I am trying to create a script which will list the name of the user, only if that specific user has a specific and available directory in his folder.
#!/bin/bash
for i in `cat /etc/passwd | cut -d : -f5\,6 | cut -d : -f2`
do
cd $i
if [ -d public_html ]
then
echo `cat /etc/passwd | cut -d : -f5,6 | cut -d : -f1`
fi
done
First, I get a list of all the user names that have home folders.
Then, for each user, I enter in his directory
If in his directory, public_html directory is found, echo the user.
When I run this in the terminal, all the users are listed:
cat /etc/passwd | cut -d : -f5,6 | cut -d : -f1
However, I need to somehow get the user i from that whole list.
Can anybody be so kind to explain what I´m doing wrong, and what to look out for?
You're getting the wrong result because you're checking for a file (with -f). Try using -d.
Using cat and cut is not bash either and can easily be replaced with script in most cases.
I'd do something like this:
while IFS=: read -r user pass uid gid desc homedir shell; do [ -d "$homedir"/public_html ] && echo "$user"; done < /etc/passwd
We set the field separator (IFS) to : (since that's what /etc/passwd uses), we read the file in and set the variables we want to use.
Then for each line we do the test and (&&) if the test is successful we echo the result.
The quotes are not really necessary since we know the formatting of the file but good practice.
If users' home directories are set up in the standard way:
cd /home
for pubdir in */public_html/ ; do
echo "${pubdir%%/*}"
done
I did it ghetto style...
#!/bin/bash
counter=0
for i in `cat /etc/passwd | cut -d : -f6`
do
if [ -r $i ]
then
cd $i
if [ -d public_html ]
then
counter2=2
for j in `cat /etc/passwd | cut -d : -f5`
do
counter2=$(($counter2 + 1))
if [ $counter -eq $counter2 ]
then
echo $j
fi
done
fi
counter=$(($counter + 1))
fi
done
#!/bin/bash
getent passwd | awk -F: '{printf "%s\t%s\n",$1, $6}'| while read -r user home
do
if [ -d ${home}/public_html ] ; then
echo ${user}
fi
done
This should work irrespective of use of /etc/passwd or ldap ...
Related
I am trying to get all users whose home directories are mentioned in /etc/passwd but are not present in /home.
The below code gives me all the home directories which are not present in /home but are present in /etc/passwd.
cut -d":" -f6 /etc/passwd | grep home | sort |
while read dir; do [ -e "$dir" ] || echo Missing $dir; done
How do I get the list of corresponding users from the first column and create the corresponding /home directory using mkhomedir_helper(user) one by one from the list?
Other implementation of Barmar's solution:
getent passwd | sort -t: -k6 | while IFS=: read -r u _ _ _ _ d _
do
if [[ "$d" =~ ^/home/ ]] && ! [[ -d "$d" ]]
then printf 'Directory %q missing for user: %q\n' "$d" "$u"
fi
done
Using getent allow to pull the same data whenever it is a file or an NIS (Network Information Service).
[[ "$d" =~ ^/home/ ]] ensure the home directory starts with /home by matching it against Extended Regular Expression. ^/home/
You have to keep the username in the data that you read. You don't need to use cut, read can split the input into fields and assign them to variables.
grep /home/ /etc/passwd | sort -t: -k6 | while IFS=: read -r username _ _ _ _ dir _
do
if ! [ -d "$dir" ]
then
echo "Username $username missing $dir"
mkhomedir_helper "$username"
fi
done
I wrote the following script to remove users from a wordlist. The wordlist has 3 or 4 fields which are First Name, Middle Name, Last Name, and User ID. I am using awk to create a username that comprises of the user's firstname initial, lastname and last two digits of their ID. Then using the command userdel with flag r to remove users with their home directories as well.
However when I run the script it gives me an error saying the following:
Usage: userdel [options] LOGIN
Options:
-f, --force force some actions that would fail otherwise
e.g. removal of user still logged in
or files, even if not owned by the user
-h, --help display this help message and exit
-r, --remove remove home directory and mail spool
-R, --root CHROOT_DIR directory to chroot into
-Z, --selinux-user remove any SELinux user mapping for the user
The script:
#! /bin/bash
# Removing users using positional parameters
getusername(){
line=${1}
len=`echo ${line}|awk '{ FS = " " } ; { print NF}'`
if [[ ${len} -eq 3 ]]
then
initial=`echo ${line}| awk {'print $1'} |cut -c1`
lastname=`echo ${line} | awk {'print $2'}`
id=`echo ${line}| awk {'print $3'}|grep -o '..$'`
username=`echo ${initial}${lastname}${id} |tr '[:upper:]' '[:lower:]'`
elif [[ ${len} -eq 4 ]]
then
initial=`echo ${line} | awk {'print $1'} |cut -c1`
lastname=`echo ${line} | awk {'print $3'}`
id=`echo ${line}| awk {'print $4'}|grep -o '..$'`
username=`echo ${initial}${lastname}${id} |tr '[:upper:]' '[:lower:]'`
else
echo "Line ${line} is not expected as it should be considered for creating Username and Password"
fi
}
sudo userdel -r $getusername
How to invoke userdel?
userdel -r username
This deletes the account of user username, and removes that user's home directory and associated mail files.
For that, you need to use a variable instead of to invoke a function.
Otherwise userdel will complain, like it is already doing or like just typing userdel.
This script looks for all users that have the string RECHERCHE inside them. I tried running it in sudo and it worked, but then stopped at line 8 (permission denied). Even when removing the sudo from the script, this issue still happens.
#!/bin/bash
#challenge : user search and permission rewriting
echo -n "Enter string to search : "
read RECHERCHE
echo $(cat /etc/passwd | grep "/home" | cut -d: -f5 | grep -i "$RECHERCHE" | sed s/,//g)
echo "Changing permissions"
export RECHERCHE
sudo ./challenge2 $(/etc/passwd) &
The second script then changes permissions of each file belonging to each user that RECHERCHE found, in the background. If you could help me figure out what this isn't doing right, it would be of great service. I
#!/bin/bash
while read line
do
if [-z "$(grep "/home" | cut -d: -f5 | grep -i "$RECHERCHE")" ]
then
user=$(cut -f: -f1)
file=$(find / -user $(user))
if [$(stat -c %a file) >= 700]
then
chmod 700 file 2>> /home/$(user)/challenge.log
fi
if [$(stat -c %a file) < 600]
then
chmod 600 file 2>> /home/$(user)/challenge.log
fi
umask 177 2>> /home/$(user)/challenge.log
fi
done
I have to idea what I'm doing.
the $(...) syntax means command substitution, that is: it will be replaced by the output of the command within the paranthesis.
since /etc/passwd is no command but just a text-file, you cannot execute it.
so if you want to pass the contents of /etc/passwd to your script, you would just call it:
./challenge2 < /etc/passwd
or, if you need special permissions to read the file, something like
sudo cat /etc/passwd | ./challenge2
also in your challenge2 script, you are using $(user) which is wrong as you really only want to expand the user variable: use curly braces for this, like ${user}
/etc/passwd?
not what you were asking, but you probably should not read /etc/passwd directly anyhow.
if you want to get a list of users, use the following command:
$ getent passwd
this will probably give you more users than those stored in /etc/passwd, as your system might use other PAM backends (ldap,...)
I recently sorted a bunch of files into directories by subject. I'm trying to change the date of the directory to be the same as the modification time of the newest file in the directory. I have a script that I think should work, but it doesn't.
for DIRECTORY in * ; do FILE=`ls -1 --sort=time "$DIRECTORY" | tail -n 1` ; DATE=`stat -c %y "$DIRECTORY/$FILE" | tr '.' ' ' | awk '{print $1,$2}' ` ; echo $DATE $DIRECTORY ; touch -m -d "$DATE" "$DIRECTORY" ; done
It seems like it should work, but it doesn't. Any help?
Simply run this command :
latestFile=$(ls -1t | sed q)
for i in *; do touch -r "$latestFile"; done
! The command is not compatible with files with spaces !
I run bash scripts from time to time on my servers, I am trying to write a script that monitors log folders and compress log files if folder exceeds defined capacity. I know there are better ways of doing what I am currently trying to do, your suggestions are more than welcome. The script below is throwing an error "unexpected end of file" .Below is my script.
dir_base=$1
size_ok=5000000
cd $dir_base
curr_size=du -s -D | awk '{print $1}' | sed 's/%//g' zipname=archivedate +%Y%m%d
if (( $curr_size > $size_ok ))
then
echo "Compressing and archiving files, Logs folder has grown above 5G"
echo "oldest to newest selected."
targfiles=( `ls -1rt` )
echo "rocess files."
for tfile in ${targfiles[#]}
do
let `du -s -D | awk '{print $1}' | sed 's/%//g' | tail -1`
if [ $curr_size -lt $size_ok ];
then
echo "$size_ok has been reached. Stopping processes"
break
else if [ $curr_size -gt $size_ok ];
then
zip -r $zipname $tfile
rm -f $tfile
echo "Added ' $tfile ' to archive'date +%Y%m%d`'.zip and removed"
else [ $curr_size -le $size_ok ];
echo "files in $dir_base are less than 5G, not archiving"
fi
Look into logrotate. Here is an example of putting it to use.
With what you give us, you lack a "done" to end the for loop and a "fi" to end the main if. Please reformat your code and You will get more precise answers ...
EDIT :
Looking at your reformatted script, it is as said : The "unexpected end of file" comes from the fact you have not closed your "for" loop neither your "if"
As it seems that you mimick the logrotate behaviour, check it as suggested by #Hank...
my2c
My du -s -D does not show % sign. So you can just do.
curr_size=$(du -s -D)
set -- $curr_size
curr_size=$1
saves you a few overheads instead of du -s -D | awk '{print $1}' | sed 's/%//g.
If it does show % sign, you can get rid of it like this
du -s -D | awk '{print $1+0}'. No need to use sed.
Use $() syntax instead of backticks whenever possible
For targfiles=(ls -1rt) , you can omit the -1. So it can be
targfiles=( $(ls -rt) )
Use quotes around your variables whenever possible. eg "$zipname" , "$tfile"