Question

如何从此cmd代码中减去1天：

%date:~10,4%%date:~4,2%%date:~7,2%

我在这里看到的每个帖子都说我需要使用电源外壳等等。我只想在这里删除1天。

cmd返回：

20180227

我想：

20180226

Answer 1

这是两个代码选项。我在两者中都使用函数来获取当前的日期和时间。如果您不需要，可以将其删除。

这是两个选项中最快的。它将日期转换为Julian日期以进行数学运算，然后将其转换回正常日期。

@echo off
setlocal

REM Get the current date
REM Will return variables YY, YYYY, MM, DD, HH, Min and Sec
Call :GetDateTime

REM Add or Subtract from the current date
REM Must use + or - symbol
REM Revised date will be assigned to RetVar
REM LeapDay
Call :AddSubDate 2016 03 01 -1 LeapDay
REM Yesterday
Call :AddSubDate %YYYY% %MM% %DD% -2 past
REM Tomorrow
Call :AddSubDate %YYYY% %MM% %DD% +2 future

echo LeapDay  : %LeapDay%
echo Past     : %past%
echo Today    : %YYYY%%MM%%DD%
echo Future   : %future%
pause
GOTO :EOF

:AddSubDate Year Month Day <+/-Days> RetVar
setlocal & set a=%4
set "yy=%~1"&set "mm=%~2"&set "dd=%~3"
set /a "yy=10000%yy% %%10000,mm=100%mm% %% 100,dd=100%dd% %% 100"
if %yy% LSS 100 set /a yy+=2000 &rem Adds 2000 to two digit years
set /a JD=dd-32075+1461*(yy+4800+(mm-14)/12)/4+367*(mm-2-(mm-14)/12*12)/12-3*((yy+4900+(mm-14)/12)/100)/4
if %a:~0,1% equ + (set /a JD=%JD%+%a:~1%) else set /a JD=%JD%-%a:~1%
set /a L= %JD%+68569,     N= 4*L/146097, L= L-(146097*N+3)/4, I= 4000*(L+1)/1461001
set /a L= L-1461*I/4+31, J= 80*L/2447,  K= L-2447*J/80,      L= J/11
set /a J= J+2-12*L,      I= 100*(N-49)+I+L
set /a YYYY= I, MM=100+J, DD=100+K
set MM=%MM:~-2% & set DD=%DD:~-2%
set ret=%YYYY: =%%MM: =%%DD: =%
endlocal & set %~5=%ret%
exit /b

:GetDateTime Year Month Day Hour Minute Second
@echo off & setlocal
for /f "tokens=2 delims==" %%a in ('wmic OS Get localdatetime /value') do set "dt=%%a"
set "YY=%dt:~2,2%" & set "YYYY=%dt:~0,4%" & set "MM=%dt:~4,2%" & set "DD=%dt:~6,2%"
set "HH=%dt:~8,2%" & set "Min=%dt:~10,2%" & set "Sec=%dt:~12,2%"
( ENDLOCAL
   set "YY=%YY%" 
   set "YYYY=%YYYY%" 
   set "MM=%MM%" 
   set "DD=%DD%"
   set "HH=%HH%" 
   set "Min=%Min%"
   set "Sec=%Sec%"
)
exit /b

第二个选项使用带有xcopy命令的技巧来检查日期是否为有效日期。所以从技术上来说，没有做任何日期数学，就像之前的解决方案一样，转换到Julian日期来做那个日期数学。此选项仅进行减法。第一个选项可以加或减。

@echo off
setlocal

REM set the number of days to substract
SET DAYS=180

REM Call function to check if the date is valid.
CALL :validdate "%days%" subdate
echo Old date: %subdate%

pause

endlocal
GOTO :EOF

:validdate
setlocal
set "day=%~1"
set rand=%random%
md "%temp%\dummy%rand%\empty%rand%"

REM Get todays date
for /f "tokens=2 delims==" %%a in ('wmic OS Get localdatetime /value') do set "dt=%%a"

REM set year month and day into its own variables.
set /a y=%dt:~0,4%
set /a m=1%dt:~4,2%
set /a d=1%dt:~6,2%

:loop
if "%day%"=="0" (
    rd /s /q "%temp%\dummy%rand%"
    endlocal &set "%~2=%y%%m:~-2%%d:~-2%"
    GOTO :EOF
)
set /a d-=1

if %d% lss 101 (
    set d=131
    set /a m-=1

    if %m% lss 101 (
        set m=112
        set /a y-=1
    )
)

xcopy /d:%m:~-2%-%d:~-2%-%y% /t "%temp%\dummy%rand%\empty%rand%" "%temp%\dummy%rand%" >nul 2>&1 && (set /a day-=1 & goto loop) || goto loop

GOTO :EOF

如何从批处理文件中减去 - 天？

1 个答案: