In Latex it´s possible to renew the date command in different formats and to take the language into account but I couldn´t find how you calculate with the date parts like \the\day+1 in Latex.
\leadingzero{\month}/\the\year -> \leadingzero{\month}[+4]/\the\year[-1]
Is there a way to do it?
The advdate package has built-in conditions to deal with the advancement (forward or backward) or date components. Specifically, \AdvMonth{<num>} and \AdvYear{<num>} can be used to adjust the month and year from \today.
Here's an example:
\documentclass{article}
\usepackage{advdate}
\begin{document}
\today
\AdvMonth{4}\AdvYear{-1}
\today
\end{document}
Consider the order of advancement. It's better to advance from least to most significant date component; that is, day, then month, then year, since day advancement may have to consider leap years.
One can also set \today to some fixed date using \SetDate{dd/mm/yyyy}.
Related
I see the time/date variables available to VS Code User Defined Snippets are:
CURRENT_YEAR The current year
CURRENT_YEAR_SHORT The current year's last two digits
CURRENT_MONTH The month as two digits (example '02')
CURRENT_MONTH_NAME The full name of the month (example 'July')
CURRENT_MONTH_NAME_SHORT The short name of the month (example 'Jul')
CURRENT_DATE The day of the month as two digits (example '08')
CURRENT_DAY_NAME The name of day (example 'Monday')
CURRENT_DAY_NAME_SHORT The short name of the day (example 'Mon')
CURRENT_HOUR The current hour in 24-hour clock format
CURRENT_MINUTE The current minute as two digits
CURRENT_SECOND The current second as two digits
CURRENT_SECONDS_UNIX The number of seconds since the Unix epoch
Unfortunately, without TIMEZONE or TIMEZONE OFFSET from UTC, you can't automatically create an accurate, time-aware timestamp using these variables.
Example:
${CURRENT_YEAR}-${CURRENT_MONTH}-${CURRENT_DATE}T${CURRENT_HOUR}:${CURRENT_MINUTE}:${CURRENT_SECOND}
Evaluates to: 2022-06-03T07:32:09, but thats not accurate enough. Without a TIMEZONE identifier, that "timestamp" is +/- 23:59:59 reflection of "actual time".
I need:
2022-06-03T07:32:09-07:00 and I want the timezone offset to adjust for daylight savings time to 2022-11-04T07:32:09-08:00 as appropriate.
Must be human readable. Not going to use Unix time.
How can I do this without spinning up a whole Extension?
I came here looking for the same thing. Its not a snippet, but I use the Insert Date String extension to handle this. You can define the default format as iso, and it comes with a key binding of ctrl-shift-alt-I. Works well enough, but I too would prefer a snippet.
Working with Google Sheets, I'm currently making a year schedule. The sheet should be recycled every year.
Column A has several dates.
D2 is the year, e.g. 2015, 2016 or 2017.
I'm trying to apply conditional formatting to column A. Every month to get its own colour. January coldblue, February blue, March little bit green, et cetera.
The easy way is to use date before .., like 'date before 1-2-2015' for January. But then, the conditional formatting should be altered every year.
How can I use custom formulae (in conditional formatting) using the year from D2 + before start of the next month?
I would not recommend that many different colours (if only because each colour after the first needs a separate rule!) and don't really see the point of D2 for the CF - unless you plan to have a separate set of 12 colours for each year? However, I think what you request is a plethora of formulae like:
=and(year(A1)=D$2,month(A1)=1)
to be repeated with the last 1 changed in integer steps all the way to 12, with a different colour for each such value.
I want to calculate the number of days passed between past date and a current date. My past date is in the format dd/mm/yyyy format. I have used below mentioned formulas but giving the proper output.
=DAYS360(A2,TODAY())
=MINUS(D2,TODAY())
In the above formula A2 = 4/12/2012 (dd/mm/yyyy) and I am not sure whether TODAY returns in dd/mm/yyyy format or not. I have tried using 123 button on the tool bar, but no luck.
The following seemed to work well for me:
=DATEDIF(B2, Today(), "D")
DAYS360 does not calculate what you want, i.e. the number of days passed between the two dates – see the end of this post for details.
MINUS() should work fine, just not how you tried but the other way round:
=MINUS(TODAY(),D2)
You may also use simple subtraction (-):
=TODAY()-D2
I made an updated copy of #DrCord’s sample spreadsheet to illustrate this.
Are you SURE you want DAYS360? That is a specialized function used in the
financial sector to simplify calculations for bonds. It assumes a 360 day
year, with 12 months of 30 days each. If you really want actual days, you'll
lose 6 days each year.
[source]
Since this is the top Google answer for this, and it was way easier than I expected, here is the simple answer. Just subtract date1 from date2.
If this is your spreadsheet dates
A B
1 10/11/2017 12/1/2017
=(B1)-(A1)
results in 51, which is the number of days between a past date and a current date in Google spreadsheet
As long as it is a date format Google Sheets recognizes, you can directly subtract them and it will be correct.
To do it for a current date, just use the =TODAY() function.
=TODAY()-A1
While today works great, you can't use a date directly in the formula, you should referencing a cell that contains a date.
=(12/1/2017)-(10/1/2017) results in 0.0009915716411, not 61.
I used your idea, and found the difference and then just divided by 365 days. Worked a treat.
=MINUS(F2,TODAY())/365
Then I shifted my cell properties to not display decimals.
If you are using the two formulas at the same time, it will not work...
Here is a simple spreadsheet with it working:
https://docs.google.com/spreadsheet/ccc?key=0AiOy0YDBXjt4dDJSQWg1Qlp6TEw5SzNqZENGOWgwbGc
If you are still getting problems I would need to know what type of erroneous result you are getting.
Today() returns a numeric integer value: Returns the current computer system date. The value is updated when your document recalculates. TODAY is a function without arguments.
The following worked for me. Kindly note that TODAY() must NOT be the first argument in the function otherwise it will not work.
=DATEDIF( W2, TODAY(), "d")
Today() does return value in DATE format.
Select your "Days left field" and paste this formula in the field
=DAYS360(today(),C2)
Go to Format > Number > More formats >Custom number format and select the number with no decimal numbers.
I tested, it works, at least in new version of Sheets, March 2015.
In MySql I can use YEARWEEK() to receive the week and the related year of this week in one string. (E.g. SELECT YEARWEEK('1987-01-01'); which leads to "198653").
Is there anything like that in Oracle10g?
I only know about the TO_CHAR function. But if I use TO_CHAR(sysdate, 'YYYYIW'); I receive 198753 and not 198653. So, how I am able to calculate this correctly?
Does using IYYYIW format with TO_CHAR() make any difference? Note the "I" in the beginning instead of first "Y", it is for 4-digit year based on the ISO standard.
I can't reproduce your example that Oracle returns 198753.
select TO_CHAR(DATE '1987-01-01', 'YYYYIW') from dual returns 198701 for me which is correct according to the ISO definition of week numbers.
Oracle has another format mask WW (instead of IW) that uses the week where the first day of the year is in as week #1 - which again would return week number 1 for the January 1st.
Have a look here: http://en.wikipedia.org/wiki/Week_number#Week_numbering
I find MySQL's week number a bit strange actually, because no week numbering scheme I know would return week 53 for January 1st, 1987 (but that doesn't mean very much though...)
Is there a function in Haskell that will allow me to enter component of a date (like a string representation or day month year components) that I can get information from (like day of week, days in a month, etc.)?
I've looked online and it looks like there are a lot of custom libraries, but I'm hoping there's one in the standard prelude library of ghci 10.6.4 that's just not well documented?
Are Data.Time.Calendar and Data.Time.Format in the time library sufficient?
You can parse a string representation of a date and get the length of a month using gregorianMonthLength. Not sure about day of the week, though you could format the date as a string using a format that just displays the week day.
A quick Google search turns up this, which may be what you want. It lets you parse strings representing dates and extract information from them.
You can find the day of the week with mondayStartWeek or sundayStartWeek, depending on whether you think a week starts on Monday, or on Sunday. Both functions are in Data.Time.Calendar.OrdinalDate.
λ> snd $ mondayStartWeek $ fromGregorian 2017 10 3
2
In the above example, the return value is 2, which indicates the second day of the week. Since the function is called mondayStartWeek, Monday is the first day, so 2 corresponds to Tuesday. This is true of October 3, 2017.
A warning regarding week numbers
Both functions return a tuple, where the second element is the week day. As far as I can tell, that should be trustworthy.
The first element, however, is the week number of the year. Be careful with that, because the rules for week numbering are political. If I remember correctly, in USA, week 1 is the week that contains January 1. That's not the case in Denmark, where I live. Here, week 1 is the first week where Thursday falls in the new year. This can mean that December 31 can fall in week 1 of the next year. IIRC, this is the rule for many other European countries. Some years, the American and the European week numbers align, but some years, they don't.