While loop Bash script that outputs how many users use a certain shell

Question

So I am stuck on trying to create a bash script while loop function that takes the input from /etc/passwd and outputs how many of users use bash, nolongin, csh, and ksh. I believe I am supposed to use cat and cut. And I know that the type of shell a user uses is on the 7th field.

Am I correct in thinking that I should create a variable for each type of shell and increment it each time it finds a user that uses that particular shell?

So far this is the code I have but for some reason it only outputs the final if statement..

declare -i bash;
declare -i nologin;
declare -i csh;
declare -i ksh;

bash=0;
nologin=0;
csh=0;
ksh=0;
cat /etc/passwd | cut -d: -f7 $line | while read line;
do
    if [ $line == "bash" ]
            then
                    bash=bash+1;
    else  fi [ $line == "/sbin/nologin" ]
            then
                    nologin=nologin+1;
    else fi [ $line == "chs" ]
            then
                    chs=chs+1;
    else
            then
                    ksh=ksh+1;
    fi
done

Update: So far I have changed the code and updated what I have. It loops through and gives out the correct portion of the if statement, but I am having trouble trying to increment the variables I have.

Answer 1

Here's my version:

#!/bin/bash

bash=0;
nologin=0;
csh=0;
ksh=0;

while read line;
do
    innerline=$(echo "$line"| cut -d: -f7)

    if [[ $innerline == "/bin/bash" ]]
    then
                    bash=$(echo "$bash+1"|bc);
    elif [[ $innerline == "/sbin/nologin" ]]
    then
                    nologin=&(echo "$nologin+1"|bc);
    elif [[ $innerline == "/bin/csh" ]]
    then
                    csh=$(echo "$csh+1"|bc);
    else
                    ksh=$(echo "$ksh+1"|bc);
    fi
done < /etc/passwd

echo "$bash $nologin $csh $ksh"

To increment the values I pass the calculation job to bc which is small calculator utility.

Let me know if it works for you.

To be honest I must say I really dislike this solution. While loop is unnecessary and I prefer @Nat solution, however since you clearly stated that while loop must be present - here it is.

PS. This is just terrible idea:

else
                        ksh=$(echo "$ksh+1"|bc);

to assume that every unknown/unmatched shell is ksh.

Answer 2

I'm not sure if this is homework so this might not fit your requirements but you don't actually need a whole script to do this. This can be done with a one-liner.

cat /etc/passwd | cut -d ':' -f 7 | sort | uniq -c

Here's what the output looks like

  4 /bin/bash
  6 /sbin/nologin
  4 /usr/bin/csh
 13 /usr/bin/ksh

How this works is that uniq(1) "Filters adjacent matching lines from INPUT" and so with the -c option it will count adjacent lines which is why we need to sort the lines first so that all the matching lines will be adjacent.

Answer 3

Some tought

 cat /etc/passwd | while read line;

will place into line

 tmpfoo:x:1004:1005:utilisateur test sudo:/home/tmpfoo:/bin/bash

then,

  cut -d: -f7 $line == "bash"

will evaluate to

 cut -d: -f7 tmpfoo:x:1004:1005:utilisateur test sudo:/home/tmpfoo:/bin/bash == "bash"

cut will search in the 3 files tmpfoo:x:1004:1005:utilisateur, test and sudo:/home/tmpfoo:/bin/bash which is probably not what you want.

as a gift, try this sample loop

cat /etc/passwd |  cut -d: -f7 | while read line;
do
    echo $line
done

Answer 4

I believe I am supposed to use cat and cut.

It is something I am working on for a Unix class so I am supposed to create a while loop function.

Of course, files where the records are variable-length delimited by newlines and the fields are variable-length delimited by a single delimiter character, like the UNIX Version 7 password file, are things that awk excels at. So for the benefit of those people coming here without the aforegiven constraints, here's the awk one-liner:

awk -F : '{x[$7]++;} END {for (k in x) print k": "x[k]}' /etc/passwd

The lessons to learn here are:

• UNIX has specialized tools tailored to its historical database file formats. This particular type of file format is a breeze to handle in awk.
• You do not have to use cat, despite what instructors may say. ☺ There is such a thing as overuse of command pipelines. In the while loop case simple redirection will do, as in mnmnc's answer. In the case of awk, the files to process are simply given as arguments.
• The while loop here is, of course, inside the awk program. Its primary function is to sequentially read and process all records in its input file(s), which it does with a loop.

• Am I correct in thinking that I should create a variable for each type of shell and increment it each time it finds a user that uses that particular shell?

  Yes, and the awk one-liner here uses what is called an associative array, the x variable. The second part of the awk script shows the usual way to iterate over the contents of such an associative array. The first part is simply an increment (the ++ postfix operator) of the value of one element in the array, indexed by field seven of the record (which awk makes available as $7).

  Note that you can also do associative arrays in Bourne Again shell scripts and adapt this technique to the script for the constrained answer. You may be constrained to using a while loop; but no-one has constrained you to using if statements.

Further reading

• https://askubuntu.com/questions/385528/