I have a date type field and I need to add 7 days to it.
If I was to use mySQL I'd be looking at something like:
SELECT DATE_ADD(DRAWING_COMPLETION_DATE, INTERVAL 7 DAY);
However, I need to translate this to PostgreSQL. Can anyone help?
SELECT DRAWING_COMPLETION_DATE + INTERVAL '7' DAY
from the_table;
alternatively, because the default unit is days for date arithmetics:
SELECT DRAWING_COMPLETION_DATE + 7
from the_table;
Related
pdt.startTime is datetime
s_first.FromTimeOfDay is a time
I want to subtract the time drom the datetime. When i run the code below, Snowflake gives me this error invalid type [CAST(S_FIRST.FROMTIMEOFDAY AS TIMESTAMP_NTZ(9))] for parameter 'TO_TIMESTAMP_NTZ'
select (pdt.StartTime - (SELECT s_first.FromTimeOfDay::datetime FROM Shift s_first))
from RAW_CPMS_AAR.POWERBI_DowntimeTable AS PDT
When i try this:
select (pdt.StartTime::TIMESTAMP_NTZ(9) - (SELECT s_first.FromTimeOfDay::TIMESTAMP_NTZ(9) FROM Shift s_first))
from RAW_CPMS_AAR.POWERBI_DowntimeTable AS PDT
I get more or less the same error: invalid type [CAST(S_FIRST.FROMTIMEOFDAY AS TIMESTAMP_NTZ(9))] for parameter 'TO_TIMESTAMP_NTZ'
How do I convert the time into a datetime format so that I can subtract the two. It doesnt seem to me that there is a clear way to convert time into datetime in snowflake.
Is this what you're after?
select current_timestamp() as sample_timestamp
, time(sample_timestamp) as sample_time
, date(sample_timestamp) as sample_date;
A user pointed me in the right direction. i didnt realize i could use "dateadd" to also subtract time.
dateadd(HOUR, - (HOUR(current_timestamp())), temp.DateTime)
Select distinct Format(DateAdd(""s""," & columnname & ",""1/1/1980 12:00:00 AM""), 'dd-MMM-yyyy') as A
I have assumed that the seconds to add and the original date are hard coded values below whilst awaiting clarifications requested in the comments.
To add a number of seconds to a date you can use:
select datetime('1980-01-01 00:00:00', "345000 seconds");
This gives the result: 1980-01-04 23:50:00
The example above is just under 4 days in seconds, if you want to truncate the result to just the date as implied by the query in your questions then you can wrap this inside a date function. However, this would give the result in the format "YYYY-MM-DD" rather than "DD-MMM-YYYY" as your access query does.
Unfortunately I cannot find any native SQLite function to convert a numeric month value to mmm format. You can do this manually with replace (similar to the answer to this question), but this is a bit messy.
If you are happy to live with the numeric months then you can simply use:
select strftime('%d-%m-%Y', '1980-01-01 00:00:00', "345000 seconds");
This gives the result: 04-01-1980
More information on the SQLite date / time functions can be found here.
I am using SQLite and want to show the day of the month in a number format. I need to use the current day. I was trying to sysdate and get the day of the month from that, but that is not working.
Here is what I was trying:
SUBSTR(sysdate,1,2)
Anyone have any ideas?
Are you looking for this?
local sysdate = "28-12-2013"
local day = sysdate:sub(1,2)
print("Day from string is: "..day)
Keep Coding..................... :)
Ok, I got the first part of my question answered, so here's the second part. :-) In a PLSQL query, I have criteria that looks like this:
where Doc3.clinicalDate >= ml.convert_date_to_id(:DateBegin)
and Doc3.clinicalDate < ml.convert_date_to_id(:DateEnd)
Now, I don't want to use :DateEnd itself -- I want to add 1 day so that when it compares the datetime to midnight, I get midnight of the next day. Unfortunately, when I do
where Doc3.clinicalDate >= ml.convert_date_to_id(:DateBegin)
and Doc3.clinicalDate < ml.convert_date_to_id(:DateEnd + 1)
I get "ORA-06553: PLS-306: wrong number or types of arguments in call to 'CONVERT_DATE_TO_ID'". ":DateEnd + interval '1' day" gives me "ORA-30081: invalid data type for datetime/interval arithmetic" (where :DateEnd is bound to 31-MAY-2012). If I do "convert_date_to_id(add_months(:DateEnd, 1))", it works fine. Any thoughts? Thanks.
ETA: I should clarify that this is an SSRS 2008 R2 project, and DateBegin and DateEnd are defined in the report parameters as DateTime parameters. My current workaround involves setting the :DateEnd query parameter equal to the #DateEnd report parameter + 1, but I'm worried that someday I'll forget to document this properly and confuse the heck out of whomever's trying to maintain the report (and it might be me). I don't want to pass string parameters, as suggested before.
I'm thinking that the parameters being DateTime is root of the problem. Microsoft DateTime datatypes are way more granular than Oracle's in that it supports fractional seconds and Oracle DATE format does not (Oracle TIMESTAMP does however).
Since ADD_MONTHS just spits back whatever it's passed in DATE datatype (i.e. passed TIMESTAMP becomes DATE). So maybe you can convert the parameter and add the day that way:
where Doc3.clinicalDate >= ml.convert_date_to_id(:DateBegin)
and Doc3.clinicalDate < ml.convert_date_to_id(CAST(:DateEnd as DATE)+1)
Alternatively, forget about conversions and date arithmetic on the parameters and subtract a day from the second clinicalDate:
where Doc3.clinicalDate >= ml.convert_date_to_id(:DateBegin)
and Doc3.clinicalDate - 1 < ml.convert_date_to_id(:DateEnd)
Assuming that ml.convert_date_to_id takes a DATE as an input parameter rather than a VARCHAR2 that represents a date, and assuming that the :DateEnd bind variable is a VARCHAR2, you would need something like
ml.convert_date_to_id( to_date( :DateEnd, 'DD-MON-YYYY' ) + 1 )
or
ml.convert_date_to_id( to_date( :DateEnd, 'DD-MON-YYYY' ) + interval '1' day )
Use to_date to convert the value. For example:
select &date + 1 from dual
Informing to_date('29052012','ddmmyyyy') works fine
Informing '29-may-2012' gives ORA-01722: invalid number
Is it possible to write a SQL query that sorts data set by day of week starting from a specific day?
For example, if today is Thursday, the results are sorted from THU-FRI-SAT-...-MON-TUE-WED.
If today is Tuesday, the results would be sorted from TUE-WED-THU-...-SAT-SUN-MON.
Days are stored as integers.
Assuming you have the day of the week stored into field WD where value 1 means MON, 2 means TUE etc and SW is the "start of the week" index (again 1:MON, 2:TUE,...) then something like
CASE WHEN WD < SW THEN WD + 7 ELSE WD END
should give you a value to order by. I don't use sqlite so I'm not sure can you put it right into the ORDER BY or do you have to use it as a field and then order by that field.
in mysql: ORDER BY DATE_FORMAT(date,%w)
see: http://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html#function_date-format
Microsoft databases suport this (not shure)
... ORDER by DATENAME ( dw , table.datefield )
Check out DATEPART:
http://www.tizag.com/sqlTutorial/sqldatepart.php