Systemctl 命令显示正在运行的服务的摘要

Question

systemctl我将使用哪个选项或命令来显示当前运行的所有服务的摘要？

Answer 1

您可以使用systemctl's 的一些选项：

-t, --type=
       The argument should be a comma-separated list of unit types such as
       service and socket.

       If one of the arguments is a unit type, when listing units, limit
       display to certain unit types. Otherwise, units of all types will
       be shown.

       As a special case, if one of the arguments is help, a list of
       allowed values will be printed and the program will exit.

   --state=
       The argument should be a comma-separated list of unit LOAD, SUB, or
       ACTIVE states. When listing units, show only those in the specified
       states. Use --state=failed to show only failed units.

       As a special case, if one of the arguments is help, a list of
       allowed values will be printed and the program will exit.

所以可能你想要：

systemctl --type=service --state=active list-units

其中列出了所有活动的服务，包括那些已经退出的服务。如果您只关注此时正在运行的那些，您可以使用：

systemctl --type=service --state=running list-units

Answer 2

它是（见man 1 systemctl）：

systemctl list-units | grep -E 'service.*running'

或（另见man 8 service）

service --status-all

其中[+]表示实际运行的服务。

Answer 3

在环顾四周超过必要的时间后，我想出了这种确定正在运行的服务的略有不同的方法。它还展示了如何计算正在运行的服务的数量。这种方式确保它不会意外地捕获服务名称本身中带有 running 或 service 一词的内容。

# Output all active services:
systemctl -t service --state=active --no-pager --no-legend

# Count of all active services:
systemctl -t service --state=active --no-pager --no-legend | grep -c -

# Output all running services:
systemctl -t service --state=active --no-pager --no-legend | egrep '^*\.service.*running'

# Count of all running services:
systemctl -t service --state=active --no-pager --no-legend | egrep '^*\.service.*running' -c -

# Output only the service and its description:
systemctl -t service --state=active --no-pager --no-legend | egrep '^*\.service.*running' | awk 'BEGIN { FS = " ";} {for (i = 2; i <= 4; i++) { $i = "" }; print}'