I need to get all users from file, containing information about all loggins within some time interval. (The delimiter is : )
So, I need to get all users from output of command last -f.
I tried to do this:
last -f file| cut -d ":" -f1
but in the output aren't just the usernames. It seems to me like some record take more than just one line and therefore it can't distinguish the records. I don't know.
Could you help me please? I would be grateful for any advice.
You could say:
last -f file | awk '{print $1}'
If you want to use cut, say:
last -f file| cut -d " " -f1
Related
I'm writing a script that checks a list of all the users connected to the server (using who) and writes to the file Information the list of usernames of only those having letters a, b, c or d. This is what I have so far:
who | grep '[a-d]' >> Information
However, the command who displays this:
username pts/148 2019-01-29 16:09 (IP address)
What I don't understand is why my grep search is also displaying the pts/148, date, time, and IP address. I just want it to send the username to the file Information.
Any help is appreciated.
Another way is to use the command cut to get the first part of the string only.
who | cut -f 1 -d ' ' | grep '[a-d]' >> Information
Using awk to output records where the first clumn matches [a-d]:
$ who | awk '$1~/[a-d]/' >> Information
Using grep to search for lines with [a-d] before the first space:
$ who | grep -o "^[^ ]*[a-d][^ ]*" >> Information
You need to get the first word, otherwise grep will display the entire line that has the matching text. You could use awk:
who | awk '{ if (substr($1,1,1) ~ /^[a-d]/ ) print $1 }' >>Information
How - given USER ID as parameter, find out what is his name? The problem is to write a Bash script, and somehow use etc/passwd file.
The uid is the 3rd field in /etc/passwd, based on that, you can use:
awk -v val=$1 -F ":" '$3==val{print $1}' /etc/passwd
4 ways to achieve what you need:
http://www.digitalinternals.com/unix/linux-get-username-from-uid/475/
Try this:
grep ":$1:" /etc/passwd | cut -f 1 -d ":"
This greps for the UID within /etc/passwd.
Alternatively you can use the getent command:
getent passwd "$1" | cut -f 1 -d ":"
It then does a cut and takes the first field, delimited by a colon. This first field is the username.
You might find the SS64 pages for cut and grep useful:
http://ss64.com/bash/grep.html
http://ss64.com/bash/cut.html
is there a posibility to find a word in a file and than to copy the following word?
Example:
abc="def"
bla="no_need"
line_i_need="information_i_need"
still_no_use="blablabla"
so the third line, is exactly the line i need!
is it possible to find this word with shell orders?
thanks for your support
Using an awk with custom field separator it is much simpler:
awk -F '[="]+' '$1=="line_i_need"{print $2}' file
information_i_need
-F '[="]+' sets field separator as 1 or more of = or "
Use grep:
grep file_name line_i_need
It will print:
line_i_need="information_i_need"
This finds the line with grep an cuts the second column using " separator
grep file_name line_i_need | cut -d '"' -f2
I'm having a bit of an issue cutting the output up from egrep. I have output like:
From: First Last
From: First Last
From: First Last
I want to cut out the "From: " (essentially leaving the "First Last").
I tried
cut -d ":" -f 7
but the output is just a bunch of blank lines.
I would appreciate any help.
Here's the full code that I am trying to use if it helps:
egrep '^From:' $file | cut -d ":" -f 7
NOTE: I've already tested the egrep portion of the code and it works as expected.
The cut command lines in your question specify colon-separated fields and that you want the output to consist only of field 7; since there is no 7th field in your input, the result you're getting isn't what you intend.
Since the "From:" prefix appears to be identical across all lines, you can simply cut from the 7th character onward:
egrep '^From:' $file | cut -c7-
and get the result you intend.
you were really close.
I think you only need to replace ":" with " " as separator and add "-" after the "7": like this:
cut -d " " -f 2-
I tested and works pretty well.
The -f argument is for what fields. Since there is only one : in the line, there's only two fields. So changing -f 7 to -f 2- will give you want you want. Albeit with a leading space.
You can combine the egrep and cut parts into one command with sed:
sed -n 's/^From: //gp' $file
sed -n turns off printing by default, and then I am using p in the sed command explicitly to print the lines I want.
You can use sed:
sed 's/^From: *//'
OR awk:
awk -F ': *' '$1=="From"{print $2}'
OR grep -oP
grep -oP '^From: *\K.*'
Here is a Bash one-liner:
grep ^From file.txt | while read -a cols; do echo ${cols[#]:1}; done
See: Handling positional parameters at wiki.bash-hackers.org
cut itself is a very handy tool in bash
cut -d (delimiter character) -f (fields that you want as output)
a single field is given directly as -f 3 ,
range of fields can be selected as -f 5-9
so in your this particular case code would be
egrep '^From:' $file | cut -d\ -f 2-3
the delimiter is space here and can be escaped using a \
-f 1 corresponds to " From " and 2-3 corresponds to " First Last "
I am working with an idea using the unix who command. As we all know, there does not seem to be a direct switch that gives just the username and line (terminal) info without the date and screen info... eg: the output is mneedham tty7...2012-02-19 11:26 (:0)
What I am trying to get is just the mneedham tty7 part. The solution needs to work no matter how long the username and terminal information.
I tried using tr -s ' ' (one space) like who | tr -s ' ' and that gave me one space between everything. Not quite what I was seeking. Tried cut -d" " -f1 gets the username only. So I am hopeful someone can help me find the right command to get both bits of information.
Thanks.
Using cut:
who | cut -d " " -f1,2
Using awk:
who | awk '{ print $1, $2 }'