Reformat with awk and sed from STDIN and execute - linux

This is just an example of what I run into a lot:
I would like to copy all .bash_histories to one directory.
grep "/bin/bash" /etc/passwd | awk -F: '{ print "cp " $6"/.bash_history /backup" $6 ".bash_history" }
Output:
cp /home/peter/.bash_history /backup/home/peter/.bash_history
cp /home/john/.bash_history /backup/home/john/.bash_history
What I would like is an output like this:
cp /home/peter/.bash_history /backup/_home_peter_.bash_history
cp /home/john/.bash_history /backup/_home_john_.bash_history
And that this output will be executed.
(It's not specifically about this issue, but just in general how to reformat with awk and sed and execute the new created command line, without really creating a script for it)

The awk script to obtain a similar output will be
grep "/bin/bash" /etc/passwd |head -2 | awk -F: '{ print "cp " $6 "/.bash_history backup/_home_"$1".bash_history" }'
giving an output like
cp /root/.bash_history backup/_home_root.bash_history
cp /home/xxx/.bash_history backup/_home_xxx.bash_history
Now inorder to excecute the commands, the system() function within the awk would be helpfull
system(command) would excecute any command, and return value being the exit status of the command.
The above script can be modified as
grep "/bin/bash" /etc/passwd |head -2 | awk -F: '{ system("cp " $6 "/.bash_history backup/_home_"$1".bash_history;") }'
Test run:
$ grep "/bin/bash" /etc/passwd |head -2 | awk -F: '{ system("cp " $6 "/.bash_history backup/_home_"$1".bash_history;") }'
$ ls backup/
_home_xxx.bash_history _home_root.bash_history
PS: It is not recommend to create directories in your root folder. So i intentionally replaced /backup in your script to backup.
Also inorder for the script to be successful, the backup folder must be created before hand.

getent passwd | grep \/bin\/bash | cut -d ":" -f 6 | while read a; do eval "cp $a/.bash_history /backup/$(echo $a | sed 's#/#_#g')_.bash_history"; done
This uses getent to fetch the passwd file and cut gets the 6th field like your awk statement did, then it reads each entry line by line and builds the string and executes it with eval.

getent passwd | grep \/bin\/bash | cut -d ":" -f 6 | while read a; do eval "cp $a/.bash_history /backup/$(echo $a | sed 's#/#_#g')_.bash_history"; done
Worked perfectly! Issue solved!

Related

Script : check 8th field of /etc/shadow

I want to check if the 8th field of /etc/shadow of a username has no entry.
This is my code:
#!/bin/bash
for i in $(cat < "users.txt")
do
sudo grep -w $i /etc/shadow | awk -F: "$8 == ' '" | cut -d: -f1 ;
done
But this is the error that i get when i execute the script
awk: line 1: syntax error at or near ==
awk syntax for this purpose would be:
awk -F 'delim' '$n1 == "text" {print $n2}'
sudo grep -w $i /etc/shadow | awk -F':' '$8 == " " {print $0}'
FYI: /etc/shadow does not contain spaces between colons. so if cat shows
bin:*:1:1:::::::
you should run $8 == ""
note that there is no space.
If n2 is 0, you'd return the entire row. Hope this helps!
Your approach can be greatly simplified, typically using grep and awk on one line indicates that you're overthinking things. Invoking one command instead of three ...
#!/bin/bash
for i in $(cat < "users.txt")
do
sudo awk -F: -v user=$i '$1==user && $8==""{print $1}' /etc/shadow
done

Linux usernames /etc/passwd listing

I want to print the longest and shortest username found in /etc/passwd. If I run the code below it works fine for the shortest (head -1), but doesn't run for (sort -n |tail -1 | awk '{print $2}). Can anyone help me figure out what's wrong?
#!/bin/bash
grep -Eo '^([^:]+)' /etc/passwd |
while read NAME
do
echo ${#NAME} ${NAME}
done |
sort -n |head -1 | awk '{print $2}'
sort -n |tail -1 | awk '{print $2}'
Here the issue is:
Piping finishes with the first sort -n |head -1 | awk '{print $2}' command. So, input to first command is provided through piping and output is obtained.
For the second command, no input is given. So, it waits for the input from STDIN which is the keyboard and you can feed the input through keyboard and press ctrl+D to obtain output.
Please run the code like below to get desired output:
#!/bin/bash
grep -Eo '^([^:]+)' /etc/passwd |
while read NAME
do
echo ${#NAME} ${NAME}
done |
sort -n |head -1 | awk '{print $2}'
grep -Eo '^([^:]+)' /etc/passwd |
while read NAME
do
echo ${#NAME} ${NAME}
done |
sort -n |tail -1 | awk '{print $2}
'
All you need is:
$ awk -F: '
NR==1 { min=max=$1 }
length($1) > length(max) { max=$1 }
length($1) < length(min) { min=$1 }
END { print min ORS max }
' /etc/passwd
No explicit loops or pipelines or multiple commands required.
The problem is that you only have two pipelines, when you really need one. So you have grep | while read do ... done | sort | head | awk and sort | tail | awk: the first sort has an input (i.e., the while loop) - the second sort doesn't. So the script is hanging because your second sort doesn't have an input: or rather it does, but it's STDIN.
There's various ways to resolve:
save the output of the while loop to a temporary file and use that as an input to both sort commands
repeat your while loop
use awk to do both the head and tail
The first two involve iterating over the password file twice, which may be okay - depends what you're ultimately trying to do. But using a small awk script, this can give you both the first and last line by way of the BEGIN and END blocks.
While you already have good answers, you can also use POSIX shell to accomplish your goal without any pipe at all using the parameter expansion and string length provided by the shell itself (see: POSIX shell specifiction). For example you could do the following:
#!/bin/sh
sl=32;ll=0;sn=;ln=; ## short len, long len, short name, long name
while read -r line; do ## read each line
u=${line%%:*} ## get user
len=${#u} ## get length
[ "$len" -lt "$sl" ] && { sl="$len"; sn="$u"; } ## if shorter, save len, name
[ "$len" -gt "$ll" ] && { ll="$len"; ln="$u"; } ## if longer, save len, name
done </etc/passwd
printf "shortest (%2d): %s\nlongest (%2d): %s\n" $sl "$sn" $ll "$ln"
Example Use/Output
$ sh cketcpw.sh
shortest ( 2): at
longest (17): systemd-bus-proxy
Using either pipe/head/tail/awk or the shell itself is fine. It's good to have alternatives.
(note: if you have multiple users of the same length, this just picks the first, you can use a temp file if you want to save all names and use -le and -ge for the comparison.)
If you want both the head and the tail from the same input, you may want something like sed -e 1b -e '$!d' after you sort the data to get the top and bottom lines using sed.
So your script would be:
#!/bin/bash
grep -Eo '^([^:]+)' /etc/passwd |
while read NAME
do
echo ${#NAME} ${NAME}
done |
sort -n | sed -e 1b -e '$!d'
Alternatively, a shorter way:
cut -d":" -f1 /etc/passwd | awk '{ print length, $0 }' | sort -n | cut -d" " -f2- | sed -e 1b -e '$!d'

Shell script to recursively print full directory tree using ls

Assignment: I have to create a shell script using diff and sort, and a pipeline using ls -l, grep '^d', and awk '{print $9}' to print a full directory tree.
I wrote a C program to display what I am looking for. Here is the output:
ryan#chrx:~/Documents/OS-Projects/Project5_DirectoryTree$ ./a.out
TestRoot/
[Folder1]
[FolderC]
[FolderB]
[FolderA]
[Folder2]
[FolderD]
[FolderF]
[FolderE]
[Folder3]
[FolderI]
[FolderG]
[FolderH]
I wrote this so far:
ls -R -l $1 | grep '^d' | awk '{print $9}'
to print the directory tree but now I need a way to sort it by folder depth and possibly indent but not required. Any suggestions? I can't use find or tree commands.
EDIT: The original assignment & restrictions were mistaken and changed at a later date. The current answers are good solutions if you disregard the restrictions so please leave them for any people with similar issues. As for the the new assignment in case anybody was wondering. I was to recursively print all sub directories, sort them, then compare them with my program to make sure they have similar results. Here was my solution:
#!/bin/bash
echo Program:
./a.out $1 | sort
echo Shell Script:
ls -R -l $1 | grep '^d' | awk '{print $9}' | sort
diff <(./a.out $1 | sort) <(ls -R -l $1 | grep '^d' | awk '{print $9}' | sort)
DIFF=$?
if [[ $DIFF -eq 0 ]]
then
echo "The outputs are similar!"
fi
You don't need neither ls nor grep nor awk for getting the tree. The Simple recursive bash function will be enouh, like:
#!/bin/bash
walk() {
local indent="${2:-0}"
printf "%*s%s\n" $indent '' "$1"
for entry in "$1"/*; do
[[ -d "$entry" ]] && walk "$entry" $((indent+4))
done
}
walk "$1"
If you run it as bash script.sh /etc it will print the dir-tree like:
/etc
/etc/apache2
/etc/apache2/extra
/etc/apache2/original
/etc/apache2/original/extra
/etc/apache2/other
/etc/apache2/users
/etc/asl
/etc/cups
/etc/cups/certs
/etc/cups/interfaces
/etc/cups/ppd
/etc/defaults
/etc/emond.d
/etc/emond.d/rules
/etc/mach_init.d
/etc/mach_init_per_login_session.d
/etc/mach_init_per_user.d
/etc/manpaths.d
/etc/newsyslog.d
/etc/openldap
/etc/openldap/schema
/etc/pam.d
/etc/paths.d
/etc/periodic
/etc/periodic/daily
/etc/periodic/monthly
/etc/periodic/weekly
/etc/pf.anchors
/etc/postfix
/etc/postfix/postfix-files.d
/etc/ppp
/etc/racoon
/etc/security
/etc/snmp
/etc/ssh
/etc/ssl
/etc/ssl/certs
/etc/sudoers.d
Borrowing from #jm666's idea of running it on /etc:
$ find /etc -type d -print | awk -F'/' '{printf "%*s[%s]\n", 4*(NF-2), "", $0}'
[/etc]
[/etc/alternatives]
[/etc/bash_completion.d]
[/etc/defaults]
[/etc/defaults/etc]
[/etc/defaults/etc/pki]
[/etc/defaults/etc/pki/ca-trust]
[/etc/defaults/etc/pki/nssdb]
[/etc/defaults/etc/profile.d]
[/etc/defaults/etc/skel]
[/etc/fonts]
[/etc/fonts/conf.d]
[/etc/fstab.d]
[/etc/ImageMagick]
[/etc/ImageMagick-6]
[/etc/pango]
[/etc/pkcs11]
[/etc/pki]
[/etc/pki/ca-trust]
[/etc/pki/ca-trust/extracted]
[/etc/pki/ca-trust/extracted/java]
[/etc/pki/ca-trust/extracted/openssl]
[/etc/pki/ca-trust/extracted/pem]
[/etc/pki/ca-trust/source]
[/etc/pki/ca-trust/source/anchors]
[/etc/pki/ca-trust/source/blacklist]
[/etc/pki/nssdb]
[/etc/pki/tls]
[/etc/postinstall]
[/etc/preremove]
[/etc/profile.d]
[/etc/sasl2]
[/etc/setup]
[/etc/skel]
[/etc/ssl]
[/etc/texmf]
[/etc/texmf/tlmgr]
[/etc/texmf/web2c]
[/etc/xml]
Sorry, I couldn't find a sensible way to use the other tools you mentioned so it may not help you but maybe it'll help others with the same question but without the requirement to use specific tools.

How to sleep for 1 second between each xargs command?

For example, if I execute
ps aux | awk '{print $1}' | xargs -I {} echo {}
I want to let the shell sleep for 1 second between each echo.
How can I change my shell command?
You can use the following syntax:
ps aux | awk '{print $1}' | xargs -I % sh -c '{ echo %; sleep 1; }'
Be careful with spaces and semicolons though. After every command in between brackets, semicolon is required (even after the last one).
Replace echo by some shell script named sleepecho containing
#!/bin/sh
sleep 1
echo $*
If your awk supports it:
ps aux | awk '{ system("sleep 1"); print $1 }' | xargs -I {} echo {}q
or skip awk and xargs altogether
ps aux | while read -r user rest;
echo $user
sleep 1;
done

Get N line from unzip -l

I have a jar file, i need to execute the files in it in Linux.
So I need to get the result of the unzip -l command line by line.
I have managed to extract the files names with this command :
unzip -l package.jar | awk '{print $NF}' | grep com/tests/[A-Za-Z] | cut -d "/" -f3 ;
But i can't figure out how to obtain the file names one after another to execute them.
How can i do it please ?
Thanks a lot.
If all you need the first row in a column, add a pipe and get the first line using head -1
So your one liner will look like :
unzip -l package.jar | awk '{print $NF}' | grep com/tests/[A-Za-Z] | cut -d "/" -f3 |head -1;
That will give you first line
now, club head and tail to get second line.
unzip -l package.jar | awk '{print $NF}' | grep com/tests/[A-Za-Z] | cut -d "/" -f3 |head -2 | tail -1;
to get second line.
But from scripting piont of view this is not a good approach. What you need is a loop as below:
for class in `unzip -l el-api.jar | awk '{print $NF}' | grep javax/el/[A-Za-Z] | cut -d "/" -f3`; do echo $class; done;
you can replace echo $class with whatever command you wish - and use $class to get the current class name.
HTH
Here is my attempt, which also take into account Daddou's request to remove the .class extension:
unzip -l package.jar | \
awk -F'/' '/com\/tests\/[A-Za-z]/ {sub(/\.class/, "", $NF); print $NF}' | \
while read baseName
do
echo " $baseName"
done
Notes:
The awk command also handles the tasks of grep and cut
The awk command also handles the removal of the .class extension
The result of the awk command is piped into the while read... command
baseName represents the name of the class file, with the .class extension removed
Now, you can do something with that $baseName

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