Accessing variables ($var and ${var}) in Bash Scripting - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to print the value of the variable name.

Bash Scripting

echo [1]

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

A${name}

Bname

C$name

D&name

Attempts:

3 left

💡 Hint

Common Mistakes

Forgetting the dollar sign before the variable name.

Using the variable name without any prefix.

In bash, to access the value of a variable, you use the dollar sign followed by the variable name, like $name.

2fill in blank

medium

Complete the code to print the value of the variable path using braces.

Bash Scripting

echo [1]

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

A${path}

B$path

Cpath

D$[path]

Attempts:

3 left

💡 Hint

Common Mistakes

Using square brackets instead of braces.

Omitting the dollar sign.

Using braces like ${path} helps to clearly mark the variable name, especially when concatenating with other text.

3fill in blank

hard

Fix the error in the code to correctly print the variable file followed by the string .txt.

Bash Scripting

echo [1]

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

A${file}.txt

B${filetxt}

C$filetxt

D$file.txt

Attempts:

3 left

💡 Hint

Common Mistakes

Writing $file.txt which looks for variable file.txt.

Concatenating without braces causing wrong variable name.

To append text after a variable without confusion, use braces: ${file}.txt ensures only file is treated as the variable.

4fill in blank

hard

Fill both blanks to correctly print the variable user and then the string _data.

Bash Scripting

echo [1][2]

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

A${user}

B$user

C_data

Ddata

Attempts:

3 left

💡 Hint

Common Mistakes

Using $user_data which looks for variable user_data.

Forgetting to use braces to separate variable and text.

Using ${user} ensures the variable ends before the appended string _data.

5fill in blank

hard

Fill all three blanks to create a variable count with value 5, then print it with braces and add the string items.

Bash Scripting

[1]=5
echo [2][3]

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

Acount

B${count}

C items

D$count

Attempts:

3 left

💡 Hint

Common Mistakes

Using $count without braces when appending text.

Forgetting to assign the variable before printing.

First assign 5 to count. Then print it with braces ${count} to safely append the string items.

Practice

1. In bash scripting, what is the main reason to use ${var} instead of just $var?

easy

A. To clearly separate the variable name from surrounding text

B. To declare a new variable

C. To comment out the variable

D. To convert the variable to uppercase

Solution

1. Step 1: Understand variable access syntax

   In bash, $var accesses the value of a variable named var.

2. Step 2: Purpose of braces

   Using ${var} helps separate the variable name from other text, avoiding confusion when concatenating.

3. Final Answer:

   To clearly separate the variable name from surrounding text -> Option A

4. Quick Check:

   Braces separate variable name from text = A [OK]

Hint: Use braces to avoid mixing variable name with text [OK]

Common Mistakes:

• Thinking braces declare variables
• Using braces to comment variables
• Assuming braces change variable case

2. Which of the following is the correct way to print the value of variable name followed immediately by the string _end in bash?

easy

A. echo $name_end

B. echo $name _end

C. echo ${name_end}

D. echo ${name}_end

Solution

1. Step 1: Understand variable and string concatenation

   To print variable name followed by _end, we must separate the variable name from the string.

2. Step 2: Use braces to separate variable name

   Using ${name}_end correctly prints the variable value plus the string _end. Without braces, $name_end looks for a variable named name_end.

3. Final Answer:

   echo ${name}_end -> Option D

4. Quick Check:

   Braces separate variable name from text = C [OK]

Hint: Use braces when adding text right after variable [OK]

Common Mistakes:

• Forgetting braces and getting wrong variable
• Adding space breaks concatenation
• Using wrong variable name inside braces

3. What will be the output of this bash script?

var="world"
echo "Hello $var!"
echo "Hello ${var}!"
echo "Hello $var123"
echo "Hello ${var}123"

medium

A. Hello world!
Hello world!
Hello world123
Hello 123

B. Hello world!
Hello world!
Hello
Hello world123

C. Hello world!
Hello world!
Hello
Hello 123

D. Hello world!
Hello world!
Hello world123
Hello world123

Solution

1. Step 1: Analyze each echo line

   First two lines print "Hello world!" because $var and ${var} both expand to "world". Third line prints "Hello " because $var123 looks for variable named var123 which is empty. Fourth line prints "Hello world123" because ${var} is "world" plus "123".

2. Step 2: Combine outputs

   Outputs are:
   Hello world!
   Hello world!
   Hello
   Hello world123

3. Final Answer:

   Hello world! Hello world! Hello Hello world123 -> Option B

4. Quick Check:

   Braces separate variable name correctly = D [OK]

Hint: Braces avoid variable name confusion with suffixes [OK]

Common Mistakes:

• Assuming $var123 expands var plus 123
• Ignoring empty variable var123
• Confusing output lines order

4. Given this script snippet:

filename="report"
echo "File: $filename.txt"

Why does it not print File: report.txt as expected?

medium

A. Because echo cannot print variables with dots

B. Because .txt is treated as a command

C. Because $filename.txt looks for variable named filename.txt which is undefined

D. Because variable names cannot contain letters

Solution

1. Step 1: Understand variable expansion with suffix

   Using $filename.txt makes bash look for variable named filename.txt, which does not exist.

2. Step 2: Fix with braces

   Using ${filename}.txt correctly expands filename and appends .txt.

3. Final Answer:

   Because $filename.txt looks for variable named filename.txt which is undefined -> Option C

4. Quick Check:

   Braces needed to separate variable name from suffix = A [OK]

Hint: Use braces when adding dot or special chars after variable [OK]

Common Mistakes:

• Thinking dot is part of variable name
• Assuming echo can't print dots
• Ignoring variable name rules

5. You want to create a bash script that prints the value of variable user followed by the string _backup only if user is not empty. Which is the best way to write the echo command?

hard

A. echo "${user:+${user}_backup}"

B. echo "${user}_backup"

C. echo "${user}_backup" && [ -n "$user" ]

D. echo "$user_backup"

Solution

1. Step 1: Understand conditional variable expansion

   The expression ${user:+${user}_backup} prints ${user}_backup only if user is not empty.

2. Step 2: Compare options

   echo "${user}_backup" always prints _backup even if user is empty. echo "$user_backup" looks for variable user_backup. echo "${user}_backup" && [ -n "$user" ] has wrong syntax mixing echo and condition.

3. Final Answer:

   echo "${user:+${user}_backup}" -> Option A

4. Quick Check:

   Use ${var:+value} to print only if var set = B [OK]

Hint: Use ${var:+value} to print only if var is set [OK]

Common Mistakes:

• Using $user_backup instead of ${user}_backup
• Printing _backup even if user empty
• Mixing echo with condition incorrectly