如何退出子Subshell

Question

基本上，我试图退出一个包含循环的子subshell。这是代码：`

stop=0
(    # subshell start
    while true           # Loop start
    do
        sleep 1           # Wait a second
        echo 1 >> /tmp/output         # Add a line to a test file

        if [ $stop = 1 ]; then exit; fi         # This should exit the subshell if $stop is 1
    done   # Loop done

) |         # Do I need this pipe?

    while true   
    do
        zenity --title="Test" --ok-label="Stop" --cancel-label="Refresh" --text-info --filename=/tmp/output --font=couriernew        # This opens Zenity and shows the file.  It returns 0 when I click stop.

      if [ "$?" = 0 ]        # If Zenity returns 0, then
      then
         let stop=1       # This should close the subshell, and
         break        # This should close this loop
      fi
    done        # This loop end
    echo Done

这不管用。从来没说过已经完成了。当我按“停止”时，它只会关闭对话框，但会一直写到文件中。

编辑:我需要能够将一个变量从子shell传递给父shell。然而，我需要继续写到文件，并保持Zenity对话框即将出现。我该怎么做？

Answer 1

当您生成子shell时，它会创建当前shell的子进程。这意味着，如果在一个shell中编辑一个变量，它将不会反映在另一个shell中，因为它们是不同的进程。我建议您将子send发送到后台，并使用$!获取它的PID，然后在准备就绪时使用该PID杀死子send。看起来是这样的：

(                               # subshell start
    while true                  # Loop start
    do
        sleep 1                 # Wait a second
        echo 1 >> /tmp/output   # Add a line to a test file
    done                        # Loop done

) &                             # Send the subshell to the background

SUBSHELL_PID=$!                 # Get the PID of the backgrounded subshell

while true   
do
  zenity --title="Test" --ok-label="Stop" --cancel-label="Refresh" --text-info --filename=/tmp/output --font=couriernew        # This opens Zenity and shows the file.  It returns 0 when I click stop.

  if [ "$?" = 0 ]               # If Zenity returns 0, then
  then
     kill $SUBSHELL_PID         # This will kill the subshell
     break                      # This should close this loop
  fi
done                            # This loop end

echo Done