I am looking to continue an increment from a date field and start the day of year from this particular date field.
For example (SQL SEVER):
(DATEPART(DAYOFYEAR, CUSTOM_Date) -58) -- Decremented 58 Days to start
I am expecting then 'Jan 01' to continue numbering but instead it goes negative?
You're getting exactly what it is supposed to do. Jan 1 is the first day of a year. That means that the dayofyear for that date will be 1. If you subtract 59(or 58 in your case) from that number you'll get -57.
If you want it to keep incrementing, you'll need to have a start date and use DATEDIFF instead.
Use a variable with the true start date (I called it #startdate) and do the following:
(DATEDIFF(DAY, #startdate, CUSTOM_date) -59)
NOTE: I think in your screenshot you actually did -58, and not -59.
Related
I am trying to write a variable using the dateadd and datediff that shows the last two days of previous month. One variable will be the second to last day of the previous month, the one I am having trouble with. The other will be the last day of the previous month, the one I was able to get. I am using SQL Server.
I've tried looking for it on Stack and I have only seen the last day of the previous month given and NOT the second to last day. I tried learning the dateadd and datediff, (which I still want to do).
This is what I tried so far:
Declare #CurrentMonth as date = '3/1/2019'
Declare #SecLastDayPrevMonth as date = DATEADD(MONTH, DATEDIFF(MONTH, 0, #currentmonth), -2)
Declare #LastDayPrevMonth as date = DATEADD(MONTH, DATEDIFF(MONTH, 0, #currentmonth), -1)
The results I am getting for the seclastdayPrevMonth is 2/28/2019. Instead I would want 2/27/2019
I am also getting 2/28/2019 for lastdayprevmonth which is what I want.
I am writing variables because the current month will change every month, and instead of having to update the other days I need within my query, I want to use variables so I am only updating the current month and everything else is flowing through.
And explanation as to why my dateadd/datediff is wrong and an explanation for why the correct dateadd/datediff is the way it is, will be very helpful
Why not refer to the last day when calculating the second last day? Also, your usage of DATEADD is very weird. The syntax is DATEADD(interval, increment, datetime)
Declare #recmonth as date = '3/1/2019'
Declare #LastDayPrevMonth as date = EOMONTH(DATEADD(MONTH, -1, #RecMonth))
Declare #SecLastDayPrevMonth as date = DATEADD(DAY, -1, #LastDayPrevMonth)
SELECT #SecLastDayPrevMonth, #LastDayPrevMonth
So we can calculate the last day of the previous month by subtracting one month from a date and then calling EOMONTH, which returns the last day of a given month. Then the second last day is just subtracting one day from that.
Yields:
SecLastDayPrevMonth LastDayPrevMonth
------------------- ----------------
2019-02-27 2019-02-28
As to "why", DATEDIFF() takes 3 arguments: datepart (string representation of a specific date part), startdate, enddate (both of which must be convertible to a date-ish object).
0 is essentially SQL's epoch, which, in this case is 1/1/1900. So the difference in months between 0 and 3/1/2019 is (119*12)+2 (+2 because we exclude March, since we aren't calculating a full month) = 1430 months difference.
Then, we are trying to add 2 months to our value. DATEADD() takes 3 arguments: datepart, number, date. But, in the example, you are adding 1430 months to whatever date -2 gets converted to (in this case, I believe it would be 12/30/1899, or 2 days before epoch). So, 1430 months after 12/30/1899 would be 2/30/2019, but February only has 28 days in 2019, so it returns 2/28/2019. In a Leap Year, it probably would return 2/29/2019.
To get your #LastDayPrevMonth and #SecLastDayPrevMonth with only DATEDIFF() and DATEADD(), you just need to change your calculations a little.
First thing you want to do is find the First Day of your Given Month. That can be done with DATEADD(month,DATEDIFF(month,0,#CurrentDate),0). We're essentially using the same thing we used above to calculate the number of months since epoch, but then we are adding those months back to epoch.
Now that we know the First Day of the Given Month, all we have to do is subtract days to get a day from the prior month.
So,
DECLARE #CurrentDate date = '2019-03-15' ; -- Changed it to something in the middle of the month.
DECLARE #FirstDayOfGivenMonth = DATEADD(month,DATEDIFF(month,0,#CurrentDate),0) ; -- 3/1/2019
DECLARE #LastDayOfPrevMonth date = DATEADD(day,-1,#FirstDayOfThisMonth) ; -- 2/28/2019
DECLARE #SecLastDayOfPrevMonth date = DATEADD(day, -2, #FirstDayOfThisMonth) ; -- 2/27/2019
SELECT #LastDayOfPrevMonth AS LDPM, #SecLastDayOfPrevMonth AS SLDPM ;
DECLARE #FourDaysLeftInPrevMonth date = DATEADD(day, -4, #FirstDayOfThisMonth) ; -- 2/25/2019
SELECT #FourDaysLeftInPrevMonth AS FourDaysLeftPrev ;
Granted, since SQL 2012, this can all be accomplished much easier with the EOM() function to get to the last day of a month. But if you only could use the two original functions from your original question, this would be one way to get to your needed values.
I have 70.000 rows of data, including a date time column (YYYY-MM-DD HH24-MM-SS.).
I want to split this data into 3 separate columns; Hour, day and Week number.
The date time column name is 'REGISTRATIONDATE' from the table 'CONTRACTS'.
This is what I have so far for the day and hour columns:
SELECT substr(REGISTRATIONDATE, 0, 10) AS "Date",
substr(REGISTRATIONDATE, 11, 9) AS "Hour"
FROM CONTRACTS;
I have seen the options to get a week number for specific dates, this assignment concerns 70.000 dates so this is not an option.
You (the OP) still have to explain what week number to assign to the first few days in a year, until the first Monday of the year. Do you assign a week number for the prior calendar year? In a Comment I asked about January 1, 2017, as an example; that was a Sunday. The week from January 2 to January 8 of 2017 is "week 1" according to your definition; what week number do you assign to Sunday, January 1, 2017?
The straightforward calculation below assigns to it week number 0. Other than that, the computation is trivial.
Notes: To find the Monday of the week for any given date dt, we can use trunc(dt, 'iw'). iw stands for ISO Week, standard week which starts on Monday and ends on Sunday.
Then: To find the first Monday of the year, we can start with the date January 7 and ask for the Monday of the week in which January 7 falls. (I won't explain that one - it's easy logic and it has nothing to do with programming.)
To input a fixed date, the best way is with the date literal syntax: date '2017-01-07' for January 7. Please check the Oracle documentation for "date literals" if you are not familiar with it.
So: to find the week number for any date dt, compute
1 + ( trunc(dt, 'iw') - trunc(date '2017-01-07', 'iw') ) / 7
This formula finds the Monday of the ISO Week of dt and subtracts the first Monday of the year - using Oracle date arithmetic, where the difference between two dates is the number of days between them. So to find the number of weeks we divide by 7; and to have the first Monday be assigned the number 1, instead of 0, we need to add 1 to the result of dividing by 7.
The other issue you will have to address is to convert your strings into dates. The best solution would be to fix the data model itself (change the data type of the column so that it is DATE instead of VARCHAR2); then all the bits of data you need could be extracted more easily, you would make sure you don't have dates like '2017-02-29 12:30:00' in your data (currently, if you do, you will have a very hard time making any date calculations work), queries will be a lot faster, etc. Anyway, that's an entirely different issue so I'll leave it out of this discussion.
Assuming your REGISTRATIONDATE if formatted as 'MM/DD/YYYY'
the simples (and the faster ) query is based ond to to_char(to_date(REGISTRATIONDATE,'MM/DD/YYYY'),'WW')
(otherwise convert you column in a proper date and perform the conversio to week number)
SELECT substr(REGISTRATIONDATE, 0, 10) AS "Date",
substr(REGISTRATIONDATE, 11, 9) AS "Hour",
to_char(to_date(REGISTRATIONDATE,'MM/DD/YYYY'),'WW') as "Week"
FROM CONTRACTS;
This is messy, but it looks like it works:
to_char(
to_date(RegistrationDate,'YYYY-MM-DD HH24-MI-SS') +
to_number(to_char(trunc(to_date(RegistrationDate,'YYYY-MM-DD HH24-MI-SS'),'YEAR'),'D'))
- 2,
'WW')
On the outside you have the solution previous given by others but using the correct date format. In the middle there is an adjustment of a certain number of days to adjust for where the 1st Jan falls. The trunc part gets the first of Jan from the date, the 'D' gets the weekday of 1st Jan. Since 1 represents Sunday, we have to use -2 to get what we need.
EDIT: I may delete this answer later, but it looks to me that the one from #mathguy is the best. See also the comments on that answer for how to extend to a general solution.
But first you need to:
Decide what to do dates in Jan before the first Monday, and
Resolve the underlying problems in the date which prevent it being converted to dates.
On point 1, if assigning week 0 is not acceptable (you want week 52/53) it gets a bit more complicated, but we'll still be able to help.
As I see it, on point 2, either there is something systematically wrong (perhaps they are timestamps and include fractions of a second) or there are isolated cases of invalid data.
Either the length, or the format, or the specific values don't compute. The error message you got suggests that at least some of the data is "too long", and the code in my comment should help you locate that.
I know this one is pretty easy but I've always had a nightmare when it comes to comparing dates in SQL please can someone help me out with this, thanks.
I need to get the month and year of now then compare it to a date stored in a DB.
Time Format in the DB:
2015-08-17 11:10:14.000
I need to compare the month and year with now and if its > 12 months old I will increment a count. I just need the number of rows where this argument is true.
I assume you have a datetime field.
You can use the DATEDIFF function, which takes the kind of "crossed boundaries", the start date and the end date.
Your boundary is the month because you are only interested in year and month, not days, so you can use the month macro.
Your start time is the value stored in the table's row.
Your end time is now. You can get system time selecting SYSDATETIME function.
So, assuming your table is called mtable and the datetime object is stored in its date field, you simply have to query:
SELECT COUNT(*) FROM mtable where DATEDIFF(month, mtable.date, (SELECT SYSDATETIME())) > 12
I have a table described here: http://sqlfiddle.com/#!3/f8852/3
The date_time field for when the time is 00:00 is wrong. For example:
5/24/2013 00:00
This should really be:
5/23/2013 24:00
So hour 00:00 corresponds to the last hour of the previous day (I didn't create this table but have to work with it). Is there way quick way when I do a select I can replace all dates with 00:00 as the time with 24:00 the previous day? I can do it easily in python in a for loop but not quite sure how to structure it in sql. Appreciate the help.
All datetimes are instants in time, not spans of a finite length, and they can exist in only one day. The instant that represents Midnight is by definition, in the next day, the day in which it is the start of the day, i.e., a day is closed on its beginning and open at its end, or, to phrase it again, valid allowable time values within a single calendar date vary from 00:00:00.00000, to 23:59:59.9999.
This would be analogous to asking that the minute value within an hour be allowed to vary from 1 to 60, instead of from 0 to 59, and that the value of 60 was the last minute of the previous hour.
What you are talking about is only a display issue. Even if you could enter a date as 1 Jan 2013 24:00, (24:00:00 is not a legal time of day) it would be entered as a datetime at the start of the date 2 Jan, not at the end of 1 Jan.
One thing that illustrates this, is to notice that, because of rounding (SQL can only resolve datetimes to within about 300 milleseconds), if you create a datetime that is only a few milleseconds before midnight, it will round up to midnight and move to the next day, as can be seen by running the following in enterprise manager...
Select cast ('1 Jan 2013 23:59:59.999' as datetime)
SQL server stoers all datetimes as two integers, one that represents the number days since 1 Jan 1900, and the other the number of ticks (1 tick is 1/300th of a second, about 3.33 ms), since midnight. If it has been zero time interval since Midnight, it is stll the same day, not the previous day.
If you have been inserting data assuming that midnight 00:00:00 means the end of the day, you need to fix that.
If you need to correct your existing data, you need to add one day to every date in your database that has midnight as it's time component, (i.e., has a zero time component).
Update tbale set
date_time = dateAdd(day, 1, date_time)
Where date_time = dateadd(day, datediff(day, 0, date_time), 0)
I want to find first day month of month and also like 3rd day or 5th day ,15th day or any day of the month .So how to find through query.I know how to find first day and last day of month.Mainy I want find other days.
For those of you following along who may not know how to get the First Day of the month in SQL Server you can do so with something like this. This will also give you the 5th, 10th or whatever you need.
DECLARE #FirstDay DATETIME
SET #FirstDay = (DATEADD(MONTH, DATEDIFF(MONTH, -1, GETDATE()) - 1, -1) + 1)
SELECT GETDATE() AS CurrentDay
, #FirstDay AS FirstDay
, DATEADD(d, 10, #FirstDay-1) AS TenthDay
The -1 after the #FirstDay in the DateAdd is because the DateAdd will add the numbers of days onto the firstday, which will give you the 11th in that example. Of course you could just add one less day to make it work without the -1 but I prefer including it. Suit yourself.
If you know how to find the first day of a month, you can add the 2-day, the 4-day or the 14-day interval to the first day of the month to get, respectively, the 3rd, the 5th or the 15th day of the month.
Similarly you can get any day of the month by simply adding the proper number of days.
Different RDBMSs may offer different syntax to achieve the goal. Assuming #MonthBeginning to be a date or datetime value representing the first day of a month, here's how you can get, for example, the 5th day of the same month in Microsoft SQL Server:
SELECT DATEADD(day, 4, #MonthBeginning) AS FifthDay
Again, it may not be the way you should do that if your RDBMS is not MS SQL Server.