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.
Related
i'm using autohotkey to work with date, i need to catch the day a week ago
example
if today is the 28th then I have to take the 21st of last week
calendar
in the following script I take the current date
FormatTime, date, , dd/MM/yyyy
MsgBox %date%
I even thought of a logic, to take the current day subtract by 7 that will take the day a week ago. I need help to create a better script
28 - 7 = 21
if anyone can help me thanks :)
Just subtracting numbers would be bad when you encountered a change between months.
Would have to create custom logic for that.
Luckily AutoHotkey's += operator supports date/time math.
So this is all you need:
;we're starting off the date1 variable as blank,
;which means the current time will be used.
date1 += -7, days
FormatTime, finalDate, % date1, dd/MM/yyyy ;format the result to our desired format
MsgBox, % finalDate
I did it that way
FormatTime, date_, , dd
sub += date_-7
FormatTime date, , /MM/yyyy
MsgBox,%sub%%date%
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.
I want to assign the current year in a YY format to either a macro or data set variable.
I am able to use the automatic macro variables &sysdate or &sysdate9 to get the current date. However, extracting the year in a YY format is proving to be a nightmare. Below are some examples of what I've been trying.
There exists the YEARw. format. But when I try to use it I get errors or weird results. For instance, running
data _null_;
yy = year(input("&sysdate9.", year2.));
put yy=;
run;
produces the error
ERROR 48-59: The informat YEAR was not found or could not be loaded.
If I try to format the variable in the output, I get 1965 instead of the current year. The following
data _null_;
yy = year(input("&sysdate9.", date9.));
put yy= yy year2.;
run;
outputs
yy=2016 65
Please help.
This works to get you the 2-digit year number of the current year:
DATA _NULL_;
YEAR = PUT(TODAY(),YEAR2.);
PUT YEAR;
RUN;
/* Returns: 16 */
To breakdown what I am doing here:
I use TODAY() to get the current date as a DATE type. &SASDATE needs to be converted to a DATE, but also it is the date that the SAS session started. TODAY() is the current date.
PUT allows us to pass in a non-character (numeric/date) value, which is why it is used with TODAY() as opposed to INPUT.
I think it is worth exploring the issues here in more detail.
First, Formats are patterns for converting numeric values to a human readable format. That's what you want to do here: convert a date value to a human readable format, in this case to a year.
Informats, on the other hand, convert human readable information to numeric values. That's not what you're doing here; you have a value already.
Second, put matches with Formats, and input matches with informats, exclusively.
Third, you get close in your last try: but you misuse the year format. Formats are basically value mappings, so they map every possible numeric value in their range (sometimes "all values" is the range, sometimes not) to a display value (string). You need to know what kind of value is expected on the input. YEARw. expects a date value as input, not a year value: meaning input is "number of days from 1/1/1960", mapped to "year". So you cannot take a value you've already mapped to a year value and map it again with that method; it will not make any sense.
Let's look at it:
data _null_;
yy = year(input("&sysdate9.", date9.));
put yy= yy year2.;
run;
yy contains the result of the year function - 2016. Good so far. Now, you need the 2 digit year (16); you can get that through mod function, if you like, or put/substr/input:
data _null_;
yy = input(substr(put(year(input("&sysdate9.", date9.)),4.),3,2),2.);
put yy=;
run;
mod is probably easier though since it's a number. But of course you could've used year:
data _null_;
yy = put(input("&sysdate9.", date9.),year2.);
put yy=;
run;
Now, yy is character, so you could wrap that with input(...,2.) or leave it character depending on your purposes.
Finally - a use note on &sysdate9.. You can easily make this a date without input:
"&sysdate9."d
So:
yy = put("&sysdate9."d,year2.);
That's called a date literal (and "..."dt and "..."t also work for datetime,time). They require things in the standard SAS formats to work properly.
And as pointed out in Nicarus' answer, today() is a bit better than &sysdate9 since it is guaranteed to be today. If you're running this in batch or restart your session daily, this won't matter, but it will if you have a long-running session.
Apply the year function to the date variable
Convert to string
Take last 2 digits
EDIT: change input to PUT
Year = substr(put(year(today()), 4.), 3);
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.
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.)