[英]batch - how can I close a batch file that was started from another batch file
I have a problem understanding how I can automatically close a batch file that was started from another batch file. 我在理解如何自动关闭从另一个批处理文件启动的批处理文件时遇到问题。
My first batch file looks like this: 我的第一个批处理文件如下所示:
@echo off
:logs
start C:\Users\Kenturrac\Desktop\copy.bat
timeout 2
goto logs
So as you can see there, I start another batch file called "copy.bat" which looks like that: 因此,如您所见,我启动了另一个名为“ copy.bat”的批处理文件,如下所示:
::set arma path
@set ARMAPATH=C:\Users\Administrator.WIN-B8I65OIU3DJ\Desktop\server\dayz_epoch_1
::set log path
@set LOGPATH=%ARMAPATH%\_LOGS
@set MOVEORCOPY=copy
:: Berechne Zeitstempel
@set DAY=%DATE:~0,2%
@set MONTH=%DATE:~3,2%
@set YEAR=%DATE:~6,4%
@set DATE=%YEAR%_%MONTH%_%DAY%
@set HH=%TIME:~0,2%
@IF "%TIME:~0,1%"==" " (
set HH=0%TIME:~1,1%
)
@set MIN=%TIME:~3,2%
@set SEC=%TIME:~6,2%
@set TIME=%HH%%MIN%
@IF NOT EXIST "%LOGPATH%\%DATE%_%TIME%\" (
@set TEMPPATH=%DATE%_-_%TIME%
) ELSE (
@set TEMPPATH=%DATE%_-_%TIME%_%SEC%
)
@set TARGETPATH=%LOGPATH%\%TEMPPATH%\
@echo %TARGETPATH%
Exit /b
The problem is that it doesn't close the additional cmd window it started for the "copy.bat". 问题在于它不会关闭为“ copy.bat”启动的其他cmd窗口。 If I use "call" instead of "start", it will mess up the "temppath" variable.
如果我使用“ call”而不是“ start”,它将弄乱“ temppath”变量。
假设第二个批处理文件除了将环境变量设置为将在新进程中启动该批处理文件时将被丢弃的值之外,还要做其他事情,并且如果唯一不使用call
而不是start
是保护变量的值在第一个过程中,在第二个批处理中使用setlocal / endlocal
指令,以避免文件关闭和变量值受干扰的问题。
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