - myTime1)
Problem is myTime & myTime1 is String Need to convert them into Time, Later use XLMOD Funtion.
Induhar,
As an addendum to this, I thought I'd add this in also: If you have two valid DateTime objects you might consider using a class which I put together a few years ago
shown on a page of my website here.
To use it, just instantiate with two DateTime objects (order doesn't matter, it'll figure it out) and you'll then have access to the public properties. For this example, I'm just showing the .ToString method:
Option Strict On
Option Explicit On
Option Infer Off
Public Class Form1
Private Sub Form1_Load(ByVal sender As System.Object, _
ByVal e As System.EventArgs) _
Handles MyBase.Load
Dim date1 As DateTime = Now
Dim date2 As DateTime = #1/1/1970 2:35:00 PM#
Dim howOld As New Age(date1, date2)
MessageBox.Show(howOld.ToString, "Age")
Stop
End Sub
End Class
I hope that helps, if not now then maybe at some point in the future. :)
Still lost in code, just at a little higher level.
Thanx frank, can use this in Future....
Similar Messages
-
Function Module to find the Difference between two times.
Hi All,
Wud you plz let me know the Function Module to find the Difference between two times.
Input Time1( Hours:Minutes) Time2 ( Hours:Minutes)
Need Output in Hours:Minutes only . ( No seconds Needed )
Ex :
Input :
06:00 to 18:00 Output : 12:00
and 20:00 to 06:00 Output: 10:00 with +ve sign only. No -ve sign.
Thanks,
N.L.Narayanacheck this .
data : p_timel like sy-uzeit,
p_timeh like sy-uzeit,
diff like sy-uzeit,
di(8) type c .
p_timel = '200000'.
p_timeh = '060000'.
diff = p_timeh - p_timel.
concatenate diff+0(2) ':' diff+2(2) into di.
write:/ di.
also check for this.
p_timel = '060000'.
p_timeh = '180000'.
see if this can be implemented in ur code .
or else u can try with Fm L_TO_TIME_DIFF passing startdate enddate starttime endtime with UOM as MIN
hope this helps regards,
vijay -
Help I need to subtract times to find the difference between two times?
Hello there, I really need help.
I'm trying to make this attendance tracking system so that when a user enters his or her ID the system wll remember his/her check in time. However I also need to find out either how late the person is or how early to store their over time and late.
So I get the time by using a timestamp, so I'm left with a timestamp that I store into the db. But the problem comes when trying to compute for the late or over time. To do that I have to get the normal login/checkin time from the employee schedules table in the db and either subtract that time from the checkInTime to compute for how late or subtract the checkIn time from it to find early.
The thing is, the schedule is stored in datetime format in the db since the db is sql. Also I have no idea on how to subtract the two dates via java. The month, date and year stored in the sql db are just fillers, all I need computed is the time difference of the hours and minutes between the checkInTime and the schedule.
I've tried to use datediff: SELECT e.employeeNumber, datediff('hh', startTime, checkInTime) as lateTime from employeeschedules e, timandattendance, contractualemployees c where c.scheduleNumber=e.scheduleNumber and t.employeeNumber=c.employeeNumber and t.checkInNumber=1;
But it keeps on giving out various errors: You have an error in your sql syntax check your manual that corresponds to your sql server version. for the right syntax to use near , checkInTime) as lateTime from employeeschedules e, timandattendance, contractualemployees c where c.scheduleNumber=e.scheduleNumber and t.employeeNumber=c.employeeNumber and t.checkInNumber=1;
My sql server is My SQL 5.0.5
Now I'm thinking of useing Gregorian instead, but I don't know how to use Gregorian and dont know the various methods. Can someone tell me what to do, provide some sample codes and references that I can use to accomplish what I need to do?NewtonsApple_2 wrote:
Now I'm thinking of useing Gregorian instead, but I don't know how to use Gregorian and dont know the various methods. Can someone tell me what to do, provide some sample codes and references that I can use to accomplish what I need to do?it may help...
GregorianCalendar -
How to do find the difference between two time stamps
Hi all,
i have a table with 2 columns called GMT time and Local time . so i need to find difference between these two time stamps.
i tried like this
select to_date(GMT_TIME,'yyyy-mm-dd hh24:hi:ss')-to_date(LOCAL_TIME,'yyyy-mm-dd hh24:hi:ss') from date_table
the result is coming as follows ..
0.291666666667
i did not under stand the out put.
In my table GMT_TIME= 2011-06-26 00:00:00 and LOCAL_TIME=2011-06-25 17:00:00 ..
please help me how to get exact hours between two dates ..
Thanks
SreedharHi Sreedhar,
Your output (0.2916666) has the unit day.
You should multiply with 24 to get the unit hour.
select ( to_date(GMT_TIME,'yyyy-mm-dd hh24:hi:ss')-to_date(LOCAL_TIME,'yyyy-mm-dd hh24:hi:ss') ) * 24 from date_table;Rgds,
Tycho -
How to find the difference between 2 time values in Java
hi all,
i have 2 time values
String time1="6:20";
String time2="21:30"
How to find the difference between 2 times in Java?
pls help
thanx in advance....Calculating Java dates: Take the time to learn how to create and use dates
Working in Java time: Learn the basics of calculating elapsed time in Java
Formatting a Date Using a Custom Format
Parsing a Date Using a Custom Format -
Need to find the Difference between two table
Hello ,
I have stucked in program as below scenario:-
I have two tables of huge data of same structure in a same schema.I need to find the difference exact values in tables.
By using MINUS we can find the difference between two table ,i need to find the what exact difference in the there values with colunm and value at that column.
Example TableA
Col1 col2 col3 col4 col5.... col50
10 ABC 2001 EE 444 TT
40 XYZ 3002 RR 445 TT3
80 DEF 6005 YY 446 YY8
TableB
Col1 col2 col3 col4 col5.... col50
10 ABC 2001 EE 444 TT
40 XYZ 3002 RR 445 TT3
81 DEF 6005 Yu 447 YY8
I need to the out put like this :-
The Diffence between two table is
TableA.COL1=80 TableB.Col1=81, Different
TableA.Col4=YY TableB.col4=Yu,Different
TableA.Col5=446TableB.col5=447,Different
Please suggest me to write the pl/sql program for the same
thanx in advance
KKThanx friends for all your efforts
I have a sample code for the same,this will compare the two tables for single row in each table .
what r the modification needed for the multiple rows of values in the two tables??
Please suggest!!
CREATE OR REPLACE PROCEDURE test_compare
IS
TYPE t_col
IS
TABLE OF VARCHAR2 (30)
INDEX BY PLS_INTEGER;
l_col t_col;
j NUMBER := 0;
l_sql VARCHAR2 (2000);
col1 VARCHAR2 (30);
col2 VARCHAR2 (30);
val1 NUMBER;
val2 NUMBER;
status VARCHAR2 (30);
CURSOR c1
IS
SELECT column_id, column_name
FROM all_tab_columns
WHERE table_name = 'TEST1';
BEGIN
FOR i IN c1
LOOP
j := j + 1;
l_col (j) := i.column_name;
END LOOP;
FOR k IN 1 .. j
LOOP
l_sql :=
'SELECT '
|| ''''
|| l_col (k)
|| ''''
|| ', '
|| 'TEST2.'
|| l_col (k)
|| ', '
|| ''''
|| l_col (k)
|| ''''
|| ', '
|| 'TEST1.'
|| l_col (k )
|| ', '
|| 'DECODE(TEST2.'
|| l_col (k)
|| ' -TEST1.'
|| l_col (k)
|| ', 0, ''NO CHANGE'', ''CHANGED'') FROM TEST2, TEST1';
EXECUTE IMMEDIATE l_sql INTO col1, val1,col2, val2, status;
IF status = 'CHANGED'
THEN
DBMS_OUTPUT.put_line( 'TEST2.'
|| col1
|| '='
|| val1
|| ', TEST1.'
|| col2
|| '='
|| val2
|| ', '
|| status);
END IF;
END LOOP;
EXCEPTION
WHEN OTHERS
THEN
DBMS_OUTPUT.put_line ('Error:- ' || SQLERRM);
END;
/ -
Function Module to get the difference between two times.
Hi All,
I want to know if there is any function module that gives me the difference between two specified times.
For Ex: Time 1: 12/01/2007 00:01 A.M
Time 2: Time 1 - 180 Seconds. This changes the time, also may even change the Date. (As in above situation, the Time 2 will be 11/30/2007 11:58 P.M)
Could some one please suggest if there is any function module for this case.
Thanks in Advance.Hi, You can use the following function module to calculate the diffrence between to dates in the unit that you want.
COPF_DETERMINE_DURATION calculates the difference between two dates and time in minutes and hours.
Parameters:
EXPORTING
I_START_DATE "The start date of the time interval
I_START_TIME "The start time of the time interval
I_END_DATE "The end date of the time interval
I_END_TIME "The end time of the time interval
I_UNIT_OF_DURATION "Time unit of the duration if not to be calculated in days.
Value ===== Meaning
' ' ===== Day (default)
D ===== Days
H ===== Hours
MIN ===== Minutes
MON ===== Months
S ===== Seconds
WK ===== Weeks
YR ===== Years
I_FACTORY_CALENDER "A factory calender if not using the Gregorian calender
IMPORTING
E_DURATION "Time difference in unit specified.
Message was edited by:
Rajesh Soman -
FormCalc Formula to tell the difference between two time fields if they are populated - Help please
Hi,
I have a cell (TotalTime1) that references two other time formatted cells (Start1 and Finish 1).
The below formula is intended to tell the difference between the Start and Finish time if they are populated.
I had it working without the HasValue condition but the formula was calculating when there was no value in Start1 and Finish1 (due to time format).
This is the code I do have, it says there is an error near the last line; I am sorry if I post this in the wrong manner, I have not posted on this forum before but here goes:
if (HasValue (Finish1) and HasValue (Start1)) then if (Time2Num(Start1.formattedValue, "HH:MM") < Time2Num (Finish1.formattedValue, "HH:MM")) then Abs (Time2Num(Start1.formattedValue, "HH:MM") - Time2Num(Finish1.formattedValue, "HH:MM")) / (60 * 60 * 1000) else 24 - Abs (Time2Num(Finish1.formattedValue, "HH:MM") - Time2Num(Start1.formattedValue, "HH:MM")) / (60 * 60 * 1000) endif
Any help would be appreciated.
Regards FetachiniThanks whyisthisme,
Your speedy response is greatly appreciated, your support has resolved my issue.
Many thanks.
Regards
Fetachini -
Find the difference between two dates for the specific month and year
Hi,
I have two dates, start date is 30/12/2012 and end date is 04/01/2013. Using datediff I found the difference of days between two dates. But I find the no of days in January 2013. ie output is 4 instead of 6. I input month and year to find the no of days
for that date. In this case I input Jan 2013. How can I sql this ?I don't understand how most of the answers provided here not analytically solving the problem with many cases possible.
First let me understand you:
You have 2 dates range and you want to calculate day range for specific month and year between the original date range.
declare @for_month int = 1 --January
declare @for_year int = 2013
declare @StartDate date = '2012-12-20'
declare @EndDate date = '2013-01-04'
SELECT
CASE
WHEN (DATEPART(MONTH, @StartDate) = @for_month and DATEPART(MONTH, @EndDate) = @for_month) and ((DATEPART(YEAR, @StartDate) = @for_year or DATEPART(YEAR, @EndDate) = @for_year)) THEN
DATEDIFF(DAY, @StartDate,@EndDate)
WHEN (@StartDate < cast(CONVERT(varchar(4), @for_year) + '-' + CONVERT(varchar(2), @for_month) + '-01' as date)) and (@EndDate between (cast(CONVERT(varchar(4), @for_year) + '-' + CONVERT(varchar(2), @for_month) + '-01' as date)) and (cast(DATEADD(d, -1, DATEADD(m, DATEDIFF(m, 0, cast( CONVERT(varchar(4), @for_year) + '-' + CONVERT(varchar(2), @for_month) + '-01' as date)) + 1, 0)) as date))) THEN
DATEDIFF(DAY, DATEADD(MONTH, DATEDIFF(MONTH, -1, @EndDate)-1, 0),@EndDate)
WHEN (@EndDate > cast(DATEADD(d, -1, DATEADD(m, DATEDIFF(m, 0, cast( CONVERT(varchar(4), @for_year) + '-' + CONVERT(varchar(2), @for_month) + '-01' as date)) + 1, 0)) as date)) and (@StartDate between (cast(CONVERT(varchar(4), @for_year) + '-' + CONVERT(varchar(2), @for_month) + '-01' as date)) and (cast(DATEADD(d, -1, DATEADD(m, DATEDIFF(m, 0, cast( CONVERT(varchar(4), @for_year) + '-' + CONVERT(varchar(2), @for_month) + '-01' as date)) + 1, 0)) as date))) THEN
DATEDIFF(DAY, @StartDate,DATEADD(d, -1, DATEADD(m, DATEDIFF(m, 0, @StartDate) + 1, 0))) + 1
WHEN ((DATEDIFF(DAY, @StartDate, cast(DATEADD(d, -1, DATEADD(m, DATEDIFF(m, 0, cast( CONVERT(varchar(4), @for_year) + '-' + CONVERT(varchar(2), @for_month) + '-01' as date)) + 1, 0)) as date)) >= 0) and (DATEDIFF(DAY, cast(CONVERT(varchar(4), @for_year) + '-' + CONVERT(varchar(2), @for_month) + '-01' as date), @EndDate) >= 0)) THEN
DATEDIFF(DAY, cast( CONVERT(varchar(4), @for_year) + '-' + CONVERT(varchar(2), @for_month) + '-01' as datetime), DATEADD(d, -1, DATEADD(m, DATEDIFF(m, 0, cast( CONVERT(varchar(4), @for_year) + '-' + CONVERT(varchar(2), @for_month) + '-01' as datetime)) + 1, 0))) + 1
ELSE
0
END as [DD]
I don't know how you calculate day range between 01/01/2013 and 04/01/2013
is 4, it is actually is 3 but if that is the case, you can add 1 from the condition. -
How do I calculate the difference between two times?
I am so embarrassed by the fact that I can't figure this out.
Cell B2- 8:00 am
Cell C2- 10:50 am
Cell D2- (How do I get this cell to calculate the difference and say 2:50?)
I know this is probably one of the most basic operations, but for the life of me I can't figure it out. Cells B2 & C2 are formatted for 24 hour clock. But if I tell the system to just subtract the two, I get "0.118". Everything I find on the forum search goes beyond what I need. Can anyone help me?
Thank you.KOENIG Yvan wrote:
Numbers states clearly in the Help and the PDF Users Guide that it doesn't know a "duration" object but a time one which is restricted to the range 00:00:0 to 23:59:59.
When I search the U.S. language Numbers User Guide for the word "duration," it is not found.
What may be more clear: _duration is not available but time is_?
Once again your response resemble to a rant againt the Help and the User Guide.
In the Help:
+date-time Any Numbers date/time value. _While you can choose to display only date or time in a cell, all Numbers date or time values contain both the date and time._+
Which wording would be more clear and precise?
TIMEVALUE
+The TIMEVALUE function converts a date, a time, or a text string to _a decimal fraction of a 24-hour day._+
Which wording would be more clear and precise?
TIME
+The TIME function converts hours, minutes, and seconds into a time format.+
+TIME(hours, minutes, seconds)+
+hours: The number of hours _(using a 24-hour clock)._+
+minutes: The number of minutes.+
+seconds: The number of seconds.+
Notes
+You can specify hour, minute, and second values greater than 23, 59, and 59, respectively. _If the hours, minutes, and seconds add up to more than 24 hours, Numbers subtracts 24 hours repeatedly until the sum is less than 24 hours._+
Which wording would be more clear and precise?
In the User Guide:
page 190
+date-time Any Numbers date/time value. _While you can choose to display_+
+_only date or time in a cell, all Numbers date or time values contain_+
+_both the date and time._+
+TIME (page 277) Converts a time to a decimal fraction of a 24-hour day.+
+TIMEVALUE (page 278) Converts a time in a string to a decimal fraction of a 24-hour day.+
TIME
+The TIME function converts the specified time to a decimal fraction of a 24-hour day.+
+TIME(hours, minutes, seconds)+
+• hours: The number of hours _(using a 24-hour clock)_.+
+• minutes: The number of minutes.+
+• seconds: The number of seconds.+
Notes
+You can specify hour, minute, and second values greater than 23, 59, and 59,+
+respectively. _If the hours, minutes, and seconds add up to more than 24 hours,_+
+_Numbers subtracts 24 hours repeatedly until the sum is less than 24 hours._+
+You can also specify fractional values for hours, minutes, or seconds.+
TIMEVALUE
+The TIMEVALUE function converts a time in a string to a decimal fraction of a 24-hour+
day.
TIMEVALUE(date-time)
+• date-time: A date, a time, or a string in any of the Numbers date and time formats.+
As you may check, the infos are exactly the same in the Help and in the Guide.
And I really don't understand how you may find them unclear.
Yvan KOENIG (from FRANCE lundi 4 août 2008 14:57:36) -
How to find the difference between two date?
Hi,
I currently writing a date comparision program. Below is the idea analogy,
Currently i need to find how many day differences between 30 July 2003 and 22 June 2004. How can i use java to code it?
Thanks.there doesn't seem to be a direct way but try this:int daysBetween = 0;
Calendar c = new GregorianCalendar(2004, Calendar.JULY, 30);
Calendar d = new GregorianCalendar(2003, Calendar.JUNE, 22);
while (c.get(Calendar.YEAR) != d.get(Calendar.YEAR)) {
daysBetween += 360;
d.add(Calendar.DAY_OF_YEAR, 360);
daysBetween += c.get(Calendar.DAY_OF_YEAR) - d.get(Calendar.DAY_OF_YEAR);This gives the correct result of 404 (= 8 days from June 22 to June 30 + 366 days between July 1 of 2003 and 2004 + 30 days from July 1 to July 30) -
Function Module to find the Diff bt Two Times : O/p : Hours:Minutes only
Hi All,
Wud you plz let me know the Function Module to find the Difference between two times.
Input Time1( Hours:Minutes:Seconds) Time2 ( Hours:Minutes:Seconds)
Need Output in Hours:Minutes only . ( No seconds Needed )
Thanks,
N.L.NarayanaHi
Please use FM
<b>SD_DATETIME_DIFFERENCE</b> -
Find the difference between loading and registering the drivers..
Dear Sir..
Could you please help me to find the difference between two activities..
1.Loading the drivers
2.Registering the drivers
What's the difference between loading and registeing the drivers and what activities take place by the JVM to do it all.Dear Sir..
Could you please help me to find the difference
between two activities..
1.Loading the drivers
2.Registering the drivers
What's the difference between loading and registeing
the drivers and what activities take place by the JVM
to do it all.You load a class - it isn't specific to a driver.
That is part of java - not JDBC.
Normally JDBC drivers register themselves when the class is loaded. This is specific to the driver and has nothing to do with a user of the driver. It is only a concern to someone who must implement a driver. -
Hi,
I am new to Labview. This question might be a joke to some of you here, but any help would be greatly appreciated. I have a spreadsheet with time stamps and power outputs from a generator. I am supposed to calculate the difference between consecutive time stamps, which will act as a delay for the next power output update that needs to be sent. For example, lets say that I have to following data:
Time Stamp Power Output
11:00:00 AM 3kW
11:00:02 AM 2.9kW
11:00:04 AM 3.2kW
11:00:06 AM 3.1kW
The above data doesn't make any sense, but it is just for the purpose of this question.
So, I have to read 11:00:00 AM and 3kW initially - 3kW is the initial request that is sent. Then I have toRepeated forum post
Please view http://forums.ni.com/ni/board/message?board.id=170&message.id=294435
Regards,
Juan Galindo
Applications Engineer
National Instruments -
Query to find the difference between the last date and the second to the last date
Hi all,
Hope all is well.
I am working on the following problem because I am trying to improve my MS SQL skills. But I am stuck at the moment and I wonder if you could provide some assistance please. Here is the issue:
Table 1: Dividends
divId
ExDate
RecordDate
PayDate
Amount
Yield
symId
1
2013-02-19
2013-02-21
2013-03-14
0.23
0.00000
3930
2
2012-11-13
2012-11-15
2012-12-13
0.23
0.00849
3930
3
2012-08-14
2012-08-16
2012-09-13
0.20
0.00664
3930
4
2012-05-15
2012-05-17
2012-06-14
0.20
0.00662
3930
5
2012-02-14
2012-02-16
2012-03-08
0.20
0.00661
3930
6
2011-11-15
2011-11-17
2011-12-08
0.20
0.00748
3930
7
2011-08-16
2011-08-18
2011-09-08
0.16
0.00631
3930
8
2011-05-17
2011-05-19
2011-06-09
0.16
0.00653
3930
9
2011-02-15
2011-02-17
2011-03-10
0.16
0.00594
3930
10
2010-11-16
2010-11-18
2010-12-09
0.16
0.00620
3930
11
2010-08-17
2010-08-19
2010-09-09
0.13
0.00526
3930
12
2010-05-18
2010-05-20
2010-06-10
0.13
0.00455
3930
13
2010-02-16
2010-02-18
2010-03-11
0.13
0.00459
3930
Table 2: Tickers
symId
Symbol
Name
Sector
Industry
1
A
Agilent Technologies Inc.
Technology
Scientific & Technical Instruments
2
AA
Alcoa, Inc.
Basic Materials
Aluminum
3
AACC
Asset Acceptance Capital Corp.
Financial
Credit Services
4
AADR
WCM/BNY Mellon Focused Growth ADR ETF
Financial
Exchange Traded Fund
5
AAIT
iShares MSCI AC Asia Information Tech
Financial
Exchange Traded Fund
6
AAME
Atlantic American Corp.
Financial
Life Insurance
7
AAN
Aaron's, Inc.
Services
Rental & Leasing Services
8
AAON
AAON Inc.
Industrial Goods
General Building Materials
9
AAP
Advance Auto Parts Inc.
Services
Auto Parts Stores
10
AAPL
Apple Inc.
Technology
Personal Computers
11
AAT
American Assets Trust, Inc.
Financial
REIT - Office
12
AAU
Almaden Minerals Ltd.
Basic Materials
Industrial Metals & Minerals
I am trying to check the last date (i.e. max date) and also check the penultimate date (i.e. the second to the last date). And then find the difference between the two (i.e. last date minus penultimate
date).
I would like to do that for each of the companies listed in Table 2: Tickers. I am able to do it for just one company (MSFT) using the queries below:
SELECT
[First] = MIN(ExDate),
[Last] = MAX(ExDate),
[Diff] = DATEDIFF(DAY, MIN(ExDate), MAX(ExDate))
FROM (
SELECT TOP 2 Dividends.ExDate
FROM Dividends, Tickers
WHERE Dividends.symId=Tickers.symId
AND Tickers.Symbol='MSFT'
ORDER BY ExDate DESC
) AS X
Outputs the following result:
First
Last
Diff
2012-11-13
2013-02-19
98
But what I would like instead is to be able to output something like this:
Symbol
First
Last
Diff
MSFT
2012-11-13
2013-02-19
98
AAN
2012-11-13
2012-12-14
1
X
2012-11-13
2012-12-14
1
Can anyone please let me know what do I need to add on my query in order to achieve the desired output?
Any help would be greatly appreciated.
Thanks in advance.Could you try this?
create table Ticker (SymbolId int identity primary key, Symbol varchar(4))
insert into Ticker (Symbol) values ('MSFT'), ('ORCL'), ('GOOG')
create table Dividend (DividendId int identity, SymbolId int constraint FK_Dividend foreign key references Ticker(SymbolId), ExDate datetime, Amount decimal(18,4))
insert into Dividend (SymbolId, ExDate, Amount) values
(1, '2012-10-1', 10),
(1, '2012-10-3', 1),
(1, '2012-10-7', 7),
(1, '2012-10-12', 2),
(1, '2012-10-23', 8),
(1, '2012-10-30', 5),
(2, '2012-10-1', 10),
(2, '2012-10-6', 1),
(2, '2012-10-29', 7),
(3, '2012-10-1', 22),
(3, '2012-10-3', 21),
(3, '2012-10-7', 3),
(3, '2012-10-12', 9)
WITH cte
AS (SELECT t.Symbol,
d.ExDate,
d.Amount,
ROW_NUMBER()
OVER (
partition BY Symbol
ORDER BY ExDate DESC) AS rownum
FROM Ticker AS t
INNER JOIN Dividend AS d
ON t.SymbolId = d.SymbolId),
ctedate
AS (SELECT Symbol,
[1] AS maxdate,
[2] AS penultimatedate
FROM cte
PIVOT( MIN(ExDate)
FOR RowNum IN ([1],
[2]) ) AS pvtquery),
cteamount
AS (SELECT Symbol,
[1] AS maxdateamount,
[2] AS penultimatedateamount
FROM cte
PIVOT( MIN(Amount)
FOR RowNum IN ([1],
[2]) ) AS pvtquery)
SELECT d.Symbol,
MIN(MaxDate) AS maxdate,
MIN(penultimatedate) AS penultimatedate,
DATEDIFF(d, MIN(penultimatedate), MIN(MaxDate)) AS numberofdays,
MIN(MaxDateAmount) AS maxdateamount,
MIN(penultimatedateAmount) AS penultimatedateamount,
MIN(MaxDateAmount) - MIN(penultimatedateAmount) AS delta
FROM ctedate AS d
INNER JOIN cteamount AS a
ON d.Symbol = a.symbol
GROUP BY d.Symbol
ORDER BY d.Symbol
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Thanks!
Aalam | Blog (http://aalamrangi.wordpress.com)
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