你的脚本应该是这样的:

prog1 &
prog2 &
.
.
progn &
wait
progn+1 &
progn+2 &
.
.

假设你的系统一次可以处理n个任务。使用wait一次只运行n个作业。

并行运行多个程序:

prog1 &
prog2 &

如果你需要脚本等待程序完成，你可以添加:

wait

在您希望脚本等待它们的地方。

#!/bin/bash
prog1 & 2> .errorprog1.log; prog2 & 2> .errorprog2.log

将错误重定向到单独的日志。

你可以试试ppss(废弃)。PPSS非常强大——你甚至可以创建一个迷你集群。 如果您有一批令人尴尬的并行处理要做，xargs -P也很有用。

下面是我为了并行运行最多n个进程而使用的函数(示例中n=4):

max_children=4

function parallel {
  local time1=$(date +"%H:%M:%S")
  local time2=""

  # for the sake of the example, I'm using $2 as a description, you may be interested in other description
  echo "starting $2 ($time1)..."
  "$@" && time2=$(date +"%H:%M:%S") && echo "finishing $2 ($time1 -- $time2)..." &

  local my_pid=$$
  local children=$(ps -eo ppid | grep -w $my_pid | wc -w)
  children=$((children-1))
  if [[ $children -ge $max_children ]]; then
    wait -n
  fi
}

parallel sleep 5
parallel sleep 6
parallel sleep 7
parallel sleep 8
parallel sleep 9
wait

如果max_children被设置为核数，该函数将尝试避免空闲核。

由于某些原因，我不能使用等待，我想出了这个解决方案:

# create a hashmap of the tasks name -> its command
declare -A tasks=(
  ["Sleep 3 seconds"]="sleep 3"
  ["Check network"]="ping imdb.com"
  ["List dir"]="ls -la"
)

# execute each task in the background, redirecting their output to a custom file descriptor
fd=10
for task in "${!tasks[@]}"; do
    script="${tasks[${task}]}"
    eval "exec $fd< <(${script} 2>&1 || (echo $task failed with exit code \${?}! && touch tasks_failed))"
    ((fd+=1))
done

# print the outputs of the tasks and wait for them to finish
fd=10
for task in "${!tasks[@]}"; do
    cat <&$fd
    ((fd+=1))
done

# determine the exit status
#   by checking whether the file "tasks_failed" has been created
if [ -e tasks_failed ]; then
    echo "Task(s) failed!"
    exit 1
else
    echo "All tasks finished without an error!"
    exit 0
fi