I want to convert a varchar column to datetime in SQL server 2014. I've tried the following and I can get the date and minute portions fine, but can't get the hour part correctly. Is it possible to get hour part?
Here is an example:
Select
mt.DatevarcharCol,
RIGHT(mt.DatevarcharCol, LEN(mt.DatevarcharCol) - CHARINDEX(',',mt.DatevarcharCol)) AS 'Number to the right of',
DATEADD(DAY, CAST( SUBSTRING(mt.DatevarcharCol, 1, charindex(',', mt.DatevarcharCol) - 1) AS bigint),(CONVERT(datetime,'12/31/1840'))) AS 'Date part',
((RIGHT(mt.DatevarcharCol, LEN(mt.DatevarcharCol) - CHARINDEX(',',mt.DatevarcharCol))) % 86400)%3600/60 AS Minute,
?? -- calculate hour part
FROM dbo.mytable mt
Result:
DatevarcharCol Number to the right of Date part Minute hour
-------------- ---------------------- ------------------- ------ ----
63402,67524 67524 2014-08-03 00:00:00 45 ??
The hour would be:
(RIGHT(mt.DatevarcharCol, LEN(mt.DatevarcharCol)-CHARINDEX(',',mt.DatevarcharCol)))/60/60%24
The seconds would be:
(RIGHT(mt.DatevarcharCol, LEN(mt.DatevarcharCol)-CHARINDEX(',',mt.DatevarcharCol)))%60
Also, I suggest you don't store comma separated values in a single column. Instead, put each value in its own column.
Related
How to subtract two dateTime field containing dateTime in ISO format and get the result in hours?
I have tried subtracting two date fields but it has just subtracted date and not taken time into consideration
to_number(
TRUNC(to_timestamp(T1.attribute_2,'YYYY-MM-DD"T"HH24:MI:SS.ff3"Z"'))-
TRUNC(to_timestamp(T2.attribute_2,'YYYY-MM-DD"T"HH24:MI:SS.ff3"Z"'))
)
Date 1 2019-04-26 10:00pm
Date 2 2019-04-26 8:00pm
Expected Outcome: Date1- Date 2 = 2(in hrs)
Actual Outcome: Date1- Date 2 should give 0
If you want to take the hours into consideration, then don't truncate the values! TRUNC() removes the time component.
For hours, multiply the difference by 24:
(to_timestamp(T1.attribute_2,'YYYY-MM-DD"T"HH24:MI:SS.ff3"Z"')-
to_timestamp(T2.attribute_2,'YYYY-MM-DD"T"HH24:MI:SS.ff3"Z"')
) * 24
How want to use the sum of the time(n) operator so that i can calculate the overall total of the time but Sql server saying can't add the Time(n) column
i have a casted column which contain difference of two dates, and being casted as Time(n) by me. Now i want to add those column to get how much time i had used in total How much hours minute and seconds so i apply
select Sum(cast ((date1-date2) as Time(0))) from ABC_tbl
where date1 is reaching time and date2 is startingtime in Date format and i want to total of all hours
Convert the time to an integer value before you sum it (for example, seconds):
SELECT SUM(
datediff(second, '00:00:00', [TimeCol])
)
FROM
...
Replace [TimeCol] with the name of the Time(n) column. This gives you the total time in seconds, which you can then easily convert to minutes, hours, etc...
Hope this example help you.
DECLARE #A TABLE (SD TIME(0),ED TIME(0))
INSERT INTO #A VALUES
('09:01:09','17:59:09'),
('09:08:09','16:10:09'),
('08:55:05','18:00:00')
SELECT SUM(DATEDIFF(MINUTE,SD,ED)) SUM_IN_MINUTES,
SUM(DATEDIFF(HOUR,SD,ED)) SUM_IN_HOURS
FROM #A
Result:
SUM_IN_MINUTES | SUM_IN_HOURS
---------------------------------------
1505 | 25
select Sum(DATEDIFF(Minute,date1,date2)) AS TIME from ABC_tbl
u have to calculate the date difference with DATEDIFF function then use SUM function to calculate your sum of time.
you can change Minute to Second-Hour-month etc..
Try this:
DECLARE
#MidnightTime TIME = '00:00:00.0000000',
#MidnightDateTime DATETIME2 = '0001-01-01 00:00:00.0000000';
SELECT SumOfTime = DATEADD(SECOND, SUM ( DATEDIFF(SECOND, #MidnightTime, x.Col1) ), #MidnightDateTime)
FROM (VALUES
(1, CONVERT(TIME, '10:10:10.0000001')),
(2, CONVERT(TIME, '00:00:05.0000002')),
(3, CONVERT(TIME, '23:59:59.0000003'))
) x(ID, Col1)
/*
SumOfTime
---------------------------
0001-01-02 10:10:14.0000000 = 1 day (!), 10 hours, 10 minutes, 14 seconds
*/
Note: instead of SECOND you could use another precision: MINUTE, HOUR or ... NANOSECOND (see section Arguments > datepart). Using a higher precision could leads to Arithmetic overflow errors (use CONVERT(BIGINT|NUMERIC(...,0), ...).
Note #2: because the precision is SECOND the result (SumOfTime) has 0000000 nanoseconds.
Goal:
Gain datatype date with year and month
Problem:
Need to help to convert varchar value into datatype date
Unfortunatley, no day is included in the data from column. Year and month only.
I tried converting into date without using day but I failed.
Varchar(10) column data
data
-------
1997-05
1984-12
1988-11
1984-10
1984-02
1984-01
1984-04
Assuming returning the first day of the given month is ok, you can use:
SELECT CONVERT(DATE, data + '-01') FROM dbo.table;
As others have suggested, you can't have a date without a day.
If you don't care about the day, you can add an arbitrary day to the data:
CAST(data + '-01' as date)
I have a question about some of the internal workings for the Oracle DATE and INTERVAL datatypes. According to the Oracle 11.2 SQL Reference, when you subtract 2 DATE datatypes, the result will be a NUMBER datatype.
On cursory testing, this appears to be true:
CREATE TABLE test (start_date DATE);
INSERT INTO test (start_date) VALUES (date'2004-08-08');
SELECT (SYSDATE - start_date) from test;
will return a NUMBER datatype.
But now if you do:
SELECT (SYSDATE - start_date) DAY(5) TO SECOND from test;
you get an INTERVAL datatype. In other words, Oracle can convert the NUMBER from the DATE subtraction into an INTERVAL type.
So now I figured I could try putting in a NUMBER datatype directly in the brackets (instead of doing 'SYSDATE - start_date' which results in a NUMBER anyways):
SELECT (1242.12423) DAY(5) TO SECOND from test;
But this results in the error:
ORA-30083: syntax error was found in interval value expression
So my question is: what's going on here? It seems like subtracting dates should lead to a NUMBER (as demonstrated in SELECT statement #1), which CANNOT be automatically cast to INTERVAL type (as demonstrated in SELECT statement #3). But Oracle seems to be able to do that somehow if you use the DATE subtraction expression instead of putting in a raw NUMBER (SELECT statement #2).
Thanks
Ok, I don't normally answer my own questions but after a bit of tinkering, I have figured out definitively how Oracle stores the result of a DATE subtraction.
When you subtract 2 dates, the value is not a NUMBER datatype (as the Oracle 11.2 SQL Reference manual would have you believe). The internal datatype number of a DATE subtraction is 14, which is a non-documented internal datatype (NUMBER is internal datatype number 2). However, it is actually stored as 2 separate two's complement signed numbers, with the first 4 bytes used to represent the number of days and the last 4 bytes used to represent the number of seconds.
An example of a DATE subtraction resulting in a positive integer difference:
select date '2009-08-07' - date '2008-08-08' from dual;
Results in:
DATE'2009-08-07'-DATE'2008-08-08'
---------------------------------
364
select dump(date '2009-08-07' - date '2008-08-08') from dual;
DUMP(DATE'2009-08-07'-DATE'2008
-------------------------------
Typ=14 Len=8: 108,1,0,0,0,0,0,0
Recall that the result is represented as a 2 seperate two's complement signed 4 byte numbers. Since there are no decimals in this case (364 days and 0 hours exactly), the last 4 bytes are all 0s and can be ignored. For the first 4 bytes, because my CPU has a little-endian architecture, the bytes are reversed and should be read as 1,108 or 0x16c, which is decimal 364.
An example of a DATE subtraction resulting in a negative integer difference:
select date '1000-08-07' - date '2008-08-08' from dual;
Results in:
DATE'1000-08-07'-DATE'2008-08-08'
---------------------------------
-368160
select dump(date '1000-08-07' - date '2008-08-08') from dual;
DUMP(DATE'1000-08-07'-DATE'2008-08-0
------------------------------------
Typ=14 Len=8: 224,97,250,255,0,0,0,0
Again, since I am using a little-endian machine, the bytes are reversed and should be read as 255,250,97,224 which corresponds to 11111111 11111010 01100001 11011111. Now since this is in two's complement signed binary numeral encoding, we know that the number is negative because the leftmost binary digit is a 1. To convert this into a decimal number we would have to reverse the 2's complement (subtract 1 then do the one's complement) resulting in: 00000000 00000101 10011110 00100000 which equals -368160 as suspected.
An example of a DATE subtraction resulting in a decimal difference:
select to_date('08/AUG/2004 14:00:00', 'DD/MON/YYYY HH24:MI:SS'
- to_date('08/AUG/2004 8:00:00', 'DD/MON/YYYY HH24:MI:SS') from dual;
TO_DATE('08/AUG/200414:00:00','DD/MON/YYYYHH24:MI:SS')-TO_DATE('08/AUG/20048:00:
--------------------------------------------------------------------------------
.25
The difference between those 2 dates is 0.25 days or 6 hours.
select dump(to_date('08/AUG/2004 14:00:00', 'DD/MON/YYYY HH24:MI:SS')
- to_date('08/AUG/2004 8:00:00', 'DD/MON/YYYY HH24:MI:SS')) from dual;
DUMP(TO_DATE('08/AUG/200414:00:
-------------------------------
Typ=14 Len=8: 0,0,0,0,96,84,0,0
Now this time, since the difference is 0 days and 6 hours, it is expected that the first 4 bytes are 0. For the last 4 bytes, we can reverse them (because CPU is little-endian) and get 84,96 = 01010100 01100000 base 2 = 21600 in decimal. Converting 21600 seconds to hours gives you 6 hours which is the difference which we expected.
Hope this helps anyone who was wondering how a DATE subtraction is actually stored.
You get the syntax error because the date math does not return a NUMBER, but it returns an INTERVAL:
SQL> SELECT DUMP(SYSDATE - start_date) from test;
DUMP(SYSDATE-START_DATE)
--------------------------------------
Typ=14 Len=8: 188,10,0,0,223,65,1,0
You need to convert the number in your example into an INTERVAL first using the NUMTODSINTERVAL Function
For example:
SQL> SELECT (SYSDATE - start_date) DAY(5) TO SECOND from test;
(SYSDATE-START_DATE)DAY(5)TOSECOND
----------------------------------
+02748 22:50:04.000000
SQL> SELECT (SYSDATE - start_date) from test;
(SYSDATE-START_DATE)
--------------------
2748.9515
SQL> select NUMTODSINTERVAL(2748.9515, 'day') from dual;
NUMTODSINTERVAL(2748.9515,'DAY')
--------------------------------
+000002748 22:50:09.600000000
SQL>
Based on the reverse cast with the NUMTODSINTERVAL() function, it appears some rounding is lost in translation.
A few points:
Subtracting one date from another results in a number; subtracting one timestamp from another results in an interval.
Oracle converts timestamps to dates internally when performing timestamp arithmetic.
Interval constants cannot be used in either date or timestamp arithmetic.
Oracle 11gR2 SQL Reference Datetime Matrix
Use extract() function to retrieve hour / minute / seconds from interval value. See below example, how to get hours from two timestamp columns. Hope this helps!
select INS_TS, MAIL_SENT_TS, extract( hour from (INS_TS - MAIL_SENT_TS) ) hourDiff from MAIL_NTFCTN;
select TIMEDIFF (STR_TO_DATE('07:15 PM', '%h:%i %p') , STR_TO_DATE('9:58 AM', '%h:%i %p'))
Im having a problem with derived columns in SSIS. When in SSMS i can set a column to have a default value using the following code:
(dateadd(month,datediff(month,(0),getdate())-(1),(0)))
and when data is entered into the database it will give it the timestamp of the previous month in the following format for example:
2010-09-01 00:00:00.000
This strangely is what im looking for and trying to duplicate/produce similar using the Derived Column DFT.
So far i have:
DATEADD("mm",DATEDIFF("mm",GETDATE(),GETDATE()) - 1,GETDATE())
which produces the month succesfully but the GETDATE() is not a correct replacement for the 0's in the original code.
Does the 0's in the original code signify the start date in SQL?
Any help would be much appreciated.
Regards,
Lee
your first code fragment is "flooring" the datetime to the month (making the date the 1st of the given month with no time) the "-1" makes it the previous month. All the extra unnecessary parenthesis in this code fragment give me a headache, here is the equivalent:
dateadd(month,datediff(month,0,getdate())-1,0)
This is how to floor a datetime to different units:
--Floor a datetime
DECLARE #datetime datetime;
SET #datetime = '2008-09-17 12:56:53.430';
SELECT '0 None', #datetime -- none 2008-09-17 12:56:53.430
UNION SELECT '1 Second',DATEADD(second,DATEDIFF(second,'2000-01-01',#datetime),'2000-01-01') -- Second: 2008-09-17 12:56:53.000
UNION SELECT '2 Minute',DATEADD(minute,DATEDIFF(minute,0,#datetime),0) -- Minute: 2008-09-17 12:56:00.000
UNION SELECT '3 Hour', DATEADD(hour,DATEDIFF(hour,0,#datetime),0) -- Hour: 2008-09-17 12:00:00.000
UNION SELECT '4 Day', DATEADD(day,DATEDIFF(day,0,#datetime),0) -- Day: 2008-09-17 00:00:00.000
UNION SELECT '5 Month', DATEADD(month,DATEDIFF(month,0,#datetime),0) -- Month: 2008-09-01 00:00:00.000
UNION SELECT '6 Year', DATEADD(year,DATEDIFF(year,0,#datetime),0) -- Year: 2008-01-01 00:00:00.000
ORDER BY 1
PRINT' '
PRINT 'Note that when you are flooring by the second, you will often get an arithmetic overflow if you use 0. So pick a known value that is guaranteed to be lower than the datetime you are attempting to floor'
PRINT 'this always uses a date less than the given date, so there will be no arithmetic overflow'
SELECT '1 Second',DATEADD(second,DATEDIFF(second,DATEADD(day,DATEDIFF(day,0,#datetime),0)-1,#datetime),DATEADD(day,DATEDIFF(day,0,#datetime),0)-1) -- Second: 2008-09-17 12:56:53.000
the second code fragment will not properly floor the datetime to the month, it will only move the date to the previous month and use the same day and time as the given datetime.
Beyond that I'm not sure what you are really asking.
here is a breakdown of what is happening in the OP's first code fragment:
select convert(datetime,0),'first datetime "0"'
select datediff(month,0,getdate()), 'months difference between the first datetime (1900-01-01) and given "getdate()"'
select datediff(month,0,getdate())-1, 'months difference between the first datetime (1900-01-01) and month previous to given "getdate()"'
select dateadd(month,datediff(month,0,getdate())-1,0), 'takes the first datetime (1900-01-01) and adds 1328 months onto that'
OUTPUT:
----------------------- ------------------
1900-01-01 00:00:00.000 first datetime "0"
----------- --------------------------------------------------------------------------------
1329 months difference between the first datetime (1900-01-01) and given "getdate()"
----------- --------------------------------------------------------------------------------------------------
1328 months difference between the first datetime (1900-01-01) and month previous to given "getdate()"
----------------------- --------------------------------------------------------------------
2010-09-01 00:00:00.000 takes the first datetime (1900-01-01) and adds 1328 months onto that
Here's what I think you may be looking for. It is an SSIS expression that gets the first day of the previous month for a given day (GETDATE() in the example).
DATEADD("mm",DATEDIFF("mm", (DT_DBTIMESTAMP)2, GETDATE()) - 1, (DT_DBTIMESTAMP)2)
I tried to simulate your SQL version of the expression, which determined the number of months between the 0 datetime and the current datetime. And, then it added the number of months to the 0 datetime.
It's not too important what the 0 datetime is, except that you wanted the 1st day of the month. In SQL the 0 datetime is 1900-01-01 00:00:00.000, so adding months automatically gives you the first day of the month. In SSIS expressions, the 0 datetime is 1899-12-30 00:00:00.000. Since you want the first day of a month, the expression above refers to the 2 datetime. So, in the expression (DT_DBTIMESTAMP)2 casts the number 2 to 1900-01-01 00:00:00.000.
month,0 makes it round down to the beginning of the month.
Can't you use your first expression if it does what you want? Or use the second expression but switch the two GETDATE()s you've added to 0. Do you get an error if you try that?
I think what you need in your derived column is this synatx:
DATEADD("mm",-1,(DT_DBTIMESTAMP)((DT_WSTR,4)DATEPART("YYYY",GETDATE()) + "-" + (DT_WSTR,2)DATEPART("mm",GETDATE()) + "-01"))