我正在尝试为我的应用程序创建启动/关闭脚本，但我没有这方面的经验，所以我想从屏幕开始（我将其称为“test_screen”）。
首先，我想我只会在 /etc/init.d 中创建脚本。我从这里得到了灵感——https://unix.stackexchange.com/a/20361/29677。
模拟我的应用程序的基本想法是使用
screen -d -m -S test_screen启动screen -S test_screen -X quit用于关机- 以及
screen –list状态（某种）
但是当我尝试时/etc/init.d/test_screen start我得到了
Reloading systemd: [ OK ]
Starting test_screen (via systemctl): Failed to start test_screen.service: Unit not found.
[FAILED]
所以看来我必须创建单位。
我尝试过https://access.redhat.com/documentation/en-US/Red_Hat_Enterprise_Linux/7/html/System_Administrators_Guide/sect-Managing_Services_with_systemd-Unit_Files.html，所以我的单位文件位于/etc/systemd/system：
# cat test_screen.service
[Unit]
Description=Testing `screen` service
[Service]
Type=simple
ExecStart=/bin/screen -d -m -S test_screen
ExecStop=/bin/screen -S test_screen -X quit
Environment=
Restart=always
[Install]
WantedBy=default.target
第一个问题是，我应该有/etc/init.d/test_screen start还是ExecStart没有？肯定不会是一班人...
无论如何，它没有运行。在/var/log/messages我看到
Sep 19 10:54:58 somehostname systemd: Started Testing `screen` service.
Sep 19 10:54:58 somehostname systemd: Starting Testing `screen` service...
Sep 19 10:54:58 somehostname systemd: test_screen.service holdoff time over, scheduling restart.
Sep 19 10:54:58 somehostname systemd: Started Testing `screen` service.
Sep 19 10:54:58 somehostname systemd: Starting Testing `screen` service...
Sep 19 10:54:58 somehostname systemd: test_screen.service holdoff time over, scheduling restart.
Sep 19 10:54:58 somehostname systemd: Started Testing `screen` service.
Sep 19 10:54:58 somehostname systemd: Starting Testing `screen` service...
Sep 19 10:54:59 somehostname systemd: test_screen.service holdoff time over, scheduling restart.
Sep 19 10:54:59 somehostname systemd: Started Testing `screen` service.
Sep 19 10:54:59 somehostname systemd: Starting Testing `screen` service...
Sep 19 10:54:59 somehostname systemd: test_screen.service holdoff time over, scheduling restart.
Sep 19 10:54:59 somehostname systemd: Started Testing `screen` service.
Sep 19 10:54:59 somehostname systemd: Starting Testing `screen` service...
Sep 19 10:54:59 somehostname systemd: test_screen.service holdoff time over, scheduling restart.
Sep 19 10:54:59 somehostname systemd: start request repeated too quickly for test_screen.service
Sep 19 10:54:59 somehostname systemd: Failed to start Testing `screen` service.
Sep 19 10:54:59 somehostname systemd: Unit test_screen.service entered failed state.
Sep 19 10:54:59 somehostname systemd: test_screen.service failed.
如何找到进入失败状态的原因？欢迎所有建议。
答案1
看来，我解决了它screen......
首先我删除了该Restart=always参数。
为了使其工作，我还必须添加RemainAfterExit=True.
对于旧的初始化脚本，代码的位置很清楚。我真的不知道放置其余代码的合适位置。我应该简单地从 调用我的脚本吗ExecStart=？
我在学习过程中的额外发现。
地位
有一个很好的问题（我的意思是我也感兴趣），如何实现状态检查 -systemd 自定义状态消息？
简短的回答是，它（以某种方式）开箱即用=您不需要太在意。
因此，对于我的 test_screen 服务，我可以调用systemctl status test_screen.service，我会得到
● test_screen.service - Testing `screen` service
Loaded: loaded (/etc/systemd/system/test_screen.service; disabled; vendor preset: disabled)
Active: active (running) since Wed 2017-09-20 12:48:34 CEST; 1s ago
Process: 36633 ExecStart=/bin/screen -d -m -S test_screen (code=exited, status=0/SUCCESS)
Main PID: 36634 (screen)
CGroup: /system.slice/test_screen.service
├─36634 /bin/SCREEN -d -m -S test_screen
└─36635 /bin/sh
Sep 20 12:48:34 somehostname systemd[1]: Starting Testing `screen` service...
Sep 20 12:48:34 somehostname systemd[1]: Started Testing `screen` service.
...奇怪的是，当我停止它时，我的状态失败了：
● test_screen.service - Testing `screen` service
Loaded: loaded (/etc/systemd/system/test_screen.service; disabled; vendor preset: disabled)
Active: failed (Result: exit-code) since Wed 2017-09-20 12:51:00 CEST; 1s ago
Process: 36805 ExecStop=/bin/screen -S test_screen -X quit (code=exited, status=0/SUCCESS)
Process: 36633 ExecStart=/bin/screen -d -m -S test_screen (code=exited, status=0/SUCCESS)
Main PID: 36634 (code=exited, status=1/FAILURE)
Sep 20 12:48:34 somehostname systemd[1]: Starting Testing `screen` service...
Sep 20 12:48:34 somehostname systemd[1]: Started Testing `screen` service.
Sep 20 12:51:00 somehostname systemd[1]: Stopping Testing `screen` service...
Sep 20 12:51:00 somehostname systemd[1]: test_screen.service: main process exited, code=exited, status=1/FAILURE
Sep 20 12:51:00 somehostname systemd[1]: Stopped Testing `screen` service.
Sep 20 12:51:00 somehostname systemd[1]: Unit test_screen.service entered failed state.
Sep 20 12:51:00 somehostname systemd[1]: test_screen.service failed.
克服我必须拥有的问题RemainAfterExit=True（因为我尝试了该建议Type=forking并注释掉了该RemainAfterExit选项）。
列表单元文件
我遇到了一个问题，我现在无法重现 - 当我尝试时，systemctl start响应类似于“没有单元文件”，所以我想知道是否需要以某种方式注册它。
不，你不需要。你可以执行systemctl list-unit-files --type=service，你应该在那里看到你的单位。我的问题是，该ExecStart参数错误。当我现在尝试同样的操作时，我更容易理解消息：
$ systemctl start test_screen.service
Job for test_screen.service failed because the control process exited with error code. See "systemctl status test_screen.service" and "journalctl -xe" for details.
$ systemctl status test_screen.service
● test_screen.service - Testing `screen` service
Loaded: loaded (/etc/systemd/system/test_screen.service; disabled; vendor preset: disabled)
Active: failed (Result: exit-code) since Wed 2017-09-20 12:55:53 CEST; 1min 3s ago
Process: 37344 ExecStart=/bin/screen2 -d -m -S test_screen (code=exited, status=203/EXEC)
Main PID: 36634 (code=exited, status=1/FAILURE)
Sep 20 12:55:53 somehostname systemd[1]: Starting Testing `screen` service...
Sep 20 12:55:53 somehostname systemd[37344]: Failed at step EXEC spawning /bin/screen2: No such file or directory
Sep 20 12:55:53 somehostname systemd[1]: test_screen.service: control process exited, code=exited status=203
Sep 20 12:55:53 somehostname systemd[1]: Failed to start Testing `screen` service.
Sep 20 12:55:53 somehostname systemd[1]: Unit test_screen.service entered failed state.
Sep 20 12:55:53 somehostname systemd[1]: test_screen.service failed.
执行脚本
我尝试修改我的 test_screen 来调用脚本
$ cat /etc/systemd/system/test_screen_script.service
[Unit]
Description=Testing `screen` service
[Service]
Type=forking
ExecStart=/root/test_screen_start.sh
ExecStop=/root/test_screen_stop.sh
#Environment=
#Restart=always
RemainAfterExit=True
[Install]
WantedBy=default.target
而脚本只是为之前的调用而包装
$ cat /root/test_screen_start.sh
/bin/screen -d -m -S test_screen
$ cat /root/test_screen_stop.sh
/bin/screen -S test_screen -X quit
当我这样做时，它没有启动：
$ systemctl start test_screen_script
Job for test_screen_script.service failed because the control process exited with error code. See "systemctl status test_screen_script.service" and "journalctl -xe" for details.
$ systemctl status test_screen_script.service
● test_screen_script.service - Testing `screen` service
Loaded: loaded (/etc/systemd/system/test_screen_script.service; disabled; vendor preset: disabled)
Active: failed (Result: exit-code) since Wed 2017-09-20 15:47:59 CEST; 8s ago
Process: 63582 ExecStart=/root/test_screen_start.sh (code=exited, status=203/EXEC)
Main PID: 60698 (code=exited, status=0/SUCCESS)
Sep 20 15:47:59 somehostname systemd[1]: Starting Testing `screen` service...
Sep 20 15:47:59 somehostname systemd[63582]: Failed at step EXEC spawning /root/test_screen_start.sh: Exec format error
Sep 20 15:47:59 somehostname systemd[1]: test_screen_script.service: control process exited, code=exited status=203
Sep 20 15:47:59 somehostname systemd[1]: Failed to start Testing `screen` service.
Sep 20 15:47:59 somehostname systemd[1]: Unit test_screen_script.service entered failed state.
Sep 20 15:47:59 somehostname systemd[1]: test_screen_script.service failed.
最好有人能描述一下原因。解决这个问题的方法是添加#!/bin/bash.