下面是一种在单行中获取日期时间的方法:

for /f "tokens=2,3,4,5,6 usebackq delims=:/ " %a in ('%date% %time%') do echo %c-%a-%b %d%e

在美国，这将输出“yyyy-mm-dd hhmm”。不同的区域设置将导致不同的%date%输出，但是您可以修改令牌顺序。

如果需要不同的格式，可以通过重新排列令牌或使用不同(或不使用)分隔符来修改echo语句。

我知道已经提到了很多方法。但下面是我的方法来分解它，以了解它是如何做到的。希望对喜欢循序渐进法的人有所帮助。

:: Check your local date format
echo %date%

    :: Output is Mon 08/15/2016

:: get day (start index, number of characters)
::         (index starts with zero)
set myday=%DATE:~0,4%
echo %myday%
    :: output is Mon 

:: get month
set mymonth=%DATE:~4,2%
echo %mymonth%
    :: output is 08

:: get date 
set mydate=%DATE:~7,2% 
echo %mydate%
    :: output is 15

:: get year
set myyear=%DATE:~10,4%
echo %myyear%
    :: output is 2016

另一种方法(credit):

@For /F "tokens=2,3,4 delims=/ " %%A in ('Date /t') do @( 
    Set Month=%%A
    Set Day=%%B
    Set Year=%%C
)

@echo DAY = %Day%
@echo Month = %Month%
@echo Year = %Year%

请注意，我在这里的答案仍然依赖于日期和月份的顺序，由区域设置决定-不知道如何解决这个问题。

我用vMax的批处理文件更改了答案，所以它也适用于荷兰语。 荷兰语(尽管我们是持久性的)在%date%、date/t和date中有一些更改，这些更改破坏了原始的批处理文件。

如果有些人也能检查其他Windows locale，并报告结果，那就太好了。 如果批处理文件在您的位置失败，那么请在命令提示符中包含这两个语句的输出: 回音:^ |日期 日期/ t

这是你应该从批处理文件中得到的输出示例:

C:\temp>set-date-cmd.bat
Today is Year: [2011] Month: [01] Day: [03]
20110103

以下是修改后的代码及其原因:

:: https://stackoverflow.com/questions/203090/how-to-get-current-datetime-on-windows-command-line-in-a-suitable-format-for-usi
:: Works on any NT/2k machine independent of regional date settings
::
:: 20110103 - adapted by jeroen@pluimers.com for Dutch locale
:: Dutch will get jj as year from echo:^|date, so the '%%c' trick does not work as it will fill 'jj', but we want 'yy'
:: luckily, all countries seem to have year at the end: http://en.wikipedia.org/wiki/Calendar_date
::            set '%%c'=%%k
::            set 'yy'=%%k
::
:: In addition, date will display the current date before the input prompt using dashes
:: in Dutch, but using slashes in English, so there will be two occurances of the outer loop in Dutch
:: and one occurence in English.
:: This skips the first iteration:
::        if "%%a" GEQ "A"
::
:: echo:^|date
:: Huidige datum: ma 03-01-2011
:: Voer de nieuwe datum in: (dd-mm-jj)
:: The current date is: Mon 01/03/2011
:: Enter the new date: (mm-dd-yy)
::
:: date/t
:: ma 03-01-2011
:: Mon 01/03/2011
::
:: The assumption in this batch-file is that echo:^|date will return the date format
:: using either mm and dd or dd and mm in the first two valid tokens on the second line, and the year as the last token.
::
:: The outer loop will get the right tokens, the inner loop assigns the variables depending on the tokens.
:: That will resolve the order of the tokens.
::
@ECHO off
    set v_day=
    set v_month=
    set v_year=

    SETLOCAL ENABLEEXTENSIONS
    if "%date%A" LSS "A" (set toks=1-3) else (set toks=2-4)
::DEBUG echo toks=%toks%
      for /f "tokens=2-4 delims=(-)" %%a in ('echo:^|date') do (
::DEBUG echo first token=%%a
        if "%%a" GEQ "A" (
          for /f "tokens=%toks% delims=.-/ " %%i in ('date/t') do (
            set '%%a'=%%i
            set '%%b'=%%j
            set 'yy'=%%k
          )
        )
      )
      if %'yy'% LSS 100 set 'yy'=20%'yy'%
      set Today=%'yy'%-%'mm'%-%'dd'%

    ENDLOCAL & SET v_year=%'yy'%& SET v_month=%'mm'%& SET v_day=%'dd'%

    ECHO Today is Year: [%V_Year%] Month: [%V_Month%] Day: [%V_Day%]
    set datestring=%V_Year%%V_Month%%V_Day%
    echo %datestring%

    :EOF

——基珀斯

"d:\Program Files\7-Zip\7z.exe" a -r code_%date:~10,4%-%date:~4,2%-%date:~7,2%.zip

这是我使用的:

::Date Variables - replace characters that are not legal as part of filesystem file names (to produce name like "backup_04.15.08.7z")
SET DT=%date%
SET DT=%DT:/=.%
SET DT=%DT:-=.%

如果你想进一步的想法自动备份到7-Zip档案，我有一个免费/开放的项目，你可以使用或审查的想法:http://wittman.org/ziparcy/