I'm looking for a way to use the ouput of a command (say command1) as an argument for another command (say command2).
I encountered this problem when trying to grep
the output of who
command but using a pattern given by another set of command (actually tty
piped to sed
).
Context:
If tty
displays:
/dev/pts/5
And who
displays:
root pts/4 2012-01-15 16:01 (xxxx)
root pts/5 2012-02-25 10:02 (yyyy)
root pts/2 2012-03-09 12:03 (zzzz)
Goal:
I want only the line(s) regarding "pts/5"
So I piped tty
to sed
as follows:
$ tty | sed 's/\/dev\///'
pts/5
Test:
The attempted following command doesn't work:
$ who | grep $(echo $(tty) | sed 's/\/dev\///')"
Possible solution:
I've found out that the following works just fine:
$ eval "who | grep $(echo $(tty) | sed 's/\/dev\///')"
But I'm sure the use of eval
could be avoided.
As a final side node: I've noticed that the "-m" argument to who
gives me exactly what I want (get only the line of who
that is linked to current user). But I'm still curious on how I could make this combination of pipes and command nesting to work...
You can do it like this:
You can do this without resorting to sed with the help of Bash variable mangling, although as @ruakh points out this won't work in the single line version (without the semicolon separating the commands). I'm leaving this first approach up because I think it's interesting that it doesn't work in a single line:
This first puts the output of
tty
into a variable, then erases the leading/dev/
on grep's use of it. But without the semicolonTTY
is not in the environment by the moment bash does the variable expansion/mangling for grep.Here's a version that does work because it spawns a subshell with the already modified environment (that has
TTY
):The result is left in
$WHOLINE
.@Eduardo's answer is correct (and as I was writing this, a couple of other good answers have appeared), but I'd like to explain why the original command is failing. As usual,
set -x
is very useful to see what's actually happening:It's not completely explicit in the above, but what's happening is that
tty
is outputting "not a tty". This is because it's part of the pipeline being fed the output ofwho
, so its stdin is indeed not a tty. This is the real reason everyone else's answers work: they gettty
out of the pipeline, so it can see your actual terminal.BTW, your proposed command is basically correct (except for the pipeline issue), but unnecessarily complex. Don't use
echo $(tty)
, it's essentially the same as justtty
.One usually uses xargs to make the output of one command an option to another command. For example:
But ... while that was your question, it's not what you really want to know.
If you don't mind storing your tty in a variable, you can use bash variable mangling to do your substitution:
(You want the -w because if you're on pts/6 you shouldn't see pts/60's logins.)
You're limited to doing this in a variable, because if you try to put the
tty
command into a pipe, it thinks that it's not running associated with a terminal anymore.Note that nothing in this answer so far is specific to bash. Since you're using bash, another way around this problem is to use process substitution. For example, while this does not work:
This does: