Here is to count the number of sessions by the same login user.
I could run the direct command if I know the specific user name, such as usera, as the following:
who | grep usera | wc -l
And if I don't know the current user, I need to user parameter.
But the following codes don't work:
currentuser=`whoami`
sessionnumber=`who | grep "$currentuser" | wc -l`
What's the error?
Thanks!
Grep has the -c flag so the wc -l plus the additional pipe is not needed.
who | grep -c -- "$USER"
"$LOGNAME" is also an option instead of "$USER", which one is bash specific? I don't know, all I know is that they are both on Linux and FreeBSD system. The -- is just a habit just in case the user starts with a dash grep will not interpret it as an option.
sessionnumber=`who | grep "$currentuser" | wc -l`
You are assigning the result of the who | ... command to a variable and to see its value you can use echo $sessionnumber
Looks like you are confused about parameters and variables.
What you are trying to get is likely
who | grep $(whoami) | wc -l
The $() is equivalent to the backticks you used.
When you write
sessionnumber=``
this will run whatever is within the backticks and save the output to a variable. You can then access the variable using the dollar notation:
echo "$sessionnumber"
Related
I want to have a single command, not a script, that I can use to define a variable as a command output and then put that output into other commands. My best guess as to what that would look like is this:
LIST=$(ls) | head -1 | echo "${LIST}"
One reason that I want to do this is so that I can create a command that can find the pid of a program and then kill that pid. My best try doing that is:
DiscPid=$(ps -e | grep Discord | cut -b 1-5 | head -1 \ ) | kill "${DiscPid}"
But I'm not able to get this to work
I'm still relatively new to Linux and Bash so any help would be appreciated
Use a semicolon after setting the variable: var=foo; echo $var
You don't need piping here
DiscPid=$(code) | kill "${DiscPid}"
Change '|' to ';' and this one '\' also spare
DiscPid=$(ps -e | grep Discord | cut -b 1-5 | head -1); kill "${DiscPid}"
I created a script called monitornsuaccounts.sh that should append its output file to useraccountstatus.log. useraccountstatus.log is in the directory /var/local/nsu/logs/.
The output of this script should state every username and the following information about each username: username, last login, user home directory and associated groups. Preferably there should be columns with each information.
The command I use for the usernames is sudo cat /etc/passwd | grep ‘/home’. Last is to find the last login of each user. Groups is to the find the group of each user. When I run the command, the output file only shows the data I need for my current user rather than all users. Any recommendations that anyone has would be greatly appreciated.
#!/bin/bash
usernames=sudo cat /etc/passwd | grep ‘/home’
echo “$usernames” > /home/daniel/names.txt
mlast=$(cat names.txt | xargs -n1 last)
mgroup=$(cat names.txt | xargs -n1 groups)
cat names.txt > /var/local/nsu/logs/useraccountstatus.log
echo “$mlast” >>/var/local/nsu/logs/useraccountstatus.log
echo “$mgroup” >>/var/local/nsu/logs/useraccountstatus.log
There are a lot of issues in your script.
Your definition of users. Are you sure that this is what you want? For example: root does not have a directory under /home.
Watch your quotes. cat /etc/passwd | grep ‘/home’ returns nothing, while cat /etc/passwd | grep 'home' returns a list of stanzas in /etc/passwd
You'll probably want just a list of usernames, not a list of stanzas. Something along the line of
cat /etc/passwd | grep 'home' | sed 's/:.*//'
Why sudo in sudo cat /etc/passwd?
Look at your assignment in the
usernames=sudo cat /etc/passwd | grep ‘/home’
This does not make sense. You might try to do a
usernames=`sudo cat /etc/passwd | grep '/home'| sed 's/:.*//'`
And that is just the first line of the script.
Anyway, if your script does not work as intended, you will need to do some debugging. First question, especially if you are inexperienced, is "do the commands that I write give the result that I expect?" So in your case, you should have tried cat /etc/passwd | grep ‘/home’ and you would have seen that it does not give you the expected results. Even with the correct quotes, you'll get a list of stanzas, which is also not what you expected. Have you looked at /home/daniel/names.txt and was the content of the file what you wanted? I guess not: it was empty.
Just a quick hint, to get you started in the right direction (although there are still some issues and pepole might object to the backtics)
#!/bin/bash
usernames=`sudo cat /etc/passwd | grep '/home'| sed 's/:.*//'`
mlast=`echo $usernames | xargs -n1 last`
mgroup=`echo $usernames| xargs -n1 groups`
echo $usernames > /var/local/nsu/logs/useraccountstatus.log
echo "$mlast" >>/var/local/nsu/logs/useraccountstatus.log
echo "$mgroup" >>/var/local/nsu/logs/useraccountstatus.log
You will want to polish this and make the output more useful.
Is there way of counting the number of processes being run by a user in the unix/linux/os x terminal?
For instance, top -u taha lists my processes. I want to be able to count these.
This will show all of the users with their counts (I believe this would be close enough for you. :)
ps -u "$(echo $(w -h | cut -d ' ' -f1 | sort -u))" o user= | sort | uniq -c | sort -rn
You can use ps to output it and count the number using wc, as:
ps -u user | sed 1d | wc -l
You can also dump top output and grep it, something like:
top -u user -n1 | grep user | wc -l
I'm somewhat new to *nix, so perhaps I did not fully understand the context of your question, but here is a possible solution:
jobs | wc -l
The output of the above command is a count of all the processes reported by the jobs command. You can manipulate the parameters of the jobs command to change which processes get reported.
EDIT: Just FYI, this would only work if interested in commands originating from a particular shell. If you want more control in looking at system-wide processes you probably want to use ps as others have suggested. However, if you use wc to do your counting, make sure you take into account any extraneous white space jobs, ps or top may have generated as that will affect the output of wc.
I have a problem with one script in bash with CAT command.
This works:
#!/bin/bash
fil="| grep LSmonitor";
log="/var/log/sys.log ";
lines=`cat $log | grep LSmonitor | wc -l`;
echo $lines;
Output: 139
This does not:
#!/bin/bash
fil="| grep LSmonitor";
log="/var/log/sys.log ";
string="cat $log $fil | wc -l";
echo $string;
`$string`;
Output:
cat /var/log/sys.log | grep LSmonitor | wc -l
cat: opcion invalida -- 'l'
Pruebe 'cat --help' para mas informacion.
$fil is a parameter in this example static, but in real script, parameter is get from html form POST, and if I print I can see that the content of $fil is correct.
In this case, since you're building a pipeline as a string, you would need:
eval "$string"
But DON'T DO THIS!!!! -- someone can easily enter the filter
; rm -rf *
and then you're hosed.
If you want a regex-based filter, get the user to just enter the regex, and then you'll do:
grep "$fil" "$log" | wc -l
Firstly, allow me to say that this sounds like a really bad idea:
[…] in real script, parameter is get from html form POST, […]
You should not be allowing the content of POST requests to be run by your shell. This is a massive attack vector, and whatever mechanisms you have in place to try to protect it are probably not as effective as you think.
Secondly, | inside variables are not treated as special. This isn't specific to backticks. Parameter expansion (e.g., replacing $fil with | grep LSmonitor) happens after the command is parsed and mostly processed. There's a little bit of post-processing that's done on the results of parameter expansion (including "word splitting", which is why $fil is equivalent to the three arguments '|' grep LSmonitor rather than to the single argument '| grep LSmonitor'), but nothing as dramatic as you describe. So, for example, this:
pipe='|'
echo $pipe cat
prints this:
| cat
Since your use-case is so frightening, I'm half-tempted to not explain how you can do what you want — I think you'll be better off not doing this — but since Stack Overflow answers are intended to be useful for more people than just the original poster, an example of how you can do this is below. I encourage the OP not to read on.
fil='| grep LSmonitor'
log=/var/log/sys.log
string="cat $log $fil | wc -l"
lines="$(eval "$string")"
echo "$lines"
Try using eval (taken from https://stackoverflow.com/a/11531431/2687324).
It looks like it's interpreting | as a string, not a pipe, so when it reaches -l, it treats it as if you're trying to pass in -l to cat instead of wc.
The other answers outline why you shouldn't do it this way.
grep LSmonitor /var/log/syslog | wc -l will do what you're looking for.
an Unix shell script with only purpose - count the number of running processes of qmail (could be anything else). Easy thing, but there must be some bug in code:
#!/bin/bash
rows=`ps aux | grep qmail | wc -l`
echo $rows
Because
echo $rows
always shows greater number of rows (11) than if I just count rows in
ps aux | grep qmail
There are just 8 rows. Does it work this way on your system too?
Nowadays with linux, there is pgrep. If you have it on your system, you can skip grep -v grep
$ var=$(pgrep bash) # or `pgrep bash | wc -l`
$ echo $var
2110 2127 2144 2161 2178 2195 2212 2229
$ set -- $var; echo ${#}
8
also, if your ps command has -C option, another way
$ ps -C bash -o pid= | wc -l
if not, you can set a character class in your grep pattern
$ ps aux|grep [q]mail | wc -l
It appears that you're counting the grep process itself and the header line that ps normally prints before its output.
I'd suggest something more like:
qprocs=$(ps auxwww | grep -c "[q]mail")
... note that GNU grep has a "-c" switch to have it print a "count" of matches rather than the lines themselves. The trick with the regular expression here is to match qmail without matching the literal string that's on the grep command line. So we take any single character in the string and wrap it in square brackets such that it is a single character "class." The regexp: [q]mail matches the string qmail without matching the string [q]mail.
Note that even with this regex you may still find some false positive matches. If you really want to be more precise then you should supply a custom output format string to your ps command (see the man pages) or you should feed it through a pipemill or you should parse the output of the ps command based on fields (using awk or cut or a while read loop). (The -o option to ps is by far the easiest among these).
No, since I'm not running qmail. However, you will want to, at a bare minimum, exclude the process running your grep:
ps aux | grep qmail | grep -v grep
For debugging, you may want to do:
rows=`ps aux | grep qmail`
echo $rows >debug.input
od -xcb debug.input
(to see your input to the script in great detail) and then rewrite your script temporarily as:
#!/bin/bash
rows=`cat debug.input | wc -l`
echo $rows
That way, you can see the input and figure out what effect it's having on your code, even as you debug it.
A good debugger will eventually learn to only change one variable at a time. If your changing your code to get it working, that's the variable - don't let the input to your code change as well.
Use
$ /sbin/pidof qmail
A few ways...
ps -e | grep ' [q]mail' | wc -l
ps -C qmail -opid= | wc -l
pidof qmail | tr ' ' '\n' | wc -l
pgrep is on many Linux distributions, and I imagine available for other Unices.
[dan#khorium ~]$ whatis pgrep
pgrep (1) - look up or signal processes based on name and other attributes
[dan#khorium ~]$ pgrep mingetty
1920
1921
1922
1923
1924
In your case, pgrep qmail | wc -l should do the trick.