I was wondering how do you use the cut, sort, and uniq commands in a pipeline and give a command line that indicates how many users are using each of the shells mentioned in /etc/passwd?
i'm not sure if this is right but
cut -f1 -d':' /etc/passwd | sort -n | uniq
?
Summarizing the answers excruciatingly hidden in comments:
You were close, only
as tripleee noticed, the shell is in the seventh field
as shellter noticed, since the shells are not numbers, -n is useless
as shellter noticed, for the counting, there's uniq -c
That gives
cut -f7 -d: /etc/passwd | sort | uniq -c
Related
there is my script:
table_nm=$1
hive_db=$(echo $table_nm | cut -d'.' -f1)
hive_tb=$(echo $table_nm | cut -d'.' -f2)
At first, I got the right result:
$echo "dev.dmf_bird_cost_detail" | cut -d'.' -f1
dev #correct
$echo "dev.dmf_bird_cost_detail" | cut -d'.' -f2
dmf_bird_cost_detail #correct
but,i got something is wrong,if there is no specified character in $table_nm, I get this result：
$echo "dmf_bird_cost_detail" | cut -d'.' -f1
dmf_bird_cost_detail
$echo "dmf_bird_cost_detail" | cut -d'.' -f2
dmf_bird_cost_detail
$echo "dmf_bird_cost_detail" | cut -d'.' -f3
dmf_bird_cost_detail
The result that is not I expected, i hope it's empty, so i conducted some tests and found that if there is no specified character in the string, the command "cut" will return the original value, is that true?
At last,i know "awk" will solves my problem, but I would like to know why "cut" has the above result?
Thank you guys so much!
From POSIX cut specification:
-f list
[...] Lines with no field delimiters shall be passed through intact, unless -s is specified. [...]
why "cut" has the above result?
My guess would be that the first implementations of cut had such behavior (my guess would be it was a bug), and it was preserved and POSIX standardized existing behavior and added -s option. You can browse https://minnie.tuhs.org/cgi-bin/utree.pl for some old verison of cut.
The proper solution is probably to use a parameter expansion anyway.
hive_db=${table_nm%.*}
hive_tb=${table_nm#"$hive_db".}
If you expect more than one dot, you need some additional processing to extract the second field.
Because this uses shell built-ins, it is a lot more efficient than spawning two processes for each field you want to extract (and even then you should use proper quoting).
root:x:0:0:root:/root:/bin/bash
daemon:x:1:1:daemon:/usr/sbin:/usr/sbin/nologin
sys:x:3:1:sys:/dev:/usr/sbin/nologin
games:x:5:2:games:/usr/games:/usr/sbin/nologin
mail:x:8:5:mail:/var/mail:/usr/sbin/nologin
www-data:x:33:3:www-data:/var/www:/usr/sbin/nologin
backup:x:34:2:backup:/var/backups:/usr/sbin/nologin
nobody:x:65534:1337:nobody:/nonexistent:/usr/sbin/nologin
syslog:x:101:1000::/home/syslog:/bin/false
whoopsie:x:109:99::/nonexistent:/bin/false
user:x:1000:1000:edco8700,,,,:/home/user:/bin/bash
sshd:x:116:1337::/var/run/sshd:/usr/sbin/nologin
ntp:x:117:99::/home/ntp:/bin/false
mysql:x:118:999:MySQL Server,,,:/nonexistent:/bin/false
vboxadd:x:999:1::/var/run/vboxadd:/bin/false
this is an /etc/passwd file I need to do this command on. So far I have:
cut -d: -f1,6 testPasswd.txt | grep ???
that will display all the usernames and the folder associated, but I'm stuck on how to find only the ones that start with m,w,s and print the whole line.
I've tried grep -o '^[mws]*' and different variations of it, but none have worked.
Any suggestions?
Try variations of
cut -d: -f1,6 testPasswd.txt | grep '^m\|^w\|^s'
Or to put it more concisely,
cut -d: -f1,6 testPasswd.txt | grep '^[mws]'
That's neater especially if you have a lot of patterns to match.
But of course the awk solution is much better if doing it without constraints.
Easier to do with awk:
awk 'BEGIN{FS=OFS=":"} $1 ~ /^[mws]/{print $1, $6}' testPasswd.txt
sys:/dev
mail:/var/mail
www-data:/var/www
syslog:/home/syslog
whoopsie:/nonexistent
sshd:/var/run/sshd
mysql:/nonexistent
I'm trying to combine the below outputs into one command. The issue is that the field I'm trying to grab is in reverse order. I was told that cut doesn't support a "reverse" option and to use AWK for this purpose but it didn't end up working for my purpose. I'm trying to take the output of the ls- l against the /dev/block to return the partitions and automatically build a dd if= / of= for each outputted line based on the output of the command.
I tried piping the output to awk:
cut -d' ' -f23,25 ... | awk '{print $2,$1}'
however, the result was when using sed to input the prefix and suffix, it wasn't in the appropriate order.
I built the two statements below which individually return the expected output, just looking for the "right" way to combine both of these statements in the most efficient manner using sed / awk.
ls -l /dev/block/platform/msm_sdcc.1/by-name/ | cut -d' ' -f 25 | sed "s/^/dd if=/"
ls -l /dev/block/platform/msm_sdcc.1/by-name/ | cut -d' ' -f 23 | sed "s/.*/of=\/external_sd\/&.dsk/"
Any assistance will be appreciated.
Thank you.
If you're already using awk, I don't think you'll need cut or sed. You can probably do something like the following, though I'll have to trust you on the field numbers
ls -l /dev/block/platform/msm_sdcc.1/by-name | awk '{print "dd if=/"$25 " of=/" $23 ".dsk"}'
awk will split on all whitespace, not just the space character, so it's possible the fields will shift some, though it may be more reliable too.
Okay so lets say I have a list of addresses in a text file like this:
https://www.amazon.com
https://www.google.com
https://www.msn.com
https://www.google.com
https://www.netflix.com
https://www.amazon.com
...
There is a whole bunch of other stuff there but basically the issue I am having is that after running this:
grep "https://" addresses.txt | cut -d"/" -f3
I get amazon.com and google.com twice. I want to only get them once. I don't know how to make the search only grep for things that are unique.
Pipe your output to sort and uniq:
grep "https://" addresses.txt | cut -d"/" -f3 | sort | uniq
you can use sort for this purpose.
just add another pipe to your command and use the unique feature of sort to remove duplicates.
grep 'https://' addresses.txt | cut -d"/" -f3 | sort -u
EDIT: you can use sed instead of grep and cut which would reduce your command to
sed -n 's#https://\([^/]*\).*#\1#p' < addresses.txt | sort -u
I would filter the results post-grep.
e.g. using sort -u to sort and then produce a set of unique entries.
You can also use uniq for this, but the input has to be sorted in advance.
This is the beauty of being able to pipe these utilities together. Rather than have a single grepping/sorting/uniq(ing) tool, you get the distinct executables, and you can chain them together how you wish.
grep "https://" addresses.txt | cut -d"/" -f3 | sort | uniq is what you want
with awk you can use only one unix command instead of four with 3 pipes:
awk 'BEGIN {FS="://"}; { myfilter = match($1,/https/); if (myfilter) loggeddomains[$2]=0} END {for (mydomains in loggeddomains) {print mydomains}}' addresses.txt
how to list all of the users whom has at least one running process.
The user name should not be duplicated.
The user name should be sorted.
$ ps xau | cut -f1 -d " "| sort | uniq | tail -n +2
You may want to weed out names starting with _ as well like so :
ps xau | cut -f1 -d " "| sort | uniq | grep -v ^_ | tail -n +2
users does what is requested. From the man page:
users lists the login names of the users currently on the system, in
sorted order, space separated, on a single line.
Try this:
w -h | cut -d' ' -f1 | sort | uniq
The w -h displays all users in system, without header and some output. The cut part removes all other information without username. uniq ignores duplicate lines.