ubuntu - 对服务的 systemd 自定义命令

Question

我有一个像这样的 systemd 服务脚本:

#
# systemd unit file for Debian
#
# Put this in /lib/systemd/system
# Run:
#   - systemctl enable sidekiq
#   - systemctl {start,stop,restart} sidekiq
#
# This file corresponds to a single Sidekiq process.  Add multiple copies
# to run multiple processes (sidekiq-1, sidekiq-2, etc).
#
[Unit]
Description=sidekiq
# start sidekiq only once the network, logging subsystems are available
After=syslog.target network.target

[Service]
Type=simple
WorkingDirectory=/home/deploy/app
User=deploy
Group=deploy
UMask=0002
ExecStart=/bin/bash -lc "bundle exec sidekiq -e ${environment} -C config/sidekiq.yml -L log/sidekiq.log -P /tmp/sidekiq.pid"
ExecStop=/bin/bash -lc "bundle exec sidekiqctl stop /tmp/sidekiq.pid"

# if we crash, restart
RestartSec=1
Restart=on-failure

# output goes to /var/log/syslog
StandardOutput=syslog
StandardError=syslog

# This will default to "bundler" if we don't specify it
SyslogIdentifier=sidekiq

[Install]
WantedBy=multi-user.target

现在我可以发出如下命令:

sudo systemctl enable sidekiq

sudo systemctl start sidekiq

我想创建另一个自定义命令，使用它我可以让 sidekiq 工作人员安静下来，为了让 sidekiq 安静下来，我必须向进程发送 USR1 信号，如下所示:

sudo kill -s USR1 `cat #{sidekiq_pid}`

我想使用 systemd 服务来做这件事，所以本质上是一个像这样的命令

sudo systemctl queit sidekiq

有没有办法在 systemd 服务文件中创建自定义命令？如果是，那么该怎么做？

Answer 1

这不是“自定义”命令，但您可以使用

Sidekiq >= 5:

 systemctl kill -s TSTP --kill-who=main example.service 

Sidekiq < 5:

 systemctl kill -s USR1 --kill-who=main example.service 

发送“安静”的信号。参见 http://0pointer.de/blog/projects/systemd-for-admins-4.html以获得更多解释。