/etc/passwd and /etc/shadow in Linux CLI - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the command to display the contents of the /etc/passwd file.

Linux CLI

cat [1]

Drag options to blanks, or click blank then click option'

A/etc/shadow

B/etc/hosts

C/etc/group

D/etc/passwd

Attempts:

3 left

💡 Hint

Common Mistakes

Using /etc/shadow instead of /etc/passwd, which requires root permission to view.

Trying to view /etc/group or /etc/hosts which are different files.

The /etc/passwd file contains user account information and can be viewed with 'cat /etc/passwd'.

2fill in blank

medium

Complete the command to safely view the /etc/shadow file with root privileges.

Linux CLI

sudo [1] /etc/shadow

Drag options to blanks, or click blank then click option'

Atouch

Bls

Ccat

Decho

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'ls' which lists files but does not show contents.

Using 'echo' which prints text but not file contents.

Using 'touch' which creates or updates files.

The 'cat' command displays file contents. 'sudo cat /etc/shadow' shows the shadow file with root rights.

3fill in blank

hard

Fix the error in the command to list users from /etc/passwd by extracting the first field (username).

Linux CLI

cut -d: -f[1] /etc/passwd

Drag options to blanks, or click blank then click option'

A1

B3

C2

D4

Attempts:

3 left

💡 Hint

Common Mistakes

Choosing field 2 or 3 which contain other data like user ID or group ID.

Using wrong delimiter or field number.

Usernames are in the first field of /etc/passwd, separated by colons, so use '-f1'.

4fill in blank

hard

Fill both blanks to create a command that counts the number of user accounts in /etc/passwd.

Linux CLI

cat /etc/passwd | [1] -d: -f1 | [2] -l

Drag options to blanks, or click blank then click option'

Acut

Bwc

Csort

Duniq

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'sort' or 'uniq' instead of counting lines.

Not extracting the correct field before counting.

Use 'cut' to extract usernames and 'wc -l' to count lines (users).

5fill in blank

hard

Fill both blanks to create a command that lists usernames from /etc/passwd who have a user ID greater than 1000.

Linux CLI

awk -F: '[1] $3 [2] 1000  print $1' /etc/passwd

Drag options to blanks, or click blank then click option'

A($3

B>

C)

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Omitting parentheses causing syntax errors.

Using '<' instead of '>' which filters wrong users.

Not printing the first field ($1) which is the username.

The awk condition needs parentheses around '$3 > 1000' to filter user IDs greater than 1000 and print usernames.

Practice

1. What is the primary purpose of the /etc/passwd file in Linux?

easy

A. To store encrypted user passwords securely

B. To configure network settings

C. To store user account information accessible to all users

D. To log system errors

Solution

1. Step 1: Understand the role of /etc/passwd

   The /etc/passwd file contains user account details like username, user ID, group ID, home directory, and shell. It is readable by all users.

2. Step 2: Differentiate from /etc/shadow

   The encrypted passwords are stored separately in /etc/shadow for security, not in /etc/passwd.

3. Final Answer:

   To store user account information accessible to all users -> Option C

4. Quick Check:

   /etc/passwd = user info file [OK]

Hint: Remember: passwd = public user info, shadow = private passwords [OK]

Common Mistakes:

• Confusing /etc/passwd with /etc/shadow
• Thinking passwords are stored in /etc/passwd
• Assuming /etc/passwd stores system logs

2. Which of the following lines correctly shows the format of a typical /etc/shadow entry?

easy

A. username:x:1000:1000:User Name:/home/user:/bin/bash

B. username:$6$randomsalt$encryptedpassword:18000:0:99999:7:::

C. username:password:1000:1000:User Name:/home/user:/bin/bash

D. username:encryptedpassword:1000:1000:User Name:/home/user:/bin/bash

Solution

1. Step 1: Identify /etc/shadow format

   The /etc/shadow file stores encrypted passwords and related info. The password field looks like $6$randomsalt$encryptedpassword for SHA-512 encryption.

2. Step 2: Compare options

   The line username:$6$randomsalt$encryptedpassword:18000:0:99999:7::: matches the shadow file format with encrypted password (indicated by $6$) and password aging fields separated by colons. The other options resemble /etc/passwd format or have incorrect password fields.

3. Final Answer:

   username:$6$randomsalt$encryptedpassword:18000:0:99999:7::: -> Option B

4. Quick Check:

   /etc/shadow entries have encrypted passwords with $ signs [OK]

Hint: Look for $ signs and many colon-separated fields in shadow entries [OK]

Common Mistakes:

• Confusing /etc/passwd format with /etc/shadow
• Expecting plain text passwords in /etc/shadow
• Missing fields after encrypted password

3. Given the following command:

grep '^root:' /etc/passwd | cut -d':' -f1,3,7

What will be the output format?

medium

A. root:/bin/bash

B. root:x:0:0:root:/root:/bin/bash

C. root:0:0

D. root:0:/bin/bash

Solution

1. Step 1: Understand the command parts

   grep '^root:' /etc/passwd finds the line starting with 'root:'. cut -d':' -f1,3,7 extracts fields 1 (username), 3 (UID), and 7 (shell) separated by colons.

2. Step 2: Identify root line fields

   Typical root line: root:x:0:0:root:/root:/bin/bash. Fields: 1=root, 2=x, 3=0, 4=0, 5=root, 6=/root, 7=/bin/bash.

3. Step 3: Extract fields 1,3,7

   Fields 1=root, 3=0, 7=/bin/bash. Joined by colons: root:0:/bin/bash.

4. Final Answer:

   root:0:/bin/bash -> Option D

5. Quick Check:

   Fields 1,3,7 from /etc/passwd line = root:0:/bin/bash [OK]

Hint: Count fields carefully; cut -f1,3,7 picks username, UID, shell [OK]

Common Mistakes:

• Mixing up field numbers in cut command
• Expecting full /etc/passwd line output
• Confusing shadow file format with passwd

4. You run this command to check password expiry info:

sudo chage -l username

But it shows an error: chage: user 'username' does not exist. What is the most likely cause?

medium

A. The username is missing in /etc/passwd

B. The /etc/shadow file is empty

C. The chage command is not installed

D. The user has no password set

Solution

1. Step 1: Understand chage user check

   The chage command checks user password expiry info by looking up the username in /etc/passwd.

2. Step 2: Analyze error cause

   Error 'user does not exist' means the username is not found in /etc/passwd. If the user was missing there, chage cannot proceed.

3. Final Answer:

   The username is missing in /etc/passwd -> Option A

4. Quick Check:

   User must exist in /etc/passwd for chage to work [OK]

Hint: Check /etc/passwd for username before using chage [OK]

Common Mistakes:

• Assuming /etc/shadow emptiness causes this error
• Thinking chage command missing causes user error
• Believing no password set triggers 'user does not exist'

5. You want to write a script that lists all users who have a valid login shell (not /usr/sbin/nologin or /bin/false) by reading /etc/passwd. Which command snippet correctly filters these users?

hard

A. awk -F':' '$7 != "/usr/sbin/nologin" && $7 != "/bin/false" {print $1}' /etc/passwd

B. grep -v '/usr/sbin/nologin' /etc/shadow | cut -d':' -f1

C. cat /etc/passwd | grep '/bin/bash' | awk '{print $1}'

D. sed '/nologin/d' /etc/passwd | cut -d':' -f7

Solution

1. Step 1: Identify shell field in /etc/passwd

   The 7th field in /etc/passwd is the user's login shell. We want to exclude /usr/sbin/nologin and /bin/false.

2. Step 2: Analyze each option

   awk -F':' '$7 != "/usr/sbin/nologin" && $7 != "/bin/false" {print $1}' /etc/passwd uses awk with field separator ':' and filters out unwanted shells, printing usernames (field 1). grep -v '/usr/sbin/nologin' /etc/shadow | cut -d':' -f1 wrongly reads /etc/shadow and uses grep -v incorrectly. cat /etc/passwd | grep '/bin/bash' | awk '{print $1}' filters only /bin/bash but misses other valid shells. sed '/nologin/d' /etc/passwd | cut -d':' -f7 deletes lines with 'nologin' but then prints shell field, not usernames.

3. Final Answer:

   awk -F':' '$7 != "/usr/sbin/nologin" && $7 != "/bin/false" {print $1}' /etc/passwd -> Option A

4. Quick Check:

   Use awk on /etc/passwd to filter shell field and print usernames [OK]

Hint: Use awk with FS=':' and check field 7 for valid shells [OK]

Common Mistakes:

• Using /etc/shadow instead of /etc/passwd for shell info
• Printing wrong fields (shell instead of username)
• Using grep without proper field filtering