I am trying to create a 'Due Date' column where it takes the date from the 'DATE_OF' column and adds the dynamic variable, "DAYS".
SELECT DATEADD('DAY',TABLE1.DAYS,TABLE1.DATE_OF) AS DUE_DATE
FROM TABLE1
This is the format of the date from 'DATE_OF'
04/16/14 12:00 AM
06/24/14 12:00 AM
04/01/14 12:00 AM
DAYS uses only integers (Values between 10 and 90).
But I keep getting this for every result
12/30/99 12:00 AM
What am I doing wrong?
UPDATE
I referenced mssqltips.com and found that the abbreviation for 'DAY' is 'd'. When I changed this, it worked. I do not know why.
As you've come to figure out, there are dissimilarities between many other RDMS' SQL/functions and Access SQL/functions.
Unfortunately, 'DAY' is not a proper interval when using the DateAdd function in MS Access. It is 'd'.
For further note: If you were going to hardcode in a date, the # on either side of the date are required. More details can be found here
Access does not accept not accept the value 'DAY' in the ADDDATE() function.
A complete list of accepted values can be found here
Here are the details of the accepted values.
Value Explanation
yyyy Year
q Quarter
m Month
y Day of the year
d Day
w Weekday
ww Week
h Hour
n Minute
s Second
Related
I have a large data file containing a string type column 'YearMonthWeek'
It contains values such as '20160101' for the first week of January 2016, or '20161040' for the 40th week of the year 2016 apparently falling in October.
Now, I want to convert these strings to actual dates so that every YearMonthWeek value is converted to, say the first day of that week. (Whether that ends up being Monday or Sunday I don't really care).
I tried the following query:
PARSE_TIMESTAMP('%Y%m%W', CAST(YearMonthWeek AS STRING)) AS datefield
(See this documentation for details)
This runs without errors, but returns me the first day of the month for every single entry...
So for example '20160101' and '20160102' both get parsed as 2016-01-01 00:00:00 UTC.
Is this an issue with the PARSE_TIMESTAMP function, or am I missing something?
Try doing something like
DATE_ADD(date_expression, INTERVAL %W WEEK)
Static example:
SELECT
DATE_ADD(
DATE(PARSE_TIMESTAMP('%Y', SUBSTR(CAST('20161252' AS STRING),0,4))),
INTERVAL (CAST(SUBSTR(CAST('20160102' AS STRING),7) AS INT64)) week)
AS datefield
-
Row datefield
1 2016-01-15
You may add something as a margin to it, according to ISO 8601, the first week of the year is the one that contains January 4th. So you could have something like: 4 + 7*($week - 1)
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 am writing a query to convert a character to Date Time
The following query extracts my time stamps in Character format.
select
(to_char(TO_CHAR(MDY(month(current- 1 units month), 1,year(current- 1 units month)),'%Y-%m-%d')||' 13:00:00')),
(to_char(TO_CHAR((DATE(DATE(extend(TODAY, YEAR TO MONTH)) - 1 UNITS DAY)+1),'%d-%m-%Y')||' 13:00:00'))
from dual
Output:
`T 0÷
2015-08-01 13:00:00 01-09-2015 13:00:00
2015-08-01 13:00:00 01-09-2015 13:00:00
Now I am trying to convert the Character to Time stamp using DATETIME(2001-12-31 15:32:55) YEAR TO SECOND function. I am getting syntax error.
select
DATETIME(to_char(TO_CHAR(MDY(month(current- 1 units month), 1,year(current- 1 units month)),'%Y-%m-%d')||' 13:00:00')) YEAR TO SECOND ,
DATETIME(to_char(TO_CHAR((DATE(DATE(extend(TODAY, YEAR TO MONTH)) - 1 UNITS DAY)+1),'%d-%m-%Y')||' 13:00:00') ) YEAR TO SECOND
from dual
How ever the following is working fine:
select DATETIME(2001-12-31 15:32:55) YEAR TO SECOND
from dual
Thanks in Advance. Please do not suggest answers for Oracle. its damn easy in Oracle.
Try using a CAST to convert your output as a DATETIME YEAR TO SECOND:
select
(to_char(TO_CHAR(MDY(month(current- 1 units month), 1,year(current- 1 units month)),'%Y-%m-%d')||' 13:00:00'))::DATETIME YEAR TO SECOND ,
(to_char(TO_CHAR((DATE(DATE(extend(TODAY, YEAR TO MONTH)) - 1 UNITS DAY)+1),'%Y-%m-%d')||' 13:00:00'))::DATETIME YEAR TO SECOND
from dual
That seems to work OK, but I'd suggest you don't do this. Also, note that DATETIME is an ISO standard: YYYY-MM-DD HH:MM:SS.FFF (or part thereof). Your second example with the date in English format is not going to parse to a DATETIME.
Your first algorithm is not leap-day safe, and the second example is horribly over-complicated. You can determine 13:00 on the first day of the current month using the far simpler construction:
(TODAY-DAY(TODAY)+1)::DATETIME YEAR TO SECOND + INTERVAL(13) HOUR TO HOUR
This also has the benefit of avoiding casting back and forth between DATE and CHAR. The calculation of the same time the previous month can be written as:
MDY(MONTH(TODAY-DAY(TODAY)),1,YEAR(TODAY-DAY(TODAY)))::DATETIME YEAR TO SECOND + INTERVAL(13) HOUR TO HOUR
... which will do the right thing on Feb 29/March 1 in a leap year, which your algorithm won't.
The construction TODAY-DAY(TODAY) will always produce the last day of the prior month.
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'm pretty new to SQL so I hope this isn't a dumb question, tried to google but couldn't find anything.
I'm summing sales of departments per week in SQL and am using TD_SYSFNLIB.WEEKNUMBER_OF_YEAR (trans_dt) to get the week number.
I think everything is working except I'd like to change the format of the weeks to the start date of the week, e.g. week 1 = 1/4/15
Also, i'm not sure how to handle the very first of the year week 0 since I think that should be grouped up with week 52 of last year.
The following date math trick should get you Beginning of Week as an actual date without having to join to the SYS_CALENDAR view or using a function:
SELECT CURRENT_DATE - ((CURRENT_DATE - DATE '0001-01-07) MOD 7) AS BOW;
Starting with TD14 there's NEXT_DAY which returns the following weekday, if you subtract 7 days you get the previous day:
next_day(trans_dt - 7, 'sunday')