passing a list of dynamic names to a "PS" command in shell script?

Shell Programming and Scripting
1

Hi,

I am new to shell script. This is my first post .I have written a small script which returns list of names starts with "ram" in /etc/passwd .Here is that:-

#!/bin/ksh

NAME_LIST="name_list.txt"
cat /dev/null > $NAME_LIST

evalcmd="cat /etc/passwd | grep "^ram?*" | cut -d: -f1"
eval $evalcmd > $NAME_LIST 2>&1

echo $?

if [[ $? -ne 0 ]] then
echo "Failed to create list of names";
else
echo "List of names are created successfully";
fi

The thing is that i need to pass these dynamic names from /etc/passwd to a "ps" command like:-

ps -o user,fname -U ram,ramdev1,ramdev2,ramdev3

Since i cannot hardcode the names like ram,ramdev1,ramdev2,etc i need to pass these names in a single
command. something like:-

cat /etc/passwd | grep "^ram?*" | cut -d: -f1| ps --o user,fname -U <dynamic variable which fetches the whole name in /etc/passwd>

Since i am pretty much new to shellscript.Please do help me on this. This is really urgent to be delivered. Hence pls. do the needful.

Thanx,
Sachin

2

No replies... Pls. do help me in this. It is really urgent.

3

What shell / OS?

If you're using bash, you should be able to get away with something like:

#! /bin/bash

typeset -i n=0
typeset -a names

oldifs="$IFS"
IFS=:
 while read passwdname _; do
  [[ $passwdname == *a* ]] && names[n++]=$passwdname
 done </etc/passwd
IFS="$oldifs"

names=${names[*]}
ps --o user,fname -U ${names// /,}

Or:

ps --o user,fname | gawk '/^[^ ]*ram/{print $1}'

I'm sure there's a dozen ways of doing this better, but those are the first two things I thought of.

4

U can try (in ksh):

awk -F\: '/^ram?*/ {print $1}' /etc/passwd|xargs -i ksh -c 'ps --o user,fname |grep {}'

Regards