bash - reading user variable into bash script grep

Question

I've tried every possible combination to get this bash script working. It's part of a larger script, and it basically prompts for a username (to check if it exists) and returns the appropriate response:

#! /bin/bash
# Script to see if User exists

clear
echo -n "Enter user to check: "
read $uzer

grep -c '^${uzer}:' /etc/passwd

if [ $? -eq 0 ]; then
  echo "User does exist :)"
else
  echo "No such user"
fi

In terminal the following works fine:

grep -c '^devuser1:' /etc/passwd
RETURNS: 1
grep -c '^devuser1234:' /etc/passwd
RETURNS: 0

I've tried many combinations of passing the read variable into '^${uzer}:' with no joy. Any ideas what else I can try?

Answer 1

-c means that you want to know the number of times this user is in /etc/passwd, while $? is the exit code. Those are differents, since the number of times is printed on stdout. Use $() for getting stdout into a variable

Second problem: all your variables, like $uzer will not be substituted with their values when in single quotes. Use double quotes.

number=$(grep -c "^${uzer}:" /etc/passwd)
if [ $number -gt 0 ]; then
  echo "User does exist :)"
else
  echo "No such user"
fi

Answer 2

You need grep -q. If you don't distinguish between more exit codes than "0" and "other" there is no need to seperate grep and if:

if grep -q "^${uzer}:" /etc/passwd; then
  echo "User does exist :)"
else
  echo "No such user"
fi