当你用systemd启动一个服务时,进程只处于该activating
状态一段时间。active
一旦该过程成功启动,它就会转变为。
但是,如果我有一项复杂的服务需要一段时间才能准备好怎么办?如何延长activating
状态直到服务在内部准备就绪?
答案1
在[Service]
您的单位部分中,添加(或更改)Type=notify
。
这告诉 systemd 进程本身将在启动时发出信号。因此,当进程启动时,systemd 不会假设它已准备好。
为此,该流程必须实施sd_通知(3)。
下面是 C 语言通知服务的最小示例:
# notifier.service
[Service]
Type=notify
ExecStart=%h/bin/notifier
/* main.c */
#include <systemd/sd-daemon.h>
#include <unistd.h>
int main(void) {
/* Sleep to emulate 10s bootup time */
/* Expect status 'activating (start)' during this */
/* `systemctl start`, will block */
sleep(10);
/* Send a signal to say we've started */
sd_notify(0, "READY=1");
/* Units which are After= this unit will now start */
/* `systemctl start` will unblock now */
/* Sleep to emulate 10s run time */
/* Expect status 'active (running)' during this */
sleep(10);
/* Send a signal to say we've started the shutdown procedure */
sd_notify(0, "STOPPING=1");
/* Sleep to emulate 10s shutdown */
/* Expect status 'deactivating' during this */
sleep(10);
return 0;
/* Expect status 'inactive (dead)' at this point */
}
# makefile
a.out: main.c
gcc main.c -lsystemd
install: a.out notifier.service
install -D a.out ~/bin/notifier
install -D notifier.service ~/.config/systemd/user/notifier.service