China DOS Union

　　People often ask how to calculate the date N days ago. Because this involves the issues of big months and small months, common years and leap years, the code becomes relatively complicated and tends to scare people off.

　　Although willsort posted code like this two years ago (please refer to: (Ended) Deleting files older than N days from the DOS command line under WINDOWS), and although the idea is extremely simple—convert the date into a relative number, do the subtraction, then convert the relative number back into a date—the numbers 2472632 and 146097 and the like appear too abruptly, making it hard to figure out what's going on. To be honest, limited by my own level, even now I still find it very confusing.

　　Now I'm posting three batch files for getting the date N days ago ( the date may include the weekday, but it must be arranged in year-month-day order, and the year must be four digits, otherwise there will be errors). The key parts all have comments, so they're a bit easier to understand than willsort's:

　　Code 1： @echo off :: Get the date N days ago :: Idea: :: 　　Simulate manual calculation: subtract the number of days from the date number; :: 　　when the difference is negative, borrow 1 from the higher place value. :: Features: :: 　　The code is concise and easy to understand; speed depends on the size of the specified number of days setlocal enabledelayedexpansion :Main cls :: The date may include the weekday, but it must be arranged in year-month-day order :: The year must be four digits, otherwise there will be errors set day=%date% set days=0 echo.&echo. echo The specified date is: %day% echo. set /p input= Please specify how many days to trace back: :: Extract the date for /f "tokens=1-3 delims=-/. " %%i in ("%day%") do ( set /a sy=%%i, sm=100%%j %% 100, sd=100%%k %% 100 ) set /a sd-=input if %sd% leq 0 call :count cls echo.&echo. echo The specified date is: %day% echo. set sm=0%sm% set sd=0%sd% echo The date %input% days ago is: %sy%-%sm:~-2%-%sd:~-2% pause>nul goto Main :count set /a sm-=1 if !sm! equ 0 set /a sm=12, sy-=1 call :days set /a sd+=days if %sd% leq 0 goto count goto :eof :days :: Get the total number of days in the specified month set /a leap="^!(sy %% 4) & ^!(^!(sy %% 100)) | ^!(sy %% 400)" set /a max=28+leap set num=0 set str=31 %max% 31 30 31 30 31 31 30 31 30 31 for %%i in (%str%) do ( set /a num+=1 if %sm% equ !num! set days=%%i&goto :eof ) goto :eof
　　Code 2： @echo off :: Get the date N days ago :: Idea: :: 　　Convert the current date into a day count :: 　　Subtract N days from the current day count :: 　　Then convert the result back into a date :: Specific operation: :: 　　Calculate the day count year by year :: Features: :: The code is concise and easy to understand; relatively slow, but not affected by the size of the specified number of days setlocal enabledelayedexpansion :Main cls :: The date may include the weekday, but it must be arranged in year-month-day order :: The year must be four digits, otherwise there will be errors set day=%date% set days=0 echo.&echo. echo The specified date is: %day% echo. set /p input= Please specify how many days to trace back: rem ========Convert the specified date into a day count======== :: Extract the date for /f "tokens=1-3 delims=-/. " %%i in ("%day%") do ( set /a sy=%%i, sm=100%%j %% 100, sd=100%%k %% 100 ) :: Convert the year into a day count for /l %%i in (1,1,%sy%) do ( set /a leap="^!(%%i %% 4) & ^!(^!(%%i %% 100)) | ^!(%%i %% 400)" set /a days=days+365+leap ) :: Convert the month into a day count set /a num=0, mday=0, max=28+leap set str=0 31 %max% 31 30 31 30 31 31 30 31 30 for %%i in (%str%) do ( set /a num+=1 if !num! leq !sm! set /a mday+=%%i ) set /a days=days+mday+sd set /a days-=input rem ========Convert the day count into a date======== :: Get the year for /l %%i in (1,1,%sy%) do ( set /a leap="^!(%%i %% 4) & ^!(^!(%%i %% 100)) | ^!(%%i %% 400)" set /a days_tmp=365+leap if !days! gtr !days_tmp! ( set /a days-=days_tmp set y=%%i ) ) :: Get the month and date set /a m=1, max=28+%leap% set str=31 %max% 31 30 31 30 31 31 30 31 30 for %%i in (%str%) do ( if !days! gtr %%i ( set /a days-=%%i set /a m+=1 ) else goto next ) :next set m=0%m% set d=0%days% cls echo.&echo. echo The specified date is: %day% echo. echo The date %input% days ago is: %y%-%m:~-2%-%d:~-2% pause>nul goto Main
　　Code 3： @echo off :: Get the date N days ago :: Idea: :: 　　Convert the current date into a day count :: 　　Subtract N days from the current day count :: 　　Then convert the result back into a date :: Specific operation: : 　　First calculate using 365 days per year, then calculate the number of leap years :: 　　Finally do the conversion :: Features: :: 　　The code is complicated and not easy to understand, but very fast :: There is still a bug in the calculation; you can use 11111 to compare and test against the first two. I don't have time to update it for now setlocal enabledelayedexpansion :Main cls :: The date may include the weekday, but it must be arranged in year-month-day order :: The year must be four digits, otherwise there will be errors set day=2007-01-02 set /a days=0, sum=0 echo.&echo. echo The specified date is: %day% echo. set /p input= Please specify how many days to trace back: rem =======Convert the date into a day count======= :: Extract the date for /f "tokens=1-3 delims=/-. " %%i in ("%day%") do ( set /a sy=%%i, sm=100%%j %% 100, sd=100%%k %% 100, y=%%i ) :: Convert the year into a day count set num=0 if %sy% lss 400 goto next1 call :leap_num :next1 set /a leap_num=sy/4-num :: Convert the month into a day count set /a leap="^!(sy %% 4) & ^!(^!(sy %% 100)) | ^!(sy %% 400)" set /a max=28+leap set num=0 set str=31 %max% 31 30 31 30 31 31 30 31 30 31 for %%i in (%str%) do ( set /a num+=1 if !num! lss %sm% set /a sum+=%%i ) set /a days=365*sy+leap_num+sum+sd set /a days-=input rem =======Convert the day count into a date======= :: Get the year set /a y=days/365 set /a mod1=days %% 365 call :leap_num set /a mod2=mod1-leap_num :loop2 if %mod2% lss 0 ( set /a y-=1 set /a mod2=365+mod2 ) else goto next2 goto loop2 :: Get the month and date :next2 set /a m=1, days=mod2+1, max=28+%leap% set str=31 %max% 31 30 31 30 31 31 30 31 30 for %%i in (%str%) do ( if !days! gtr %%i ( set /a days-=%%i set /a m+=1 ) else goto next3 ) :next3 set m=0%m% set d=0%days% cls echo.&echo. echo The specified date is: %day% echo. echo The date %input% days ago is: %y%-%m:~-2%-%d:~-2% pause>nul goto Main :leap_num :: Count the years divisible by 100 but not by 400 set /a y_tmp=(%y:~0,1%%y%-%y%)/10 set num=0 :loop1 set /a mod=y_tmp %% 400 if %mod% neq 0 set /a num+=1 set /a y_tmp-=100 if %y_tmp% geq 100 goto loop1 set /a leap_num=y/4-num goto :eof

[ Last edited by namejm on 2007-6-8 at 11:15 PM ]

Using ruby is much simpler. I just want to pursue "using the shortest code, in the fastest time, with the most suitable tool, to solve it"—nothing else intended~

puts Data.today-3

[ Last edited by vkill on 2007-6-7 at 10:28 PM ]

Learning
Posting a batch file that uses VBS to calculate the date
@echo off rem Delete the folder from three days ago rem The target directory is C:\test set Target=C:\test dir /b %Target%>%temp%\FolderList.txt echo dt=date()-3>%temp%\OldDate.vbs echo s=right(year(dt),4) ^& right("0" ^& month(dt),2) ^& right("0" ^& day(dt),2)>>%temp%\OldDate.vbs echo wscript.echo s>>%temp%\OldDate.vbs for /f %%f in ('cscript /nologo %temp%\OldDate.vbs') do @set DelFlag=%%f for /f %%i in (%temp%\FolderList.txt) do ( if %%i lss %DelFlag% ( rd /s /q %Target%\%%i ) ) del %temp%\FolderList.txt del %temp%\OldDate.vbs

I want to ask: in the OP's third method, is set dat=2007-01-02 the initial time? Last night I came up with a method, but I don't know whether it's the same as the OP's. Please clarify. Also, I don't quite understand the OP's code (method three).

Below is my method:

Assume the initial time coordinate is 1900-01-01, and the input time is date1 (any time). This makes it easier to calculate the difference between the two dates (in days), tmp1. Then also input any number of days tmp2 that you want to query before or after date1. If it is N days before date1, then calculate the result as sum=tmp1+tmp2. (tmp2 can be positive or negative: positive for dates after date1, negative for dates before date1.) Then in the end you get a time difference sum measured from the initial time. Based on 1900-01-01 as the start/end time, calculating the date at a difference of sum days should be relatively simple.

This line of thought can solve the date N days before or after any given date. The start/end time can be set according to how large the calculated difference is, so as to reduce the amount of calculation.

Because my knowledge is limited, I can only provide the idea. I hope the experts can give some guidance. I don't know whether my idea is the same as the third method. To be honest, I still haven't completely understood the third method...

::Determine the day before the current date
::First directly subtract 1 day from the day count
::If the result is day 0, then subtract 1 from the month, and the day count is the last day of that month.
::If the result is month 0, then subtract 1 from the year, and the month count is the last month of that year.
@echo off&setlocal enabledelayedexpansion
set yyyy=%date:~0,4%
set mm=%date:~5,2%
set dd=%date:~8,2%
set /a nd=!dd!-1
::If it's like subtracting one day from March 1 and getting day 0, then...
if !nd!==0 call :dd0
if !mm!==0 call :mm0
set yyyymmdd=!yyyy!-!mm!-!nd!
echo Today is:%date:~0,10%
echo Yesterday was:!yyyymmdd!
pause

:dd0
set /a mm=!mm!-1
for %%a in (1 3 5 7 8 10 12)do set %%add=31
set /a pddd=!yyyy!*10/4
set pd2d=!pddd:~-1,1!
set 2dd=28
if !pd2d!==0 set 2dd=29
for %%b in (4 6 9 11)do set %%bdd=30
set nd=!%mm%dd!
goto :eof
:mm0
set /a yyyy=!yyyy!-1
set mm=12
set nd=31
goto :eof