【发布时间】:2020-12-21 11:32:46
【问题描述】:
我创建了这个脚本,灵感来自this answer,关于如何等待并行批处理脚本完成并创建了这个。
@echo off
set "lock=%temp%\wait%random%.lock"
start "1" /D .\RESULTS\1 cmd /c 9>"%lock%0" "1.exe"
start "2" /D .\RESULTS\2 cmd /c 9>"%lock%1" "2.exe"
start "3" /D .\RESULTS\3 cmd /c 9>"%lock%2" "3.exe"
start "4" /D .\RESULTS\4 cmd /c 9>"%lock%3" "4.exe"
:Wait0 for all scripts to finish (wait until lock files are no longer locked)
1>nul 2>nul ping /n 2 ::1
for %%N in (0 1 3) do (
setlocal enabledelayedexpansion
set L=!!lock!%%N!
( rem
) 9>!L! || goto :Wait0
) 2>nul
timeout /t 10 /nobreak
echo 1 done
echo 2 done
echo 3 done
echo 4 done
start "5" /D .\RESULTS\5 cmd /c 9>"%lock%4" "5.exe"
start "6" /D .\RESULTS\6 cmd /c 9>"%lock%5" "6.exe"
start "7" /D .\RESULTS\7 cmd /c 9>"%lock%6" "7.exe"
start "8" /D .\RESULTS\8 cmd /c 9>"%lock%7" "8.exe"
:Wait1 for all scripts to finish (wait until lock files are no longer locked)
1>nul 2>nul ping /n 2 ::1
for %%N in (4 1 7) do (
setlocal enabledelayedexpansion
set L=!!lock!%%N!
( rem
) 9>!L! || goto :Wait1
) 2>nul
timeout /t 10 /nobreak
echo 5 done
echo 6 done
echo 7 done
echo 8 done
start "9" /D .\RESULTS\9 cmd /c 9>"%lock%8" "9.exe"
start "10" /D .\RESULTS\10 cmd /c 9>"%lock%9" "10.exe"
:Wait2 for all scripts to finish (wait until lock files are no longer locked)
1>nul 2>nul ping /n 2 ::1
for %%N in (8 1 9) do (
setlocal enabledelayedexpansion
set L=!!lock!%%N!
( rem
) 9>!L! || goto :Wait2
) 2>nul
timeout /t 10 /nobreak
echo 9 done
echo 10 done
::delete the lock files
del "%lock%*"
一开始,1.exe2.exe3.exe4.exe 似乎一切正常,循环正在等待,而所有这些都完成工作,然后才继续。问题从5.exe6.exe7.exe8.exe 的第二个循环开始。然后在 2 或 3 个程序完成后,它的工作循环退出并且不等待最后一个程序完成它的工作。 Any1 可以看到此脚本中的错误,只是这可能不是最有效的编写方式?
【问题讨论】:
标签: windows batch-file