while loop in Bash Scripting - Mini Project: Build & Apply

Counting Down with a While Loop in Bash

📖 Scenario: You are creating a simple countdown timer script in Bash. This script will count down from a starting number to zero, showing each number on the screen.

🎯 Goal: Build a Bash script that uses a while loop to count down from a given number to zero, printing each number.

📋 What You'll Learn

Create a variable called count with the starting number 5

Create a variable called limit with the value 0

Use a while loop that runs as long as count is greater than or equal to limit

Inside the loop, print the current value of count

Decrease count by 1 in each loop iteration

💡 Why This Matters

🌍 Real World

Counting down or up is common in scripts that wait for a certain time or repeat tasks a fixed number of times.

💼 Career

Understanding loops and variables in Bash is essential for automating tasks in system administration and DevOps roles.

1

Set up the starting number

Create a variable called count and set it to 5.

Bash Scripting

# Create a variable called count and set it to 5
# Your code here

Hint

Use count=5 to create the variable.

2

Set the limit for the countdown

Create a variable called limit and set it to 0.

Bash Scripting

count=5
# Create a variable called limit and set it to 0
# Your code here

Hint

Use limit=0 to create the variable.

3

Write the while loop to count down

Write a while loop that runs while count is greater than or equal to limit. Inside the loop, print the value of count and then decrease count by 1.

Bash Scripting

count=5
limit=0
# Write a while loop that runs while count >= limit
# Inside the loop, print count and decrease count by 1
# Your code here

Hint

Use while [ "$count" -ge "$limit" ] to start the loop. Use echo "$count" to print. Use count=$((count - 1)) to decrease.

4

Run the script and show the countdown

Run the script to print the countdown from 5 to 0.

Bash Scripting

count=5
limit=0
while [ "$count" -ge "$limit" ]
do
  echo "$count"
  count=$((count - 1))
done

# Run the script to show the countdown
# Your code here

Hint

Run the script in your terminal. It should print numbers from 5 down to 0, one per line.

Practice

1. What does a while loop do in a bash script?

easy

A. Stops the script immediately

B. Runs commands only once

C. Executes commands only if a condition is false

D. Repeats commands as long as a condition is true

Solution

1. Step 1: Understand the purpose of a while loop

   A while loop runs repeatedly while its condition remains true.

2. Step 2: Compare options with the definition

   Only Repeats commands as long as a condition is true correctly describes this behavior.

3. Final Answer:

   Repeats commands as long as a condition is true -> Option D

4. Quick Check:

   while loop = repeat while true [OK]

Hint: While loops run repeatedly while condition is true [OK]

Common Mistakes:

• Thinking while loops run only once
• Confusing while with if statements
• Assuming it runs when condition is false

2. Which of the following is the correct syntax to start a while loop in bash?

easy

A. while [ condition ]; do

B. while ( condition ) {

C. while condition then

D. while { condition } do

Solution

1. Step 1: Recall bash while loop syntax

   The correct syntax uses square brackets for condition and 'do' keyword.

2. Step 2: Match options with correct syntax

   while [ condition ]; do matches the correct bash syntax: while [ condition ]; do.

3. Final Answer:

   while [ condition ]; do -> Option A

4. Quick Check:

   Correct bash while syntax = while [ condition ]; do [OK]

Hint: Use square brackets and 'do' to start while loop [OK]

Common Mistakes:

• Using parentheses instead of square brackets
• Missing 'do' keyword
• Using curly braces incorrectly

3. What will be the output of this script?

count=1
while [ $count -le 3 ]; do
  echo $count
  count=$((count + 1))
done

medium

A. 1 2 3 4

B. 1 2 3

C. 3 2 1

D. No output

Solution

1. Step 1: Analyze the loop condition and increments

   The loop starts with count=1 and runs while count is less or equal to 3, incrementing count by 1 each time.

2. Step 2: Trace the output for each iteration

   It prints 1, then 2, then 3, then stops because count becomes 4.

3. Final Answer:

   1 2 3 -> Option B

4. Quick Check:

   Loop prints 1 to 3 [OK]

Hint: Count increments until condition false, prints each value [OK]

Common Mistakes:

• Forgetting to increment count causing infinite loop
• Expecting output beyond 3
• Confusing loop condition operators

4. Identify the error in this script:

count=1
while [ $count -lt 5 ]
  echo $count
  count=$((count + 1))
done

medium

A. Wrong comparison operator '-lt'

B. Incorrect variable increment syntax

C. Missing 'do' keyword after while condition

D. Missing closing bracket in condition

Solution

1. Step 1: Check while loop syntax

   The while loop must have 'do' after the condition to start the loop body.

2. Step 2: Identify missing 'do'

   The script misses 'do' after the condition line, causing syntax error.

3. Final Answer:

   Missing 'do' keyword after while condition -> Option C

4. Quick Check:

   while loop needs 'do' [OK]

Hint: Always put 'do' after while condition [OK]

Common Mistakes:

• Omitting 'do' keyword
• Misplacing brackets
• Incorrect increment syntax

5. You want to write a bash script that prints all even numbers from 2 to 10 using a while loop. Which script correctly does this?

hard

A. num=1
while [ $num -le 10 ]; do
if [ $((num % 2)) -eq 0 ]; then echo $num; fi
num=$((num + 1))
done

B. num=2
while [ $num -le 10 ]; do
echo $num
num=$((num + 1))
done

C. num=2
while [ $num -lt 10 ]; do
echo $num
num=$((num + 2))
done

D. num=0
while [ $num -le 10 ]; do
echo $num
num=$((num + 2))
done

Solution

1. Step 1: Understand the goal

   We want to print even numbers from 2 to 10 inclusive.

2. Step 2: Check each option's logic

   num=2 while [ $num -le 10 ]; do echo $num num=$((num + 1)) done starts at 2 and increments by 1 up to 10, printing 2 3 4 5 6 7 8 9 10 - incorrect.
   num=1 while [ $num -le 10 ]; do if [ $((num % 2)) -eq 0 ]; then echo $num; fi num=$((num + 1)) done starts at 1, increments by 1, and prints only if even - correct.
   num=2 while [ $num -lt 10 ]; do echo $num num=$((num + 2)) done stops before 10, missing 10.
   num=0 while [ $num -le 10 ]; do echo $num num=$((num + 2)) done starts at 0, which is even but outside requested range (2 to 10).

3. Step 3: Choose the best option

   Only the script using modulo to explicitly check evenness prints exactly 2,4,6,8,10.

4. Final Answer:

   num=1 while [ $num -le 10 ]; do if [ $((num % 2)) -eq 0 ]; then echo $num; fi num=$((num + 1)) done -> Option A

5. Quick Check:

   Check even with modulo inside loop [OK]

Hint: Use modulo inside loop to print even numbers [OK]

Common Mistakes:

• Stopping loop before 10
• Starting from 0 when 2 is required
• Not checking evenness explicitly