I need a grep or another like command that will pull the usernames only from the /etc/passwd file in linux. Anything before the colon. I know this is doing with reg ex however I am not nearly experienced enough...
The following command will give all ACTUAL users, I need a way to pipe to grep or another line of code to only display the username portion.
awk -v LIMIT=500 -F: '{print $1}' '($3>=LIMIT) && ($3!=65534)' /etc/passwd
This should do:
awk -F: '$3>=LIMIT && $3!=65534 {print $1}' LIMIT=500 /etc/passwd
To do this in mostly plain bash:
limit=500
nfsnobody_id=65534
cut -d: -f1,3 /etc/passwd | while IFS=: read username uid; do
(( uid >= limit && uid != nfsnobody_id )) && echo $username
done
Get out of the habit of using VARNAMES_IN_CAPS: one day you'll write PATH=$(dirname $FILE) and then wonder why commands can no longer be found.
You simply need to rewrite the filter part of your awk command a bit. For instance, the following should work:
awk -v LIMIT=500 -F: '{if (($3>=LIMIT) && ($3!=65534)) print $1}' /etc/passwd
Otherwise, to answer your question strictly, if you want to use grep, the command would be
... | grep -o "^[^:]*"
but that would not be the way to go.
Related
This question already has answers here:
How do I use shell variables in an awk script?
(7 answers)
Closed 1 year ago.
good morning, I am a newbie in the world of scripts and I have this problem and I don't know why it happens to me and why I have it wrong, thanks in advance.
For example, I have this command that searches for users with X less letters:
cut -d: -f1 /etc/passwd | awk 'length($1) <= 4'
It works correctly but when I substitute a 4 for a variable with the same value it doesn't do it well:
number=4
echo -e $(cut -d: -f1 /etc/passwd | awk 'length($1) <= $number')
The same error happens to me here too, when I search for users who have an old password
awk -F: '{if($3<=18388)print$1}' < /etc/shadow
Works, but when I use the variable it stops working
variable=18388
awk -F: '{if($3<=$variable)print$1}' < /etc/shadow
Consider using awk's ability to import variables via the -v option, eg:
number=4
cut -d: -f1 /etc/passwd | awk -v num="${number}" 'length($1) <= num'
Though if the first field from /etc/passwd contains white space, and you want to consider the length of the entire field, you should replace $1 with $0, eg:
number=4
cut -d: -f1 /etc/passwd | awk -v num="${number}" 'length($0) <= num'
Even better, eliminate cut and the subprocess (due to the pipe) and use awk for the entire operation by designating the input field separator as a colon:
number=4
awk -F':' -v num="${number}" 'length($1) <= num { print $1 }' /etc/passwd
You need to use double quotes:
echo -e $(cut -d: -f1 /etc/passwd | awk "length(\$1) <= $number")
In single quotes, variables are not interpreted.
Updated: Here is an illustrative example:
> X=1; echo '$X'; echo "$X"
$X
1
Update 2: as rightfully pointed out in the comment, when using double quotes one need to make sure that the variables that are meant for awk to interpret are escaped, so they are not interpreted during script evaluation. Command updated above.
Alternatively, this task could be done using only a single GNU sed one-liner:
num=4
sed -E "s/:.*//; /..{$num}/d" /etc/passwd
Another one-liner, using only grep would be
grep -Po "^[^:]{1,$num}(?=:)" /etc/passwd
but this requires a grep supporting perl regular expressions, for the lookahead construct (?=...).
All the users in the system that dosen't have as an ending character on their names a, s, t, r, m, z must be shown in Bash.
The users names can be obtained from the /etc/passwd file, in the first column. But I can not perceive the correct approach to exclude those characters from the search.
Should I use grep? Or just a cut?
Something like
grep -o '^[^:]*[^astrmz:]:' /etc/passwd | tr -d :
or
cut -d: -f1 /etc/passwd | grep '[^astrmz]$'
[^blah] matches any character but the ones listed, the opposite of [blah].
GNU grep using a lookahead:
grep -Po '^[^:]*[^astrmz:](?=:)' /etc/passwd
Or using awk instead:
awk -F: '$1 ~ /[^astrmz]$/ { print $1 }' /etc/passwd
Or in pure bash without external commands:
while IFS=: read -r name rest; do
if [[ $name =~ [^astrmz]$ ]]; then
echo "$name"
fi
done < /etc/passwd
As you can see, there's lots of potential approaches.
Simple one liner when using bash:
compgen -u | grep -v '[astrmz]$'
The compgen -u command will produce a list of users (without all of the extra fields present in /etc/passwd); compgen is a builtin in bash where it's normally used for username completion.
This should do the trick:
cut -d: -f1 /etc/passwd | grep -vE 's$|t$|r$|m$|z$'
The cut command strips out the username from the password file.
Then grep -v (does the UNMATCHING)
grep -E does multiple matching (OR OR OR)
the $ sign indicates the last character to match
For example , on my mac, I get:
_gamecontrollerd
_ondemand
_wwwproxy
_findmydevice
_ctkd
_applepay
_hidd
_analyticsd
_fpsd
_timed
_reportmemoryexception
(You see no names end with those 5 letters).
Good Luck.
I am using grep to parse the password file.
When I use
grep -w "avahi" /etc/passwd
I get two responses
avahi and avahi-autoipd
I have not found a method to give me the unique response.
This command is part of a bigger script where the name (avahi) is actually a variable.
This does work when the name is rpc and rpcuser. So I am guessing the it has something to do with the dash (-) in the name.
Actual code:
#!/bin/ksh
getent shadow |cut -d: -f1-2|grep ':!!'| cut -d: -f1 > /tmp/pasck
while read line
do
NOLOGIN=`grep -w $line /etc/passwd | cut -d -f7|cut -d/ -f3`
if [[ $NOLOGIN != "nologin: && $NOLOGIN != "false" ]] ; then
echo "$line" "$NOLOGIN" >> /tmp/pasck.list
fi
done <?tmp/pasck
The script is trying to go through the shadow file and look for users with no passwords. Then I compare the results to the passwd file to find which of those accounts are set to /bin/false or /sbin/nologin. The remainder would be actual users with no password set but allowed on the system.
Keeping things simple - you could include colon that comes after the username in your grep statement:
$ grep "^avahi:" /etc/passwd
You can use awk and explicitly test the first colon-separated field:
awk -F: '$1 = "avahi"' /etc/passwd
Suppose I have a file input.txt with few columns and few rows, the first column is the key, and a directory dir with files which contain some of these keys. I want to find all lines in the files in dir which contain these key words. At first I tried to run the command
cat input.txt | awk '{print $1}' | xargs grep dir
This doesn't work because it thinks the keys are paths on my file system. Next I tried something like
cat input.txt | awk '{system("grep -rn dir $1")}'
But this didn't work either, eventually I have to admit that even this doesn't work
cat input.txt | awk '{system("echo $1")}'
After I tried to use \ to escape the white space and the $ sign, I came here to ask for your advice, any ideas?
Of course I can do something like
for x in `cat input.txt` ; do grep -rn $x dir ; done
This is not good enough, because it takes two commands, but I want only one. This also shows why xargs doesn't work, the parameter is not the last argument
You don't need grep with awk, and you don't need cat to open files:
awk 'NR==FNR{keys[$1]; next} {for (key in keys) if ($0 ~ key) {print FILENAME, $0; next} }' input.txt dir/*
Nor do you need xargs, or shell loops or anything else - just one simple awk command does it all.
If input.txt is not a file, then tweak the above to:
real_input_generating_command |
awk 'NR==FNR{keys[$1]; next} {for (key in keys) if ($0 ~ key) {print FILENAME, $0; next} }' - dir/*
All it's doing is creating an array of keys from the first file (or input stream) and then looking for each key from that array in every file in the dir directory.
Try following
awk '{print $1}' input.txt | xargs -n 1 -I pattern grep -rn pattern dir
First thing you should do is research this.
Next ... you don't need to grep inside awk. That's completely redundant. It's like ... stuffing your turkey with .. a turkey.
Awk can process input and do "grep" like things itself, without the need to launch the grep command. But you don't even need to do this. Adapting your first example:
awk '{print $1}' input.txt | xargs -n 1 -I % grep % dir
This uses xargs' -I option to put xargs' input into a different place on the command line it runs. In FreeBSD or OSX, you would use a -J option instead.
But I prefer your for loop idea, converted into a while loop:
while read key junk; do grep -rn "$key" dir ; done < input.txt
Use process substitution to create a keyword "file" that you can pass to grep via the -f option:
grep -f <(awk '{print $1}' input.txt) dir/*
This will search each file in dir for lines containing keywords printed by the awk command. It's equivalent to
awk '{print $1}' input.txt > tmp.txt
grep -f tmp.txt dir/*
grep requires parameters in order: [what to search] [where to search]. You need to merge keys received from awk and pass them to grep using the \| regexp operator.
For example:
arturcz#szczaw:/tmp/s$ cat words.txt
foo
bar
fubar
foobaz
arturcz#szczaw:/tmp/s$ grep 'foo\|baz' words.txt
foo
foobaz
Finally, you will finish with:
grep `commands|to|prepare|a|keywords|list` directory
In case you still want to use grep inside awk, make sure $1, $2 etc are outside quote.
eg. this works perfectly
cat file_having_query | awk '{system("grep " $1 " file_to_be_greped")}'
// notice the space after grep and before file name
I have a number of log files in a directory. I am trying to write a script to search all the log files for a string and echo the name of the files and the line number that the string is found.
I figure I will probably have to use 2 grep's - piping the output of one into the other since the -l option only returns the name of the file and nothing about the line numbers. Any insight in how I can successfully achieve this would be much appreciated.
Many thanks,
Alex
$ grep -Hn root /etc/passwd
/etc/passwd:1:root:x:0:0:root:/root:/bin/bash
combining -H and -n does what you expect.
If you want to echo the required informations without the string :
$ grep -Hn root /etc/passwd | cut -d: -f1,2
/etc/passwd:1
or with awk :
$ awk -F: '/root/{print "file=" ARGV[1] "\nline=" NR}' /etc/passwd
file=/etc/passwd
line=1
if you want to create shell variables :
$ awk -F: '/root/{print "file=" ARGV[1] "\nline=" NR}' /etc/passwd | bash
$ echo $line
1
$ echo $file
/etc/passwd
Use -H. If you are using a grep that does not have -H, specify two filenames. For example:
grep -n pattern file /dev/null
My version of grep kept returning text from the matching line, which I wasn't sure if you were after... You can also pipe the output to an awk command to have it ONLY print the file name and line number
grep -Hn "text" . | awk -F: '{print $1 ":" $2}'