Informix is not my normal environment and the way it handles datetime values is throwing me for a loop. I can't imagine this is difficult, but for the life of me I'm not yet able to figure it out.
This is the SQL:
SELECT agentid,
extension As Ext,
resourcefirstname As FirstNm,
resourcelastname As LastNm,
Min(eventdatetime) As FirstIn
FROM agentstatedetail AS asdr Join
resource As r On asdr.agentid = r.resourceid
WHERE asdr.eventdatetime BETWEEN '2016-10-20 04:00:00' AND '2016-10-21 03:59:59'
AND eventtype = 3
AND assignedteamid = 14
Group By agentid, extension, resourcefirstname, resourcelastname
Order By Min(eventdatetime)
Everything works as is, but the dates in the Between clause are currently entered manually- not optimal. I just need some way to describe "yesterday at 4:00 AM" and "Today at 4:00 AM" Will somebody please clue me in?
Using Informix version 12.10.FC6DE, I can do this:
SELECT
TODAY::DATETIME YEAR TO SECOND AS today_zerohour
, TODAY::DATETIME YEAR TO SECOND - '20:00:00'::INTERVAL HOUR TO SECOND AS yesterday_dawn
, TODAY::DATETIME YEAR TO SECOND + '04:00:00'::INTERVAL HOUR TO SECOND AS today_dawn
FROM
systables
WHERE
tabid = 1;
And it returns:
today_zerohour yesterday_dawn today_dawn
2016-10-21 00:00:00 2016-10-20 04:00:00 2016-10-21 04:00:00
So, what is happening here:
The operator TODAY returns the system date as a DATE type. The DATE type does not have the precision I want (it only has year, month and day), so I cast the value (cast operator is ::) to a DATETIME with precision from year to second (the hour, minutes and seconds are set to zero):
TODAY::DATETIME YEAR TO SECOND
In Informix, for an addition or subtraction with a DATETIME value to return another DATETIME value, I need to add or subtract an INTERVAL value. So I created 2 INTERVAL values.
One INTERVAL of 20 hours to subtract from the today value (again the cast operator :: is used, this time to cast from a string to an INTERVAL):
'20:00:00'::INTERVAL HOUR TO SECOND
One INTERVAL of 4 hours to add to the today value:
'04:00:00'::INTERVAL HOUR TO SECOND
Related
When I use the formula below the results of the EOMONTH function
returns the start of the next month for any month with 30 days instead of the last day of the specified month. The month and years are correct, so I'm pretty sure it's EOMONTH when used in another function.
For example,the results in B3 should be "11/31/1965" but it returns "12/1/1965".
=DATE(YEAR(B2),MONTH(B2)+6,DAY(TEXT(EOMONTH(MONTH(B2)+6,0))))
I have tried subtracting a day, but it returns the end-of-month -1 for months with 31 days (30). So I have the same problem in the other case.
I have also used IFS() to account for months with 30 days, and it miscalculates the date the same way.
=IFS( MONTH(B2)+6 = 4,DATE(YEAR(B2),MONTH(B2)+6,DAY(EOMONTH(MONTH(B2)+6,0))-2) ,
MONTH(B2)+6 =
6,DATE(YEAR(B2),MONTH(B2)+6,DAY(EOMONTH(MONTH(B2)+6,0))-2) ,
MONTH(B2)+6 =
9,DATE(YEAR(B2),MONTH(B2)+6,DAY(EOMONTH(MONTH(B2)+6,0))-2) ,
MONTH(B2)+6 =
11,DATE(YEAR(B2),MONTH(B2)+6,DAY(EOMONTH(MONTH(B2)+6,0))-2) ,
TRUE ,DATE(YEAR(B2),MONTH(B2)+6,DAY(EOMONTH(MONTH(B2)+6,0)) ) )
The EOMONTH function by itself where I just pass in the date as a string works correctly (column F).
Any Idea on what I'm doing wrong?
Thanks in advance.
Yes, as #player0 has explained, you can't just add something to a month and feed it into eomonth. Try putting
=eomonth(month(B2)+6,0)
into B3 (formatted as a date).
You get
1/31/1900
Why? month(b2)+6 gives 11 (which is just a number). Dates in google sheets are represented as days since 12/31/1899. So 11 formatted as a date gives 1/11/1900. Applying eomonth to that gives the last day of January 1900, which is the 31st. Feeding that into your formula would give 11/31/65, but that date doesn't exist, so you get 12/1/65.
If you want to go forward 6 months and then get the last day of the month, you need
=eomonth(date(year(B2),month(B2)+6,1),0)
You can also use the Edate function, which does not roll over into the first day of the next month:
=eomonth(edate(B2,6),0)
EOMONTH does not understand MONTH. instead, it converts it into date. to use EOMONTH you need to supply it with valid date
=EOMONTH(B2, 0)
There's the subtract() method but the documentation says it's not aware of daylight savings which makes it pretty much useless in this case, or in any other case except where the programmer doesn't know how many milliseconds there are in 24 hours.
I'm thinking of two ways:
get the day of the month, and then subtract N from it and if it's less than 1 then subtract the month and the year if appropriate and set the day for the last day of whichever month it turns out to be
OR
subtract N days from the noon of the current day and then get start of the day for the resulting day
Is there some easier/better way to do this?
You should probably try to convert both DateTimes to UTC (standardize), then call difference(). That converts it to a nice, easy Duration, which you can convert as necessary to hours, days, months, or whatever else.
DateTime one = somedatetime.toUtc();
DateTime two = someotherdatetime.toUtc();
Duration diff = one.difference(two);
//Then just convert...
return diff.inDays;
I want to get first day of previous month in Dart as DateTime.
Below code is working correctly even for x = 1 (passing 0 as month)
print(new DateTime(2016,x-1,1));
but is it by design or I should not relay on it ?
It is by design.
The DateTime constructor you are using allows overflow and underflow on both days and months. A month value less than one means a month prior to January, which is a month in a previous year. Likewise a day value of less than one is a day in a previous month.
I want to store date in my Postgres database.
The only problem is that this date can have optional day or even month.
Example:
User provides time period when he was employed - not necessary full date (day + month + year), but only start year and end year.
However there are users, who worked only from may to october in the same year so month have to be provided too.
How to handle this kind of optional date parts?
Use a proper date type anyway. Do not store text or multiple columns. That would be more expensive and less reliable.
Use the function to_date(), which is fit to deal with your requirements out of the box. For instance, if you call it with a pattern 'YYYYMMDD' and the actual string is missing characters for day, or month and day, it defaults to the first month / day of the year / month:
db=# SELECT to_date('2001', 'YYYYMMDD');
to_date
------------
2001-01-01
db=# SELECT to_date('200103', 'YYYYMMDD');
to_date
------------
2001-03-01
You could store a precision flag indicating year / month / day in addition if you need that.
While the accepted answer is a good one, there is another alternative.
ISO 8601
The ISO 8601 standard defines sensible formats for textual representations of various kinds of date-time values.
A year is represented in the obvious manner, a four-digit number: 2014
A year-month is represented with a required hyphen: 2014-01Note that in other ISO 8601 formats, the hyphen is optional. But not for year month, to avoid ambiguity.
A full date is similar: 2014-08-21 or without optional hyphens: 20140821. I recommend keeping the hyphens.
So you could store the values as text. The length of text would tell you whether it is year-only, year-month, or date.
I'm trying to execute a query in rails 3 with the following syntax: CourseOrder.where("DATE(ordered_at) = ?", date).
My problem is that rails saves all times in UTC, so in my Timezone (+2) at 0:33 its the 4th but in UTC its the 3th of the month. Is there a way to query the date part of a DateTime with Timezone?
Don't use date, do it like this instead:
time = DateTime.civil_from_format(:local, 2011, 4, 9)
CourseOrder.where("ordered_at > ? AND ordered_at < ?", time - 1.day, time)
You're storing a datetime, so query that instead of trying to use date, which doesn't account for Timezones.
solved the problem with a change in which values I compare.
I have a Date value to, and now I'm comparing a date value with a date value which is better than comparing date value with datetime value which have to fail.
thanks for leading me to the right direction