I'm trying to insert date field in MS Word that will display tomorrow's date we one opens the document.
I can insert Today's date {DATE \# "dd/MM/yyyy"}, can we insert tomorrow's date using modified formula?
Thanks
Without VBA, the calculation is possible, but not straightforward, because the Word field language has very limited support for date-related operations.
Originally I thought Word would auto-update the DATE when you open or close/re-open the document, but further experiments suggest that even the second suggestion in here will not do that.
In the specific case described (add 1 day), you should be able to use the following field coding:
{QUOTE {SET xxx { DATE }}{SET yyy {xxx \#YYYY}}{SET mmm {xxx \#M}}{SET xxx1 {={xxx \#YYYYMMDD}+1 \#0000'-'00'-'00}}{=13-{xxx1 \#M} \#"'{xxx1}';'{=mmm-11 \#'{=yyy+1}-01-01';'{yyy}-{=mmm+1 \#00'-01'}'"}'"} \#DD/MM/YYYY}
All the {} must be the special field code brace pairs that you can insert on Windows Word using ctrl-F9, and (typically) on Mac Word using cmd-F9 or fn-cmd-F9, depending on your keyboard setup. You can change the format at the end ("\#DD/MM/YYYY" ) as required.
However, that set of field codes probably will not be updated automatically by Word when you open the document, so the user would need to select the field codes and press F9.
I originally thought Word would update the date on open and/or close/re-open using the following coding, but I now believe I was wrong. The one thing it does achieve on recent versions of Windows Word is to present the Date field in a "bubble" with an option to update the field:
{DATE \#"'{QUOTE {SET xxx { DATE }}{SET yyy {xxx \#YYYY}}{SET mmm {xxx \#M}}{SET xxx1 {={xxx \#YYYYMMDD}+1 \#0000'-'00'-'00}}{=13-{xxx1 \#M} \#"'{xxx1}';'{=mmm-11 \#'{=yyy+1}-01-01';'{yyy}-{=mmm+1 \#00'-01'}'"}'"} \#DD/MM/YYYY}'"}
Here is some pseudocode for the algorithm:
Set xxx to the date.
Set yyy to the 4-digit year
Set mmm to the month
Set xxx1 to the date but with the day number incremented by 1. e.g., for 2016-12-31, that would be a string, "2106-12-32"
This is the tricky bit:
Try to extract the month from that date using { xxx1 \#M }. If the date is valid, { xxx1 \#M } will return a valid month number, i.e. in the range 1 to 12. If the date is not valid, { xxx1 \#M } will return xxx1, e.g. "2106-12-32", which the { = } field will treat as a calculation, i.e. "year-(a maximum of 12+32=44)", so it is always going to return a number larger than 12.
If xxx1 is a valid date then
result=xxx1
Else 'xxx1 is not a valid date so...
If mmm (the original month) is 12 then
result = "(yyy+1)-01-01"
Else
result = "yyy-(mmm+1)-01"
End If
End If
Apply the date format you want to "result".
NB, this also relies on the assumption that Word always correctly recognises the month and day when you specify a date in the format "YYYY-MM-DD", regardless of the locale, in other words that "2016-04-01" is always recognised as 01 April 2016, never as 04 January 2016. If anyone can provide a counter-example, then the assumption is wrong, the field coding will need to change, and will probably need to be locale-dependent.
Related
I'm looking for a way to set an arbitrary date, and every time I press a key it will print the day after it (tomorrow).
global jDate = "June 1, 1986"
^+z::
;Output our date in LongDate format
FormatTime, TimeString, %jDate%, LongDate
SendInput, %TimeString%
;Increment the date by a single day
jDate += 1, Days
Return
Unfortunately, it the code keeps starting jDate as today's current date/time rather than the past date I specify in the initial variable assignment. Not sure why. The incrementing works fine, it just increments starting from todays date rather that the 1986 date.
FormatTime is expecting any date/time input to be in the "YYYYMMDD..." format. Since what you've assigned to jDate doesn't fit that criterion, it assumes it's invalid and uses today's date. To make it work how you expect, just modify your jDate input.
jDate := "19860601" ; 1986 -> YYYY, 06 -> MM, 01 ->DD
A couple of things to note: (1) global is not needed in this context; (2) I would recommend getting out of the habit of assigning variables using the = comparator (use := assignment operator instead). It only works for legacy reasons but generates more confusion than it's worth. In the context that you're using it, the quotes would need to be removed.
What is the command for displaying the time and date in qbasic? Could the syntax for the commands be given as well? And an explanation if possible?
You can use DATE$ and TIME$
These can also set the date and time as well.
The command for printing the time(current system time) is time$
The time$ is actually a function, in this case, no parameter is needed.
And the code is...
PRINT TIME$
The time is printed in hh: mm: ss format(hour: minutes: seconds).
And therefore the output would be something like this:
14:55:28
For printing the current system date, we use date$ function which is also a string function
The code is:
PRINT DATE$
The date is printed in mm-dd-yyyy format or month-day-year(American date format).
Hence the output will be:
02-17-2018
Hope it helps...
The QB date/time functions are:
DATE$ returns the date in a string in the form MM-DD-YYYY
TIME$ returns the time in a string in the form HH:MM:SS
When used as a command the date$ and time$ can be assigned to set the system date and time, for example DATE$ = "12-10-1990" or TIME$ = "12:10:10"
If the year is a leap year then the 29th day of February could be set. Otherwise if it is not a leap year then a syntax error will occur trying to set the date in February to the 29th.
In Pig Latin, is there a built-in function to find the Month End date for a given date ? For example, if the given date is '2015-03-15', the month end date returned should be '2015-03-31' and if given date is '2015-04-15', the month end date should be '2015-04-30'.
This is how you do it:
REGISTER /usr/lib/pig/piggybank.jar;
DEFINE ISOToMonth org.apache.pig.piggybank.evaluation.datetime.truncate.ISOToMonth();
%declare END_OF_MONTH SubtractDuration(AddDuration(ToDate(ISOToMonth('2015-03-15')),'P1M'),'P1D')
A = LOAD 'DummyFileWithOneRow.txt' USING PigStorage(',') AS (f1:chararray, f2:chararray);
result = FOREACH A GENERATE
f1 AS f1,
$END_OF_MONTH AS end_of_month;
DUMP result
The result of this run is:
(1,2015-03-31T00:00:00.000Z).
You can now convert this result to your desired format.
You can do this calculation as part of the foreach on the loaded values.
The ordinary way to do such things, if you do not find that the language in question already has a built-in set of functions to "do such things," is to ... in this case:
Determine the first day of the current month. ("Month/01/Year" This is the only step that you "do by hand.")
Add "one month" to that. (There should be some kind of "DateAdd()" function in your language...)
Finally, using the same function, "subtract one day."
December 15th => December 1st => January 1st (of next year) => December 31st (of this year).
But first, look carefully. "Accountants want to do this sort of thing all the time." There is usually a pretty-good, sometimes very-good, set of functions to do date-manipulation. (And if they're not built-in to the language, there's often a contributed library of "goodies" that someone else already wrote and perfected.)
I'm trying to use joda-time to parse a date string of the form YYYY-MM-DD. I have test code like this:
DateTimeFormatter dateDecoder = DateTimeFormat.forPattern("YYYY-MM-DD");
DateTime dateTime = dateDecoder.parseDateTime("2005-07-30");
System.out.println(dateTime);
Which outputs:
2005-01-30T00:00:00.000Z
As you can see, the DateTime object produced is 30 Jan 2005, instead of 30 July 2005.
Appreciate any help. I just assumed this would work because it's one of the date formats listed here.
The confusion is with what the ISO format actually is. YYYY-MM-DD is not the ISO format, the actual resulting date is.
So 2005-07-30 is in ISO-8601 format, and the spec uses YYYY-MM-DD to describe the format. There is no connection between the use of YYYY-MM-DD as a pattern in the spec and any piece of code. The only constraint the spec places is that the result consists of a 4 digit year folowed by a dash followed by a 2 digit month followed by a dash followed by a two digit day-of-month.
As such, the spec could have used $year4-$month2-$day2, which would equally well define the output format.
You will need to search and replace any input pattern to convert "Y" to "y" and "D" to "d".
I've also added some enhanced documentation of formatting.
You're answer is in the docs: http://www.joda.org/joda-time/apidocs/org/joda/time/format/DateTimeFormat.html
The string format should be something like: "yyyy-MM-dd".
The date format described in the w3 document and JodaTime's DateTimeFormat are different.
More specifically, in DateTimeFormat, the pattern DD is for Day in year, so the value for DD of 30 is the 30th day in the year, ie. January 30th. As the formatter is reading your date String, it sets the month to 07. When it reads the day of year, it will overwrite that with 01 for January.
You need to use the pattern strings expected by DateTimeFormat, not the ones expected by the w3 dat and time formats. In this case, that would be
DateTimeFormatter dateDecoder = DateTimeFormat.forPattern("yyyy-MM-dd");
I have found a statement that compares two dates to see if one is greater than another:
If Format(Date, "m/d/yyyy") > #1/1/2000# Then MsgBox "Okay"
That's basically saying if today is later than January 1st, 2000 then pop up a message box.
This is fine if I am putting the date in the code (i.e., 1/1/2000), but I would like to get a date from the user during run-time and can't figure out how to incorporate that into the statement above.
If I provide a TextBox for the user to enter a date, I have surrounded the text with the # symbol:
If MyDate > "#" & Text1.Text & "#" then
That does not work. I've tried converting the text to Double, Integer, and Long. I have no idea how to use the enclosing # symbols with user input.
Thanks in advance for your help.
Use CDate(Text1.Text) to convert a string to a date. In reality you will want to do some checking to make sure that the value entered is a valid date, something along the lines of:
Dim myDate As Date
If IsDate(Text1.Text) Then
MyDate = CDate(Text1.Text)
Else
'Indicate the error to the user
End If
The # characters are "date quotes", if you will - they serve the same purpose for date literals as " does to delimit a string literal. You can't just concatenate # to a value to make it into a date.