Backticks and $() for command substitution in Bash Scripting - Mini Project: Build & Apply

Using Backticks and $() for Command Substitution in Bash

📖 Scenario: You are writing a simple bash script to display information about files in a directory. You want to capture the output of commands and use it inside your script.

🎯 Goal: Learn how to use backticks `command` and $(command) to run commands inside your script and store their output in variables.

📋 What You'll Learn

Create a variable with the output of a command using backticks

Create a variable with the output of a command using $() syntax

Use these variables in your script

Print the results to the terminal

💡 Why This Matters

🌍 Real World

Command substitution is used in scripts to capture the output of commands and use it for decisions, calculations, or messages.

💼 Career

Knowing how to capture command output is essential for writing effective bash scripts in system administration, automation, and DevOps roles.

1

Create a variable using backticks for command substitution

Create a variable called file_count_backticks that stores the number of files in the current directory using the command ls | wc -l inside backticks.

Bash Scripting

# Create a variable called file_count_backticks using backticks
# Your code here

Hint

Use backticks ` around the command ls | wc -l to capture its output.

2

Create a variable using $() for command substitution

Create a variable called file_count_dollar that stores the number of files in the current directory using the command ls | wc -l inside $( ).

Bash Scripting

file_count_backticks=`ls | wc -l`
# Create a variable called file_count_dollar using $( )
# Your code here

Hint

Use $(command) syntax to capture the output of ls | wc -l.

3

Use the variables in a message

Write a line that creates a variable called message with the text: "There are X files (backticks) and Y files ($()) in the directory." where X is the value of file_count_backticks and Y is the value of file_count_dollar. Use double quotes and variable substitution.

Bash Scripting

file_count_backticks=`ls | wc -l`
file_count_dollar=$(ls | wc -l)
# Create a variable message with the counts inside a sentence
# Your code here

Hint

Use double quotes and $variable to insert variable values inside the string.

4

Print the message to the terminal

Write a line to print the variable message to the terminal using echo.

Bash Scripting

file_count_backticks=`ls | wc -l`
file_count_dollar=$(ls | wc -l)
message="There are $file_count_backticks files (backticks) and $file_count_dollar files ($()) in the directory."
# Print the message variable
# Your code here

Hint

Use echo "$message" to print the message with variable values.

Practice

1. What does the command substitution syntax $(date) do in a bash script?

easy

A. Creates a new variable named date

B. Runs the date command and replaces $(date) with its output

C. Comments out the date command

D. Prints the literal string $(date) without running the command

Solution

1. Step 1: Understand command substitution

   Command substitution runs the command inside $() and replaces it with the command's output.

2. Step 2: Apply to $(date)

   The date command outputs the current date and time, so $(date) is replaced by that output.

3. Final Answer:

   Runs the date command and replaces $(date) with its output -> Option B

4. Quick Check:

   Command substitution = runs command and inserts output [OK]

Hint: Remember: $() runs command and inserts output [OK]

Common Mistakes:

• Thinking it creates a variable
• Confusing it with comments
• Expecting literal string output

2. Which of the following is the preferred syntax for command substitution in bash?

easy

A. echo `ls -l`

B. echo $(ls -l

C. echo $[ls -l]

D. echo $(ls -l)

Solution

1. Step 1: Identify valid command substitution syntax

   Command substitution uses either backticks `command` or $(command). Both the backtick version and the incomplete $() version look valid at first.

2. Step 2: Check for best practice and syntax correctness

   echo `ls -l` uses backticks which are valid but less readable and harder to nest. echo $(ls -l) uses $() which is modern and preferred. echo $[ls -l] uses invalid syntax with $[]. echo $(ls -l misses closing parenthesis.

3. Final Answer:

   echo $(ls -l) -> Option D

4. Quick Check:

   Use $() for clear command substitution [OK]

Hint: Use $() for command substitution, it's clearer and nestable [OK]

Common Mistakes:

• Using $[] which is arithmetic expansion, not command substitution
• Missing closing parenthesis
• Confusing backticks with other symbols

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

files=$(ls /tmp)
echo "$files"

medium

A. Shows an error because of missing quotes

B. Prints the literal string '$(ls /tmp)'

C. Lists all files and directories inside /tmp one per line

D. Prints nothing because variable is empty

Solution

1. Step 1: Understand command substitution with ls

   The command ls /tmp lists files in /tmp. Using files=$(ls /tmp) stores that output as a string with filenames separated by newlines.

2. Step 2: Echo the variable with quotes

   Using echo "$files" prints the content of files preserving newlines, so it lists all files and directories one per line.

3. Final Answer:

   Lists all files and directories inside /tmp one per line -> Option C

4. Quick Check:

   Command substitution stores output string = list of files [OK]

Hint: Command substitution stores output as string, echo prints it [OK]

Common Mistakes:

• Expecting literal command string output
• Confusing quotes causing errors
• Assuming variable is empty

4. Identify the error in this script snippet:

current_dir=`pwd
 echo "Current directory is: $current_dir"

medium

A. Missing closing backtick for command substitution

B. Incorrect use of double quotes

C. Variable name is invalid

D. No error, script runs fine

Solution

1. Step 1: Check command substitution syntax

   The command substitution using backticks must start and end with backticks. Here, the closing backtick is missing after pwd.

2. Step 2: Analyze impact of missing backtick

   Without the closing backtick, bash will throw a syntax error and not run the script properly.

3. Final Answer:

   Missing closing backtick for command substitution -> Option A

4. Quick Check:

   Backticks must be paired to run commands [OK]

Hint: Always close backticks to avoid syntax errors [OK]

Common Mistakes:

• Forgetting to close backticks
• Misplacing quotes inside command substitution
• Assuming variable names cause errors

5. You want to store the number of files in your home directory in a variable file_count. Which command correctly uses command substitution to do this?

hard

A. file_count=$(ls ~ | wc -l)

B. file_count=`ls ~ wc -l`

C. file_count=$(ls ~ wc -l)

D. file_count=ls ~ | wc -l

Solution

1. Step 1: Understand the command to count files

   To count files, we list them with ls ~ and pipe to wc -l to count lines (files).

2. Step 2: Use correct command substitution syntax

   file_count=$(ls ~ | wc -l) correctly uses $(ls ~ | wc -l) to run the full command and store output in file_count. The other options miss the pipe or fail to use command substitution.

3. Final Answer:

   file_count=$(ls ~ | wc -l) -> Option A

4. Quick Check:

   Use $() with full command and pipe inside [OK]

Hint: Use $() with full command including pipes for substitution [OK]

Common Mistakes:

• Missing pipe between commands
• Using backticks without pipe
• Assigning command without substitution