使用 bash 脚本查找 linux 用户

Question

i am newbie in bash-scripting and linux. I want to write a script to get a username from the terminal and say whether that username is on the passwd or not. i wrote and tried below script but it doesn't work. help me that what i can do.

users = cat /etc/passwd
echo User_Name
cat users | grep User_Name
if [ User_Name in users ];
then
    echo "User_Name FOUND"
else
    echo "User NOT FOUND!"
fi

how i can define a variable to read a string from termnial.

Answer 1

Your script could look like that:

#!/usr/bin/env bash

while read -rp "Enter user name: " user_name
do
    if [ -z "$user_name" ]
    then
        echo "Enter a non-empty username"    
        continue
    fi

    if grep -q "^$user_name:" /etc/passwd
    then
        echo "$user_name FOUND"
        exit 0
    else
        echo "$user_name NOT FOUND!" >&2
        exit 1
    fi
done

Answer 2

Here is my take on this question. Using getent to search user in the file /etc/passwd

If you just want to check if a user exists.

getent passwd "$username"

That should print the name (and all info in the Gecos Field ) of the user if it does exists, and return with zero 0 status which means true. Otherwise non-zero, see the manual for the meaning of different status.

---

Now that we know that information at hand we can use the if-statement

if getent passwd "$username"; then
  echo "$username found."
else
  echo "$username not found." >&2
fi

---

Redirect the output to /dev/null if that is not the desired result, or if you're just after the exit/return status of the test.

if getent passwd "$username" >/dev/null; then
  echo "$username found."
else
  echo "$username not found." >&2
fi

---

To add that to a full blown script.

#!/bin/sh

check_user() {
  printf '%s' "Enter user name: "
  read -r user_name
  case $user_name in
    '') printf '%s\n' "You have not entered anything!" >&2  ##: If empty input.
     return 1;;
  esac
  if getent passwd "$user_name" >/dev/null; then ##: getent has no -q option unfortunately.
    printf '%s\n' "$user_name FOUND"
    return 0
  else
    printf '%s\n' "$user_name NOT FOUND!" >&2
    return 1
  fi
}

##: Continue the loop until a match in `/etc/passwd` was given.
until check_user; do
  printf 'Please try again!\n' >&2  ##: Send message to stderr and loop again.
done

---

Instead of redirecting to /dev/null you can save the output of getent in a variable using Command Substituion and use that value, plus a little bit of Parameter Expansion .

check_user() {
  printf '%s' "Enter user name: "
  read -r user_name
  case $user_name in
    '') printf '%s\n' "You have not entered anything!" >&2
     return 1;;
  esac
  if user=$(getent passwd "$user_name"); then
    printf '%s\n' "${user%%:*} FOUND"  ##: Print only the username using P.E.
    return 0
  else
    printf '%s\n' "${user:-"$user_name"} NOT FOUND" >&2  ##: If the value of "$user" is empty (No match found by getent)
    return 1
  fi
}

until check_user; do
  printf 'Please try again!\n' >&2
done

• The until loop will continue until it receives a 0 return status.
• See getent(1) aka man 1 getent
• See Bash Parameter Expansion
• See Posix Parameter Expansion
• Tested on both GNU Linux and FreeBSD

Answer 3

To read a value as a user input, you can use the "read" command.

read User_Name


echo "User is ${User_Name}"

For convenience, most scripts use ALLCAPS on the variable names for shell scripting.

Also consider passing this value as a parameter into your script. It's readable as "$1" for a single parameter.

Hint1: You don't need to store the whole file in memory just for this check.

Hint2: To find if a user is present in the passwd file, you should check for the first column only of that file and try a match.

Hint3: If you want to use the grep command, it will return success when there's a match and failure when there's not match.

The code below would also work for the task

#!/bin/bash
read WHO; 
grep "^${WHO}:" /etc/passwd 2>/dev/null >/dev/null && echo "found" || echo "not found"

Answer 4

We use read to get input from the user at the terminal. In the below solution, we use POSIX id (present on BSDs as well as Linux) to determine if the user exists or not.

#!/bin/bash
set -o errexit -o pipefail -o nounset


function interactive_check_if_user_exists {
    local username output

    read -p 'Username to check: ' -r username

    if output="$(id -u "$username" 2>&1)"
    then
        echo "Found \`$username\` (uid: $output)" >&2
    else
        local code=$?
        if [[ $code -eq 1 ]]
        then
            echo "User \`$username\` is not present on this system." >&2
        else
            echo "Unexpected error occurred: $output" >&2
            return "$code"
        fi
    fi
}

interactive_check_if_user_exists

Answer 5

To print and read an argument from terminal, you should use echo and then use read commane and then call it with dollar-sign, finally the script know what you want. for example:

echo "Enter UserName: "
read User_Name

here is the correct form of your code:

echo "Enter the Username: "
read User_Name
cut -f 1 -d: /etc/passwd | grep -q $User_Name
if [ $? -eq 0 ];
then
    echo "User $User_Name FOUND"
else
    echo "User NOT FOUND!"
fi

---

Tips:

cut -f 1 -d  #cut the first column of the passwd list
grep -q # find username and do  not  write  anything  to  standard   output in terminal.

if [ $? -eq 0 ] #in linux terminal when we want to know whether the previous command is executed correctly we use $? and the output must be 0. and - eq means equal to 0. and when we use that, it means if previos command (grep the username) executed correctly and the username was found, then run next step and print it found.