我有一个类似这样的 bash 脚本：
# one.sh, two.sh and three.sh do not depend on anything,
# and can run simultaneously in any order,
# and the execution time for each is random
# four.sh can only run once three.sh has finished
one.sh &
ONE=$!
two.sh &
TWO=$!
three.sh &
THREE=$!
wait $THREE
four.sh
其中 one.sh、two.sh 和 Three.sh 看起来像这样：
echo -n "doing stuff..."
#some command or set of commands
if [ $? ]
then
echo $RESULT_PASS
else
echo $RESULT_FAIL
fi
我得到的输出是这样的：
doing stuff1...done
doing stuff2...
doing stuff3...
doing stuff4...done
7263
doing stuff5...done
doing stuff6...9823
doing stuff7...done
9283
因为一个脚本中的某些任务在另一脚本中的任务开始之前尚未完成。
我正在寻找的输出是这样的：
doing stuff1...done
doing stuff2...done
doing stuff3...8373
doing stuff4...done
doing stuff5...1234
doing stuff6...fail
doing stuff7...done
我将不胜感激一些指导。
Thanks!
我会用GNU 并行并行运行脚本。它非常简单、直观且灵活。因此，要并行运行三个脚本并保持输出顺序并等待所有三个脚本完成：
parallel -k ::: ./one.sh ./two.sh ./three.sh
样本输出
doing stuff1...pass
doing stuff2...pass
doing stuff3...pass
搜索示例堆栈溢出通过输入[gnu-parallel]在搜索框中 - 包含方括号。您可以并行运行任意数量的脚本、标记输出行、跨多个进程循环大文件、远程登录到其他计算机以将进程分布在网络上、如果一个脚本失败或 10% 失败则停止所有其他脚本，或者如果有任何通过的话...
或者你可以问GNU 并行用脚本名称标记每一行：
parallel -k --tag ::: ./one.sh ./two.sh ./three.sh
./one.sh doing stuff1...pass
./two.sh doing stuff2...pass
./three.sh doing stuff3...pass
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