自守护 bash 脚本
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Self-daemonizing bash script
提问by pepoluan
I want to make a script to be self-daemonizing, i.e., no need to invoke nohup $SCRIPT &>/dev/null &manually on the shell prompt.
我想让脚本自我守护，即不需要nohup $SCRIPT &>/dev/null &在 shell 提示符下手动调用。
My plan is to create a section of code like the following:
我的计划是创建一段如下所示的代码：
#!/bin/bash
SCRIPTNAME="if [[ != "--daemonize" ]]; then
"
...
# Preps are done above
if [[ "" != "--daemonize" ]]; then
nohup "$SCRIPTNAME" --daemonize "${PARAMS[@]}" &>/dev/null &
exit $?
fi
# Rest of the code are the actual procedures of the daemon
Is this wise? Do you have better alternatives?
这是明智的吗？你有更好的选择吗？
采纳答案by konsolebox
Here are things I see.
这是我看到的东西。
nohup $SCRIPTNAME --daemonize "${PARAMS[@]}" &>/dev/null &
Shouln't that be == --daemonize?
不应该是 == --daemonize?
(
Codes that run in daemon mode.
) </dev/null >/dev/null 2>&1 &
disown
Instead of calling your script again, you could just summon a subshell that's placed in a background:
无需再次调用您的脚本，您只需调用一个放置在背景中的子外壳即可：
function daemon_mode {
Codes that run in daemon mode.
}
daemon_mode </dev/null >/dev/null 2>&1 &
disown
Or
或者
##代码##