如何将新建作业转换为systemd服务？

Question

我有以下的新贵工作：

# hwclock - adjust system clock and timezone
#
# The hwclock task adjusts the system clock when the hardware clock is
# set to localtime (e.g. when dual-booting with Windows), and also
# ensures that the system timezone is set so that timestamps are written
# to FAT devices.

description     "adjust system clock and timezone"

start on starting mountall

task

script
    exec hwclock --systz --utc --noadjfile
end script

我想把这个换成systemd服务。

如何在systemd上实现start on starting mountall？

我创建了systemd服务，如下所示，但我不知道如何执行start on starting mountall。

[Unit]
Description=hwclock
After=
Before=

[Service]
ExecStart=/sbin/hwclock --systz --utc --noadjfile

Answer 1

您将需要以下几行：

Requires=
After=

如下文所述：

Requires=：此指令列出了该单元本质上依赖的任何单元。如果当前单元被激活，这里列出的单元也必须成功激活，否则此单元将失败。默认情况下，这些单元与当前单元并行启动。After=：本指令中列出的单元将在启动当前单元之前启动。这并不意味着依赖关系，如果需要，必须通过上述指令建立依赖关系。

其结构应是：

[Unit]
Description=hwclock
Requires= # mountall most happen
After= # mountall should have started before hwclock run

[Service]
Type=oneshort
ExecStart=/sbin/hwclock --systz --utc --noadjfile

来自这里：

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Upstart stanza | systemd unit file directive | systemd unit file section
               |                             |
-------------------------------------------------------------------------
start on       |    Wants, Requires, Before, |
               |    After                    |  Unit 
--------------------------------------------------------------------------

注意:这是一个Ubuntu系统，但应该是类似的。见：https://www.freedesktop.org/software/systemd/man/systemd.unit.html。