Bash 对话框取消按钮正在返回，退出代码为 0；转义键也是如此

Question

我创建了一个简单的菜单对话框，但取消按钮和退出键返回 0 作为退出代码。我在这里做错了什么？

exec 3>&1;
selection=$(dialog \
    --title "Main Menu" \
    --backtitle "$BACK_TITLE" \
    --menu "Choose an option to configure. Up and down arrows change selection. Use the Enter key to make a selection." "$HEIGHT" "$WIDTH" 5 \
    0 "Select Directory" \
    1 "Select File Extension" \
    2 "Options" \
    3 "Execute with current config" \
    4 "Exit" \
    2>&1 1>&3);
exec 3>&-;
exitStatus=$?
echo "selected: $selection exit code: $exitStatus"

当我按下退出键或取消按钮时会产生上述情况。

selected: exit code: 0

Answer 1

您正在保存上一个exec 命令的退出状态，而不是dialog。将exitStatus=$? 紧跟在selection=... 之后。

exec 3>&1;
selection=$(dialog \
    --title "Main Menu" \
    --backtitle "$BACK_TITLE" \
    --menu "Choose an option to configure. Up and down arrows change selection. Use the Enter key to make a selection." "$HEIGHT" "$WIDTH" 5 \
    0 "Select Directory" \
    1 "Select File Extension" \
    2 "Options" \
    3 "Execute with current config" \
    4 "Exit" \
    2>&1 1>&3);
exitStatus=$?
exec 3>&-;
echo "selected: $selection exit code: $exitStatus"