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 }'
Related
I just have the following cut code and would like to know what is the $DATA_STORE
x=`dd if=$FILEPATH bs=2 count=1`
log "---> x = $x"
DATA_STORE=`echo $x | od -c | awk '{print ($2 $3)}'`
log "---> Data_which_stored_is= $DATA_STORE"
i have a xml file for check start with :
.
i just want to know what exactly this command do .
Thanks for helping :).
This ugly piece of code using dd, awk and od tries to represent in a human readable way the two first bytes of a file.
You an achieve the same result by using only od:
DATA_STORE=$(od -c -N2 -An "$FILEPATH")
If necessary, you can remove the blanks from the output with some option (I can't remember) or with | tr -d ' '.
I tried to automate the process of cleaning up various wordlists I am working with. This is the following code for it:
#!/bin/bash
# Removes spaces and duplicates in a wordlist
echo "Please be in the same directory as wordlist!"
read -p "Enter Worldlist: " WORDLIST
RESULT=$( awk '{print length, $0}' $WORDLIST | sort -n | cut -d " " -f2- )
awk '!(count[$0]++)' $RESULT > better-$RESULT
This is the error I recieve after running the program:
./wordlist-cleaner.sh: fork: Cannot allocate memory
First post, I hope I formatted it correctly.
You didn't describe your intentions or desired output, but I guess this may do what you want
awk '{print length, $0}' "$WORDLIST" | sort -n | cut -d " " -f2- | uniq > better-RESULT
Notice that it's better-RESULT instead of better-$RESULT as you don't want that as a filename.
Yeah okay I got it to run successfully. I was trying to clean up wordlists I was downloading of the net. I have some knowledge of the basic variable usage in Bash, but not enough of the text manipulation commands like sed or awk. Thanks for the support.
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
I would like to grab a substring of a file to get the default password of mysql in centos.
This is the command I am using to get the password:
sudo grep 'temporary password' /var/log/mysqld.log
which result is:
2018-02-21T07:03:11.681201Z 1 [Note] A temporary password is generated for root#localhost: >KkHAt=#z6OV
Now, I am using this command to get the password only and remove the unnecessary stuff, so I can use it in a script:
sudo grep 'temporary password' /var/log/mysqld.log | cut -d ':' -f 4 | cut -d ' ' -f 2
But using 2 cuts seems very ugly. Is there another command or tool that I can use, or a more elegant way to do this?
Using awk:
$ awk '/temporary password/{print $NF}' file
>KkHAt=#z6OV
Bearing in mind that awk splits the lines in fields based on a field separator (by default whitspaces) and NF refers to the number of fields, you can print the last field with:
$ grep 'temporary password' /var/log/mysqld.log | awk '{print $NF}'
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