在Windows (Windows XP)批处理脚本中，我需要格式化当前日期和时间，以便以后在文件名等中使用。
这类似于堆栈溢出问题如何在批处理文件中添加日期，但也包含时间。
到目前为止，我有这个:
echo %DATE%
echo %TIME%
set datetimef=%date:~-4%_%date:~3,2%_%date:~0,2%__%time:~0,2%_%time:~3,2%_%time:~6,2%
echo %datetimef%
这使:
28/07/2009
8:35:31.01
2009_07_28__ 8_36_01
是否有一种方法可以允许%TIME%中的个位数小时，以便我可以得到以下结果?
2009_07_28__08_36_01
当前回答
set hourstr = %time:~0,2%
if "%time:~0,1%"==" " (set hourstr=0%time:~1,1%)
set datetimestr=%date:~0,4%%date:~5,2%%date:~8,2%-%hourstr%%time:~3,2%%time:~6,2%
其他回答
set hourstr = %time:~0,2%
if "%time:~0,1%"==" " (set hourstr=0%time:~1,1%)
set datetimestr=%date:~0,4%%date:~5,2%%date:~8,2%-%hourstr%%time:~3,2%%time:~6,2%
使用REG保存/修改/恢复对您的bat文件最有用的值。这是windows 7，对于其他版本，你可能需要一个不同的键名。
reg save "HKEY_CURRENT_USER\Control Panel\International" _tmp.reg /y
reg add "HKEY_CURRENT_USER\Control Panel\International" /v sShortDate /d "yyyy-MM-dd" /f
set file=%DATE%-%TIME: =0%
reg restore "HKEY_CURRENT_USER\Control Panel\International" _tmp.reg
set file=%file::=-%
set file=%file:.=-%
set file
::========================================================================
::== CREATE UNIQUE DATETIME STRING IN FORMAT YYYYMMDD-HHMMSS
::======= ================================================================
FOR /f %%a IN ('WMIC OS GET LocalDateTime ^| FIND "."') DO SET DTS=%%a
SET DATETIME=%DTS:~0,8%-%DTS:~8,6%
第一行输出的格式与时区无关: 20150515150941.077000 + 120 这样，您只需按照自己的愿望格式化输出即可。
@ECHO OFF
: Sets the proper date and time stamp with 24Hr Time for log file naming
: convention ('YYYYMMDD_HHMMSS')
: Scrapes the characters out of their expected permissions in the date/time
: environment variables.
: Expects a date format of '____MM_DD_YYYY'
: Expects a time format of 'HH:MM:SS' or ' H:MM:SS'
SET HOUR=%time:~0,2%
SET dtStamp9=%date:~-4%%date:~4,2%%date:~7,2%_0%time:~1,1%%time:~3,2%%time:~6,2%
SET dtStamp24=%date:~-4%%date:~4,2%%date:~7,2%_%time:~0,2%%time:~3,2%%time:~6,2%
if "%HOUR:~0,1%" == " " (SET dtStamp=%dtStamp9%) else (SET dtStamp=%dtStamp24%)
ECHO %dtStamp%
PAUSE
当我需要一个日期/时间字符串时，我通常这样做:
set dt=%DATE:~6,4%_%DATE:~3,2%_%DATE:~0,2%__%TIME:~0,2%_%TIME:~3,2%_%TIME:~6,2%
set dt=%dt: =0%
这是为德国日期/时间格式(dd.mm。yyyy hh: mm: ss)。基本上，我将子字符串连接起来，最后将所有空格替换为零。
结果字符串的格式为:yyyy_mm_dd__hh_mm_ss
子字符串如何工作的简短解释:
%VARIABLE:~num_chars_to_skip,num_chars_to_keep%
因此，要想从“29.03.2018”这样的日期中得到年份，可以使用:
%DATE:~6,4%
^-----skip 6 characters
^---keep 4 characters
