Using pipes to list the first two and last two users on the system?
up vote
2
down vote
favorite
I have to use the who command to display who's online, then use pipes to display the first and last 2 users online. the only thing I know how to do is something like:
who | head -5 | tail -2 .
That doesn't work, though.
pipe tail head who
edited Nov 24 at 20:19
Rui F Ribeiro
38.3k1475126
asked Feb 23 '17 at 2:41
user217538
111
add a comment |
up vote
2
down vote
favorite
I have to use the who command to display who's online, then use pipes to display the first and last 2 users online. the only thing I know how to do is something like:
who | head -5 | tail -2 .
That doesn't work, though.
pipe tail head who
edited Nov 24 at 20:19
Rui F Ribeiro
38.3k1475126
asked Feb 23 '17 at 2:41
user217538
111
you can combine multiple commands inside()or{}separated by;... see Command-Grouping
– Sundeep
Feb 23 '17 at 3:02
First and last users in terms of what? Just from the output?
– DarkHeart
Feb 23 '17 at 4:48
Similar: Command to display first few and last few lines of a file
– Stéphane Chazelas
Feb 23 '17 at 16:19
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have to use the who command to display who's online, then use pipes to display the first and last 2 users online. the only thing I know how to do is something like:
who | head -5 | tail -2 .
That doesn't work, though.
pipe tail head who
edited Nov 24 at 20:19
Rui F Ribeiro
38.3k1475126
asked Feb 23 '17 at 2:41
user217538
111
I have to use the who command to display who's online, then use pipes to display the first and last 2 users online. the only thing I know how to do is something like:
who | head -5 | tail -2 .
That doesn't work, though.
pipe tail head who
pipe tail head who
edited Nov 24 at 20:19
Rui F Ribeiro
38.3k1475126
asked Feb 23 '17 at 2:41
user217538
111
edited Nov 24 at 20:19
Rui F Ribeiro
38.3k1475126
asked Feb 23 '17 at 2:41
user217538
111
edited Nov 24 at 20:19
Rui F Ribeiro
38.3k1475126
edited Nov 24 at 20:19
Rui F Ribeiro
38.3k1475126
edited Nov 24 at 20:19
Rui F Ribeiro
38.3k1475126
38.3k1475126
asked Feb 23 '17 at 2:41
user217538
111
asked Feb 23 '17 at 2:41
user217538
111
asked Feb 23 '17 at 2:41
user217538
111
111
you can combine multiple commands inside()or{}separated by;... see Command-Grouping
– Sundeep
Feb 23 '17 at 3:02
First and last users in terms of what? Just from the output?
– DarkHeart
Feb 23 '17 at 4:48
Similar: Command to display first few and last few lines of a file
– Stéphane Chazelas
Feb 23 '17 at 16:19
add a comment |
you can combine multiple commands inside()or{}separated by;... see Command-Grouping
– Sundeep
Feb 23 '17 at 3:02
First and last users in terms of what? Just from the output?
– DarkHeart
Feb 23 '17 at 4:48
Similar: Command to display first few and last few lines of a file
– Stéphane Chazelas
Feb 23 '17 at 16:19
you can combine multiple commands inside
() or {} separated by ;... see Command-Grouping– Sundeep
Feb 23 '17 at 3:02
you can combine multiple commands inside
() or {} separated by ;... see Command-Grouping– Sundeep
Feb 23 '17 at 3:02
First and last users in terms of what? Just from the output?
– DarkHeart
Feb 23 '17 at 4:48
First and last users in terms of what? Just from the output?
– DarkHeart
Feb 23 '17 at 4:48
Similar: Command to display first few and last few lines of a file
– Stéphane Chazelas
Feb 23 '17 at 16:19
Similar: Command to display first few and last few lines of a file
– Stéphane Chazelas
Feb 23 '17 at 16:19
add a comment |
3 Answers
3
active
oldest
votes
up vote
4
down vote
Directly:
who | head -2
who | tail -2
answered Feb 23 '17 at 4:21
Jeff Schaller
36.6k1052121
add a comment |
up vote
0
down vote
You can use tee and a redirection to stderr
who | tee >(head -n2 1>&2 ) | tail -n2
Where tee allows you to copy stdin to both a file and stdout. Here we actually replace the file by the head command and redirect its output from stdout to stderr (via 1>&2) as all of stout is piped to the tail command and processed, so we need to circumvent tailing the head result, too. stderr itself is however still printed on the terminal.
Updates following discussion in comments
As pointed out by Stéphane, head might send a SIGPIPE signal kill the pipe prematurely so that tail might not see the actual end of the input stream.
To prevent this, one could ignore the SIGPIPE signal as described here, by using a trap with an empty command. Tested in bash only .....
who | { trap "" PIPE ; tee >( head -n 2 3>&1 >&2 ) ; } | tail -n 2
Additionally mentioned in the comments: non-bash shells might tumble the ordering of the results - see the proposed solution by Stéphane for this.
edited Apr 13 '17 at 12:36
Community♦
1
answered Feb 23 '17 at 14:19
Fiximan
3,143524
1
who -l | ( tee >(head -n2 1>&3 ) | tail -n2 ) 3>&1will ensure bothheadandtailwrite to stdout.
– roaima
Feb 23 '17 at 14:45
Depending on timing and on the number of logged-in users,teemay get a SIGPIPE whenheadterminates so thatteewould not be able to write the full output totail. Also, asheadandtailrun in parallel, the order of their output is not guaranteed.
– Stéphane Chazelas
Feb 23 '17 at 16:28
@StéphaneChazelas I can confirm the SIGPIPE problem simulating with a large enoughseqasstdin- order is never tumbled, buttaildoes not show the last two number of thesequence. However I do not understand properly what is happening. Could you elaborate?
– Fiximan
Feb 23 '17 at 16:43
1
Ifheadfinishes beforeteehas written all the data to the pipe tohead, then the next timeteewrite to that pipe it will get a SIGPIPE and die.tailwill output the last two lines it has received before that.
– Stéphane Chazelas
Feb 23 '17 at 17:37
1
Withbash, the order will be preserved becausebash(at least current versions) process substitution leaves abashprocess around with its stdout still connected to the pipe totailwaiting for theheadcommand, sotailwon't see eof until that process has terminated so not until head has terminated. Other shells likeksh93orzshdon't have that limitation, so you may see the output tumbled there. One way to guarantee the order with any shell would be withtee >(head -n 2 3>&1 >&2) | tail -n 2. Here that fd 3 holds the pipe to tail open.
– Stéphane Chazelas
Feb 23 '17 at 18:00
|
show 2 more comments
up vote
-1
down vote
It goes wrong for you when you are trying to use tail command. You are not actually using it to extract what you want from who command's output but from head command.
If you need to use head and tail combined with piping, you could do it like this for example:
(who |head -n5 && who |tail -n2)
Modify numbers to meet your needs
answered Feb 23 '17 at 3:05
jiipeezz
21125
1
This won't work becauseheadwill eat up all of stdin, leaving nothing fortail.
– roaima
Feb 23 '17 at 14:42
@roaima I see where it went wrong now. But there's one weird thing I don't uderstand at all. I tried to run following commandls -l |(head -n5 && tail -n2). I tested it 10 times now, 6 times I got the result I wanted and 4 times I didn't. Why is that?
– jiipeezz
Feb 23 '17 at 15:56
1
@jiipeezz, that very much depends on timing and the number of files in the directory. It depends whether the firstread()done byheadreads the whole output ofls -lor not.
– Stéphane Chazelas
Feb 23 '17 at 18:04
@StéphaneChazelas thanks for elaborating. Now it makes more sense. Corrected the answer to something that works.
– jiipeezz
Feb 23 '17 at 20:38
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Directly:
who | head -2
who | tail -2
answered Feb 23 '17 at 4:21
Jeff Schaller
36.6k1052121
add a comment |
up vote
4
down vote
Directly:
who | head -2
who | tail -2
answered Feb 23 '17 at 4:21
Jeff Schaller
36.6k1052121
add a comment |
up vote
4
down vote
up vote
4
down vote
Directly:
who | head -2
who | tail -2
answered Feb 23 '17 at 4:21
Jeff Schaller
36.6k1052121
Directly:
who | head -2
who | tail -2
answered Feb 23 '17 at 4:21
Jeff Schaller
36.6k1052121
answered Feb 23 '17 at 4:21
Jeff Schaller
36.6k1052121
answered Feb 23 '17 at 4:21
Jeff Schaller
36.6k1052121
answered Feb 23 '17 at 4:21
Jeff Schaller
36.6k1052121
36.6k1052121
add a comment |
add a comment |
up vote
0
down vote
You can use tee and a redirection to stderr
who | tee >(head -n2 1>&2 ) | tail -n2
Where tee allows you to copy stdin to both a file and stdout. Here we actually replace the file by the head command and redirect its output from stdout to stderr (via 1>&2) as all of stout is piped to the tail command and processed, so we need to circumvent tailing the head result, too. stderr itself is however still printed on the terminal.
Updates following discussion in comments
As pointed out by Stéphane, head might send a SIGPIPE signal kill the pipe prematurely so that tail might not see the actual end of the input stream.
To prevent this, one could ignore the SIGPIPE signal as described here, by using a trap with an empty command. Tested in bash only .....
who | { trap "" PIPE ; tee >( head -n 2 3>&1 >&2 ) ; } | tail -n 2
Additionally mentioned in the comments: non-bash shells might tumble the ordering of the results - see the proposed solution by Stéphane for this.
edited Apr 13 '17 at 12:36
Community♦
1
answered Feb 23 '17 at 14:19
Fiximan
3,143524
1
who -l | ( tee >(head -n2 1>&3 ) | tail -n2 ) 3>&1will ensure bothheadandtailwrite to stdout.
– roaima
Feb 23 '17 at 14:45
Depending on timing and on the number of logged-in users,teemay get a SIGPIPE whenheadterminates so thatteewould not be able to write the full output totail. Also, asheadandtailrun in parallel, the order of their output is not guaranteed.
– Stéphane Chazelas
Feb 23 '17 at 16:28
@StéphaneChazelas I can confirm the SIGPIPE problem simulating with a large enoughseqasstdin- order is never tumbled, buttaildoes not show the last two number of thesequence. However I do not understand properly what is happening. Could you elaborate?
– Fiximan
Feb 23 '17 at 16:43
1
Ifheadfinishes beforeteehas written all the data to the pipe tohead, then the next timeteewrite to that pipe it will get a SIGPIPE and die.tailwill output the last two lines it has received before that.
– Stéphane Chazelas
Feb 23 '17 at 17:37
1
Withbash, the order will be preserved becausebash(at least current versions) process substitution leaves abashprocess around with its stdout still connected to the pipe totailwaiting for theheadcommand, sotailwon't see eof until that process has terminated so not until head has terminated. Other shells likeksh93orzshdon't have that limitation, so you may see the output tumbled there. One way to guarantee the order with any shell would be withtee >(head -n 2 3>&1 >&2) | tail -n 2. Here that fd 3 holds the pipe to tail open.
– Stéphane Chazelas
Feb 23 '17 at 18:00
|
show 2 more comments
up vote
0
down vote
You can use tee and a redirection to stderr
who | tee >(head -n2 1>&2 ) | tail -n2
Where tee allows you to copy stdin to both a file and stdout. Here we actually replace the file by the head command and redirect its output from stdout to stderr (via 1>&2) as all of stout is piped to the tail command and processed, so we need to circumvent tailing the head result, too. stderr itself is however still printed on the terminal.
Updates following discussion in comments
As pointed out by Stéphane, head might send a SIGPIPE signal kill the pipe prematurely so that tail might not see the actual end of the input stream.
To prevent this, one could ignore the SIGPIPE signal as described here, by using a trap with an empty command. Tested in bash only .....
who | { trap "" PIPE ; tee >( head -n 2 3>&1 >&2 ) ; } | tail -n 2
Additionally mentioned in the comments: non-bash shells might tumble the ordering of the results - see the proposed solution by Stéphane for this.
edited Apr 13 '17 at 12:36
Community♦
1
answered Feb 23 '17 at 14:19
Fiximan
3,143524
1
who -l | ( tee >(head -n2 1>&3 ) | tail -n2 ) 3>&1will ensure bothheadandtailwrite to stdout.
– roaima
Feb 23 '17 at 14:45
Depending on timing and on the number of logged-in users,teemay get a SIGPIPE whenheadterminates so thatteewould not be able to write the full output totail. Also, asheadandtailrun in parallel, the order of their output is not guaranteed.
– Stéphane Chazelas
Feb 23 '17 at 16:28
@StéphaneChazelas I can confirm the SIGPIPE problem simulating with a large enoughseqasstdin- order is never tumbled, buttaildoes not show the last two number of thesequence. However I do not understand properly what is happening. Could you elaborate?
– Fiximan
Feb 23 '17 at 16:43
1
Ifheadfinishes beforeteehas written all the data to the pipe tohead, then the next timeteewrite to that pipe it will get a SIGPIPE and die.tailwill output the last two lines it has received before that.
– Stéphane Chazelas
Feb 23 '17 at 17:37
1
Withbash, the order will be preserved becausebash(at least current versions) process substitution leaves abashprocess around with its stdout still connected to the pipe totailwaiting for theheadcommand, sotailwon't see eof until that process has terminated so not until head has terminated. Other shells likeksh93orzshdon't have that limitation, so you may see the output tumbled there. One way to guarantee the order with any shell would be withtee >(head -n 2 3>&1 >&2) | tail -n 2. Here that fd 3 holds the pipe to tail open.
– Stéphane Chazelas
Feb 23 '17 at 18:00
|
show 2 more comments
up vote
0
down vote
up vote
0
down vote
You can use tee and a redirection to stderr
who | tee >(head -n2 1>&2 ) | tail -n2
Where tee allows you to copy stdin to both a file and stdout. Here we actually replace the file by the head command and redirect its output from stdout to stderr (via 1>&2) as all of stout is piped to the tail command and processed, so we need to circumvent tailing the head result, too. stderr itself is however still printed on the terminal.
Updates following discussion in comments
As pointed out by Stéphane, head might send a SIGPIPE signal kill the pipe prematurely so that tail might not see the actual end of the input stream.
To prevent this, one could ignore the SIGPIPE signal as described here, by using a trap with an empty command. Tested in bash only .....
who | { trap "" PIPE ; tee >( head -n 2 3>&1 >&2 ) ; } | tail -n 2
Additionally mentioned in the comments: non-bash shells might tumble the ordering of the results - see the proposed solution by Stéphane for this.
edited Apr 13 '17 at 12:36
Community♦
1
answered Feb 23 '17 at 14:19
Fiximan
3,143524
You can use tee and a redirection to stderr
who | tee >(head -n2 1>&2 ) | tail -n2
Where tee allows you to copy stdin to both a file and stdout. Here we actually replace the file by the head command and redirect its output from stdout to stderr (via 1>&2) as all of stout is piped to the tail command and processed, so we need to circumvent tailing the head result, too. stderr itself is however still printed on the terminal.
Updates following discussion in comments
As pointed out by Stéphane, head might send a SIGPIPE signal kill the pipe prematurely so that tail might not see the actual end of the input stream.
To prevent this, one could ignore the SIGPIPE signal as described here, by using a trap with an empty command. Tested in bash only .....
who | { trap "" PIPE ; tee >( head -n 2 3>&1 >&2 ) ; } | tail -n 2
Additionally mentioned in the comments: non-bash shells might tumble the ordering of the results - see the proposed solution by Stéphane for this.
edited Apr 13 '17 at 12:36
Community♦
1
answered Feb 23 '17 at 14:19
Fiximan
3,143524
edited Apr 13 '17 at 12:36
Community♦
1
edited Apr 13 '17 at 12:36
Community♦
1
edited Apr 13 '17 at 12:36
Community♦
1
1
answered Feb 23 '17 at 14:19
Fiximan
3,143524
answered Feb 23 '17 at 14:19
Fiximan
3,143524
answered Feb 23 '17 at 14:19
Fiximan
3,143524
3,143524
1
who -l | ( tee >(head -n2 1>&3 ) | tail -n2 ) 3>&1will ensure bothheadandtailwrite to stdout.
– roaima
Feb 23 '17 at 14:45
Depending on timing and on the number of logged-in users,teemay get a SIGPIPE whenheadterminates so thatteewould not be able to write the full output totail. Also, asheadandtailrun in parallel, the order of their output is not guaranteed.
– Stéphane Chazelas
Feb 23 '17 at 16:28
@StéphaneChazelas I can confirm the SIGPIPE problem simulating with a large enoughseqasstdin- order is never tumbled, buttaildoes not show the last two number of thesequence. However I do not understand properly what is happening. Could you elaborate?
– Fiximan
Feb 23 '17 at 16:43
1
Ifheadfinishes beforeteehas written all the data to the pipe tohead, then the next timeteewrite to that pipe it will get a SIGPIPE and die.tailwill output the last two lines it has received before that.
– Stéphane Chazelas
Feb 23 '17 at 17:37
1
Withbash, the order will be preserved becausebash(at least current versions) process substitution leaves abashprocess around with its stdout still connected to the pipe totailwaiting for theheadcommand, sotailwon't see eof until that process has terminated so not until head has terminated. Other shells likeksh93orzshdon't have that limitation, so you may see the output tumbled there. One way to guarantee the order with any shell would be withtee >(head -n 2 3>&1 >&2) | tail -n 2. Here that fd 3 holds the pipe to tail open.
– Stéphane Chazelas
Feb 23 '17 at 18:00
|
show 2 more comments
1
who -l | ( tee >(head -n2 1>&3 ) | tail -n2 ) 3>&1will ensure bothheadandtailwrite to stdout.
– roaima
Feb 23 '17 at 14:45
Depending on timing and on the number of logged-in users,teemay get a SIGPIPE whenheadterminates so thatteewould not be able to write the full output totail. Also, asheadandtailrun in parallel, the order of their output is not guaranteed.
– Stéphane Chazelas
Feb 23 '17 at 16:28
@StéphaneChazelas I can confirm the SIGPIPE problem simulating with a large enoughseqasstdin- order is never tumbled, buttaildoes not show the last two number of thesequence. However I do not understand properly what is happening. Could you elaborate?
– Fiximan
Feb 23 '17 at 16:43
1
Ifheadfinishes beforeteehas written all the data to the pipe tohead, then the next timeteewrite to that pipe it will get a SIGPIPE and die.tailwill output the last two lines it has received before that.
– Stéphane Chazelas
Feb 23 '17 at 17:37
1
Withbash, the order will be preserved becausebash(at least current versions) process substitution leaves abashprocess around with its stdout still connected to the pipe totailwaiting for theheadcommand, sotailwon't see eof until that process has terminated so not until head has terminated. Other shells likeksh93orzshdon't have that limitation, so you may see the output tumbled there. One way to guarantee the order with any shell would be withtee >(head -n 2 3>&1 >&2) | tail -n 2. Here that fd 3 holds the pipe to tail open.
– Stéphane Chazelas
Feb 23 '17 at 18:00
1
1
who -l | ( tee >(head -n2 1>&3 ) | tail -n2 ) 3>&1 will ensure both head and tail write to stdout.– roaima
Feb 23 '17 at 14:45
who -l | ( tee >(head -n2 1>&3 ) | tail -n2 ) 3>&1 will ensure both head and tail write to stdout.– roaima
Feb 23 '17 at 14:45
Depending on timing and on the number of logged-in users,
tee may get a SIGPIPE when head terminates so that tee would not be able to write the full output to tail. Also, as head and tail run in parallel, the order of their output is not guaranteed.– Stéphane Chazelas
Feb 23 '17 at 16:28
Depending on timing and on the number of logged-in users,
tee may get a SIGPIPE when head terminates so that tee would not be able to write the full output to tail. Also, as head and tail run in parallel, the order of their output is not guaranteed.– Stéphane Chazelas
Feb 23 '17 at 16:28
@StéphaneChazelas I can confirm the SIGPIPE problem simulating with a large enough
seq as stdin - order is never tumbled, but tail does not show the last two number of the sequence. However I do not understand properly what is happening. Could you elaborate?– Fiximan
Feb 23 '17 at 16:43
@StéphaneChazelas I can confirm the SIGPIPE problem simulating with a large enough
seq as stdin - order is never tumbled, but tail does not show the last two number of the sequence. However I do not understand properly what is happening. Could you elaborate?– Fiximan
Feb 23 '17 at 16:43
1
1
If
head finishes before tee has written all the data to the pipe to head, then the next time tee write to that pipe it will get a SIGPIPE and die. tail will output the last two lines it has received before that.– Stéphane Chazelas
Feb 23 '17 at 17:37
If
head finishes before tee has written all the data to the pipe to head, then the next time tee write to that pipe it will get a SIGPIPE and die. tail will output the last two lines it has received before that.– Stéphane Chazelas
Feb 23 '17 at 17:37
1
1
With
bash, the order will be preserved because bash (at least current versions) process substitution leaves a bash process around with its stdout still connected to the pipe to tail waiting for the head command, so tail won't see eof until that process has terminated so not until head has terminated. Other shells like ksh93 or zsh don't have that limitation, so you may see the output tumbled there. One way to guarantee the order with any shell would be with tee >(head -n 2 3>&1 >&2) | tail -n 2. Here that fd 3 holds the pipe to tail open.– Stéphane Chazelas
Feb 23 '17 at 18:00
With
bash, the order will be preserved because bash (at least current versions) process substitution leaves a bash process around with its stdout still connected to the pipe to tail waiting for the head command, so tail won't see eof until that process has terminated so not until head has terminated. Other shells like ksh93 or zsh don't have that limitation, so you may see the output tumbled there. One way to guarantee the order with any shell would be with tee >(head -n 2 3>&1 >&2) | tail -n 2. Here that fd 3 holds the pipe to tail open.– Stéphane Chazelas
Feb 23 '17 at 18:00
|
show 2 more comments
up vote
-1
down vote
It goes wrong for you when you are trying to use tail command. You are not actually using it to extract what you want from who command's output but from head command.
If you need to use head and tail combined with piping, you could do it like this for example:
(who |head -n5 && who |tail -n2)
Modify numbers to meet your needs
answered Feb 23 '17 at 3:05
jiipeezz
21125
1
This won't work becauseheadwill eat up all of stdin, leaving nothing fortail.
– roaima
Feb 23 '17 at 14:42
@roaima I see where it went wrong now. But there's one weird thing I don't uderstand at all. I tried to run following commandls -l |(head -n5 && tail -n2). I tested it 10 times now, 6 times I got the result I wanted and 4 times I didn't. Why is that?
– jiipeezz
Feb 23 '17 at 15:56
1
@jiipeezz, that very much depends on timing and the number of files in the directory. It depends whether the firstread()done byheadreads the whole output ofls -lor not.
– Stéphane Chazelas
Feb 23 '17 at 18:04
@StéphaneChazelas thanks for elaborating. Now it makes more sense. Corrected the answer to something that works.
– jiipeezz
Feb 23 '17 at 20:38
add a comment |
up vote
-1
down vote
It goes wrong for you when you are trying to use tail command. You are not actually using it to extract what you want from who command's output but from head command.
If you need to use head and tail combined with piping, you could do it like this for example:
(who |head -n5 && who |tail -n2)
Modify numbers to meet your needs
answered Feb 23 '17 at 3:05
jiipeezz
21125
1
This won't work becauseheadwill eat up all of stdin, leaving nothing fortail.
– roaima
Feb 23 '17 at 14:42
@roaima I see where it went wrong now. But there's one weird thing I don't uderstand at all. I tried to run following commandls -l |(head -n5 && tail -n2). I tested it 10 times now, 6 times I got the result I wanted and 4 times I didn't. Why is that?
– jiipeezz
Feb 23 '17 at 15:56
1
@jiipeezz, that very much depends on timing and the number of files in the directory. It depends whether the firstread()done byheadreads the whole output ofls -lor not.
– Stéphane Chazelas
Feb 23 '17 at 18:04
@StéphaneChazelas thanks for elaborating. Now it makes more sense. Corrected the answer to something that works.
– jiipeezz
Feb 23 '17 at 20:38
add a comment |
up vote
-1
down vote
up vote
-1
down vote
It goes wrong for you when you are trying to use tail command. You are not actually using it to extract what you want from who command's output but from head command.
If you need to use head and tail combined with piping, you could do it like this for example:
(who |head -n5 && who |tail -n2)
Modify numbers to meet your needs
answered Feb 23 '17 at 3:05
jiipeezz
21125
It goes wrong for you when you are trying to use tail command. You are not actually using it to extract what you want from who command's output but from head command.
If you need to use head and tail combined with piping, you could do it like this for example:
(who |head -n5 && who |tail -n2)
Modify numbers to meet your needs
answered Feb 23 '17 at 3:05
jiipeezz
21125
edited Feb 23 '17 at 16:02
answered Feb 23 '17 at 3:05
jiipeezz
21125
answered Feb 23 '17 at 3:05
jiipeezz
21125
answered Feb 23 '17 at 3:05
jiipeezz
21125
21125
1
This won't work becauseheadwill eat up all of stdin, leaving nothing fortail.
– roaima
Feb 23 '17 at 14:42
@roaima I see where it went wrong now. But there's one weird thing I don't uderstand at all. I tried to run following commandls -l |(head -n5 && tail -n2). I tested it 10 times now, 6 times I got the result I wanted and 4 times I didn't. Why is that?
– jiipeezz
Feb 23 '17 at 15:56
1
@jiipeezz, that very much depends on timing and the number of files in the directory. It depends whether the firstread()done byheadreads the whole output ofls -lor not.
– Stéphane Chazelas
Feb 23 '17 at 18:04
@StéphaneChazelas thanks for elaborating. Now it makes more sense. Corrected the answer to something that works.
– jiipeezz
Feb 23 '17 at 20:38
add a comment |
1
This won't work becauseheadwill eat up all of stdin, leaving nothing fortail.
– roaima
Feb 23 '17 at 14:42
@roaima I see where it went wrong now. But there's one weird thing I don't uderstand at all. I tried to run following commandls -l |(head -n5 && tail -n2). I tested it 10 times now, 6 times I got the result I wanted and 4 times I didn't. Why is that?
– jiipeezz
Feb 23 '17 at 15:56
1
@jiipeezz, that very much depends on timing and the number of files in the directory. It depends whether the firstread()done byheadreads the whole output ofls -lor not.
– Stéphane Chazelas
Feb 23 '17 at 18:04
@StéphaneChazelas thanks for elaborating. Now it makes more sense. Corrected the answer to something that works.
– jiipeezz
Feb 23 '17 at 20:38
1
1
This won't work because
head will eat up all of stdin, leaving nothing for tail.– roaima
Feb 23 '17 at 14:42
This won't work because
head will eat up all of stdin, leaving nothing for tail.– roaima
Feb 23 '17 at 14:42
@roaima I see where it went wrong now. But there's one weird thing I don't uderstand at all. I tried to run following command
ls -l |(head -n5 && tail -n2). I tested it 10 times now, 6 times I got the result I wanted and 4 times I didn't. Why is that?– jiipeezz
Feb 23 '17 at 15:56
@roaima I see where it went wrong now. But there's one weird thing I don't uderstand at all. I tried to run following command
ls -l |(head -n5 && tail -n2). I tested it 10 times now, 6 times I got the result I wanted and 4 times I didn't. Why is that?– jiipeezz
Feb 23 '17 at 15:56
1
1
@jiipeezz, that very much depends on timing and the number of files in the directory. It depends whether the first
read() done by head reads the whole output of ls -l or not.– Stéphane Chazelas
Feb 23 '17 at 18:04
@jiipeezz, that very much depends on timing and the number of files in the directory. It depends whether the first
read() done by head reads the whole output of ls -l or not.– Stéphane Chazelas
Feb 23 '17 at 18:04
@StéphaneChazelas thanks for elaborating. Now it makes more sense. Corrected the answer to something that works.
– jiipeezz
Feb 23 '17 at 20:38
@StéphaneChazelas thanks for elaborating. Now it makes more sense. Corrected the answer to something that works.
– jiipeezz
Feb 23 '17 at 20:38
add a comment |
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you can combine multiple commands inside
()or{}separated by;... see Command-Grouping– Sundeep
Feb 23 '17 at 3:02
First and last users in terms of what? Just from the output?
– DarkHeart
Feb 23 '17 at 4:48
Similar: Command to display first few and last few lines of a file
– Stéphane Chazelas
Feb 23 '17 at 16:19