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I know systemctl by itself is supposed to give me all the running services and using systemctl status (with application name, ie foobar) gives me more details on that service. However I noticed I can also run systemctl status and get an output, but it is different from systemctl by itself.

So what is it displaying?

It seems like it defaults to display degraded services:

State: degraded
 Jobs: 0 queued

Failed: 3 units
Since: Sun 2016-03-27 01:42:15 UTC; 2h 49min ago

  • What does the man page tell you? – EightBitTony Mar 27 '16 at 5:08
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    It says it shows system status. Should it be systemd status? " status [PATTERN...|PID...]] Show terse runtime status information about one or more units, followed by most recent log data from the journal. If no units are specified, show system status." Also, why would system or systemd be degraded? Thanks – BluePython Mar 27 '16 at 5:51
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If you execute systemctl without arguments, the default behavior is systemctl list-units.

From the systemctl manpage:

list-units [PATTERN...]

List known units (subject to limitations specified with -t). If one or more PATTERNs are specified, only units matching one of them are shown.

This is the default command.

If you specify the option status

status [PATTERN...|PID...]

Show terse runtime status information about one or more units, followed by most recent log data from the journal. If no units are specified, show system status. If combined with --all, also show the status of all units (subject to limitations specified with -t). If a PID is passed, show information about the unit the process belongs to.

This function is intended to generate human-readable output. If you are looking for computer-parsable output, use show instead. By default, this function only shows 10 lines of output and ellipsizes lines to fit in the terminal window. This can be changed with --lines and --full, see above. In addition, journalctl --unit=NAME or journalctl --user-unit=NAME use a similar filter for messages and might be more convenient.

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