I am trying to make a shell script that searches the /proc directory and finds the user of the process. how am i supposed to do that? i am trying different ways with grep like
NUM=`egrep -ri "$a1" /proc/ 2> error.txt `
echo "$NUM"
but nothing happen.Can someone help?
grep searches the contents of files. To search files owned by someone, try something like
find /proc -user someone -ls
Trivially, you could pipe ls through a filter to obtain only those lines where the third field is someone, as hinted at in comments;
ls -l | awk '$3=="someone"'
... but for many various reasons you should avoid parsing ls output in general.
If you know the numeric user ID, you could do something like
awk '$1=="Uid" && $2==1234 { print FILENAME }' /proc[1-9]*/status
to obtain the file names of the processes owned by UID number 1234. Something very similar could be obtained by find, as in the first example. (Maybe restrict search to only numeric directory names?)
To find the userid of a process via the /proc pseudo-filesystem, grab the "Uid" line from the "status" file:
uid_line=`grep -e ^Uid /proc/1/status`
You can parse that apart into real, effective, saved set, and filesystem uids. The second field is the real uid:
real_uid=`echo "$uid_line"|cut -f 2`
You can also convert uid into username using getent:
username=`getent passwd $real_uid|cut -d: -f 1`
I am pretty sure userids for processes are in /proc/PID/loginid.
Related
I have to write a bash script for university, the text says:
Write a bash script that allows root user, to get a list of all users
of the machine. Selecting a user, using select, will be required to
indicate a directory (indicate the absolute path). At this point in
the output will have be shown a list of all files folder owned by the
user, ranked in ascending order according to the size of file.
To check if the user is root i used:
if[ "$(id -u)" = 0 ]; then
To get the list of users of the machine I was thinking of using awk:
awk -F':' '{ print$1}' /etc/passwd
How can I use select with awk?
Is there another way without using awk?
Thank you so much in advance
Here is the way to use awk in select statement, you need finish the rest for your homework (for example, sort the result)
#!/usr/bin/env bash
select user in $(awk -F ":" '{print $1}' /etc/passwd )
do
read -p "input the absolute directory: " path
find $path -type f -user "$user" -ls
done
Another way to test the UID by arithmetic (smarter!?) is :
if((UID==0)); then
...
else
...
fi
Check http://wiki.bash-hackers.org/syntax/arith_expr
users='awk '{print $1}' /etc/passwd | sort -u'
for user in $users
do
echo " - $user"
done
this is my shell script . Problem is that show's an error.
the error is ---> users: command not found
please give me the solution frinds
With the code the way it is now I see that you're not assigning the output of the awk|sort command to the variable (maybe you wanted to use ` instead of ' ?)
This works:
#!/bin/bash
users=$(awk '{print $1}' /etc/passwd | sort -u)
for user in $users
do
echo " - $user"
done
Although you should be aware that /etc/passwd is not separated by spaces, so awk '{print $1}' won't give you the user's name (which maybe is what you wanted)
Edit:
As per #Andy Lester's comment to your question: If you save this code in a file (let's say /tmp/myscript.bash) to run it you have to type in a terminal:
/bin/bash /tmp/myscript.bash
or, since it starts with #!/bin/bash (read here) you could make it executable (using chmod u+x /tmp/myscript.bash) and then call it, just typing /tmp/myscript.bash. You can also save it in one of the PATH directories (type echo $PATH to see which are they), make it executable and then you'll be able to call it from anywhere, but I don't really recommend doing that because you may end up overwriting juicy system's commands if you're not careful. For instance, let's say you call your script with the unfortunate name of ls, save it in the first directory of the $PATH (in my case, /usr/local/sbin) Every time you type ls, you won't be listing directories, but calling your script... Which is bad.
I am creating report for usage of shared Linux nfs mount point and sending via email. I am using du Linux command to get directory size but I need also to have information about who own these files. Is there a way to get information like this:
SIZE[MB] FILENAME OWNER
631746 /logs1 user1
372477 /logs2 user2
372477 /data3 user1
191846 /data2 user2
Thank You
Not possible with du. Try writing some custom bash script
Look into combining your commands with ls -l which will give user/group and other information, then pipe it to sed or cut for example and pipe on to output.
Something along the lines of
for i in ./* # or however you get your filenames
do
printf '%s\t%s\n' "$(du -hs "$i")" "$(ls -ld "$i" | awk '{print $3}')"
done
will do it. Though the columns to be nicely aligned in (most/all) cases, it would require some more work.
Out of many results returned by grepping a particular pattern, if I want to use all the results one after the other in my script, how can I go about it?For e.g. I grep for .der in a certificate folder which returns many results. I want to use each and every .der certificate listed from the grep command. How can I use one file after the other out of the grep result?
Are you actually grepping content, or just filenames? If it's file names, you'd be better off using the find command:
find /path/to/folder -name "*.der" -exec some other commands {} ";"
It should be quicker in general.
One way is to use grep -l. This ensures you only get every file once. -l is used to print the name of each file only, not the matches.
Then, you can loop on the results:
for file in `grep ....`
do
# work on $file
done
Also note that if you have spaces in your filenames, there is a ton of possible issues. See Looping through files with spaces in the names on the Unix&Linux stackexchange.
You can use the output as part of a for loop, something like:
for cert in $(grep '\.der' *) ; do
echo ${cert} # or something else
done
Of course, if those der things are actually files (and you're using ls | grep to get them), you can directly use the files:
for cert in *.der ; do
echo ${cert} # or something else
done
In both cases, you may need to watch out for arguments with embedded spaces.
This is part of a homework problem in a beginning bash class.
I need to bring in the passwd file, which I have done with my passfile variable, then I need to be able to extract certain pieces of it and display the different fields. When I manually grep from CLI using this statement below it works fine. I'm wanting all the variables and I get them all.
grep 1000 passfile | cut -c1-
However, when I do this from the script it stops or breaks or starts over at the first 'blank space' in the users full name. John D. Doe will return 3 lines when I only want one. I see this by echoing the value of i and the following.
for i in `grep 1000 ${passfile} | cut -c1-
user=`echo $1 | cut -d : -f1`
userID=`echo $1 | cut -d : -f3`
For example, if the line reads
jdoe:x:123:1000:John D Doe:/home/jdoe:/bin/bash
I get the following:
i = jdoe:x:123:1000:John
which gives me:
User is jdoe, UID is 509
but then in the next line i starts at R.
i = R. so User is R., UID is R.
next line
i = Johnson:/home/jjohnson:/bin/bash
which returns User is Johnson, UID is /bin/bash
The passwd file holds many users so I need to use the for loop to process them all. I think if I can get it to ignore the space I can get it. But not knowing a whole lot about linux, I'm not sure if I'm even going down the right path. Thanks in Advance for guidence/help.
By default, cut splits on spaces, not colons. If you continue to use it, specify the separator.
You probably want to use IFS=: and a read statement in a while loop to get the values in:
while IFS=: read user password uid gid comment home shell
do
...whatever...
done < /etc/passwd
Or you can pipe the output of grep into the while loop.
Are you allowed to use any external program? If so, I'd recommend awk
UID=1000
awkcmd="\$4==\"$UID\" {print \"user:\",\$1}"
cat $PASSWORDFILE | awk -F ":" "$awkcmd"
when parsing structured files with specific field delimiters such as passwd file, the appropriate tool for the job is awk.
UID=1000
awk -vuid="$UID" '$4==uid{print "user: "$1}' /etc/passwd
you do not have to use grep or cut or anything else. ( Of course, you can also use pure bash while read loops as demonstrated.)