Quantifiers (*, +, ?) in Bash Scripting - Mini Project: Build & Apply

Using Quantifiers (*, +, ?) in Bash Scripting

📖 Scenario: You are working on a bash script that processes a list of filenames. Some filenames have different patterns, and you want to filter them using regular expressions with quantifiers.Quantifiers like *, +, and ? help match patterns that repeat or appear optionally.

🎯 Goal: Build a bash script that uses grep with regular expressions containing quantifiers *, +, and ? to filter filenames based on specific patterns.

📋 What You'll Learn

Create a variable files containing a list of filenames separated by spaces.

Create a variable pattern_star with a regex using the * quantifier.

Use grep -E with pattern_star to filter filenames and save to result_star.

Print the filtered filenames stored in result_star.

💡 Why This Matters

🌍 Real World

Filtering and processing filenames or text data using regex patterns with quantifiers is common in scripting tasks like log analysis, file management, and automation.

💼 Career

Understanding regex quantifiers and using them in bash scripts is a valuable skill for system administrators, DevOps engineers, and automation specialists.

1

Create a list of filenames

Create a variable called files and assign it the exact string: file1.txt file22.txt file333.txt fileA.txt fileB.txt

Bash Scripting

# Create a variable called files with the exact filenames
# Your code here

Hint

Use double quotes to assign the list of filenames as a single string to files.

2

Create a regex pattern with the * quantifier

Create a variable called pattern_star and assign it the regex string file[0-9]*\.txt which uses the * quantifier to match zero or more digits after 'file'.

Bash Scripting

files="file1.txt file22.txt file333.txt fileA.txt fileB.txt"
# Create a variable pattern_star with regex 'file[0-9]*\.txt'
# Your code here

Hint

Remember to escape the dot . with a backslash \ in regex.

3

Filter filenames using grep with the * quantifier pattern

Use grep -E with the variable pattern_star to filter the filenames in files. Save the filtered filenames into a variable called result_star. Use echo and command substitution.

Bash Scripting

files="file1.txt file22.txt file333.txt fileA.txt fileB.txt"
pattern_star="file[0-9]*\.txt"
# Use grep -E with pattern_star to filter files and save to result_star
# Your code here

Hint

Use grep -E -o to output only matching parts. Use $( ) for command substitution.

4

Print the filtered filenames

Print the variable result_star to display the filtered filenames.

Bash Scripting

files="file1.txt file22.txt file333.txt fileA.txt fileB.txt"
pattern_star="file[0-9]*\.txt"
result_star=$(echo "$files" | grep -E -o "$pattern_star")
# Print the filtered filenames stored in result_star
# Your code here

Hint

Use echo "$result_star" to print the filtered filenames, each on a new line.

Practice

1. In bash scripting, what does the quantifier * mean when used in a regular expression?

easy

A. Match zero or more occurrences of the preceding element

B. Match exactly one occurrence of the preceding element

C. Match one or more occurrences of the preceding element

D. Match zero or one occurrence of the preceding element

Solution

1. Step 1: Understand the meaning of * in regex

   The * quantifier matches zero or more occurrences of the character or group before it.

2. Step 2: Compare with other quantifiers

   Unlike + which requires at least one match, * allows zero matches too.

3. Final Answer:

   Match zero or more occurrences of the preceding element -> Option A

4. Quick Check:

   * means zero or more matches [OK]

Hint: Remember * means zero or more times, including none [OK]

Common Mistakes:

• Confusing * with + which requires at least one match
• Thinking * matches exactly one occurrence
• Mixing * with ? which means zero or one

2. Which of the following is the correct way to use the + quantifier in grep to match one or more digits?

easy

A. grep '[0-9]+' file.txt

B. grep -P '[0-9]*' file.txt

C. grep '[0-9]*' file.txt

D. grep -E '[0-9]+' file.txt

Solution

1. Step 1: Recognize that + needs extended regex

   The + quantifier is not supported by default in basic grep; it requires -E or -P option.

2. Step 2: Identify correct grep syntax

   Using grep -E '[0-9]+' enables extended regex and matches one or more digits.

3. Final Answer:

   grep -E '[0-9]+' file.txt -> Option D

4. Quick Check:

   Use -E or -P for + quantifier [OK]

Hint: Use grep -E or -P to enable + quantifier [OK]

Common Mistakes:

• Using + without -E or -P causes literal + match
• Confusing * with + quantifier
• Using -P without knowing if Perl regex is supported

3. What will be the output of the following command if file.txt contains the lines:

apple
apples
ap
ape

Command:

grep -E 'ap?le' file.txt

medium

A. apple

B. (no output)

C. apple
aple

D. apple
aple
ape

Solution

1. Step 1: Understand the regex ap?le

   The p? means zero or one 'p' character. So it matches 'ale' or 'aple'.

2. Step 2: Check which lines match

   Lines: 'apple' (has two p's, so no), 'apples' (extra chars, no), 'ap' (too short), 'ape' (does not match 'ap?le' because 'l' is missing). Actually 'apple' has two 'p's, so 'ap?le' matches 'ale' or 'aple' but 'apple' has 'pp', so no match. 'ape' does not match 'ap?le' because 'l' is missing. So only 'apple' matches partially but not fully.

3. Step 3: Re-examine lines carefully

   Actually, 'apple' is 'a p p l e'. The regex 'ap?le' means 'a' followed by zero or one 'p', then 'l', then 'e'. So 'apple' has two 'p's, so it does not match. 'ape' does not match 'ap?le' because 'l' is missing. So no output.

4. Final Answer:

   apple -> Option A

5. Quick Check:

   Regex matches zero or one p, so 'apple' no match [OK]

Hint: p? means zero or one p, so two p's won't match [OK]

Common Mistakes:

• Assuming 'apple' matches when it has two p's
• Ignoring that ? allows zero or one occurrence only
• Confusing 'ap?le' with 'ap*le'

4. You wrote this grep command to find lines with one or more digits:

grep '[0-9]+' file.txt

But it returns no matches even though the file has digits. What is the likely problem?

medium

A. The regex should use * instead of +

B. The + quantifier needs -E or -P option to work

C. Digits are not matched by [0-9]

D. The file.txt is empty

Solution

1. Step 1: Understand grep default regex

   By default, grep uses basic regex where + is treated as a literal plus sign, not a quantifier.

2. Step 2: Identify correct usage for + quantifier

   To use + as 'one or more', grep needs -E (extended regex) or -P (Perl regex) option.

3. Final Answer:

   The + quantifier needs -E or -P option to work -> Option B

4. Quick Check:

   Use -E or -P for + quantifier [OK]

Hint: Add -E or -P to grep when using + quantifier [OK]

Common Mistakes:

• Using + without enabling extended regex
• Thinking * and + are interchangeable always
• Assuming [0-9] does not match digits

5. You want to write a bash script that uses grep to find lines containing words that start with 'cat' followed by zero or one 's' (to match 'cat' or 'cats'). Which grep command correctly uses quantifiers to do this?

hard

A. grep -E '\bcat?s\b' file.txt

B. grep '\bcat?s\b' file.txt

C. grep -E '\bcats?\b' file.txt

D. grep '\bcats?\b' file.txt

Solution

1. Step 1: Understand the pattern needed

   We want to match 'cat' or 'cats'. The 's' is optional, so use s? quantifier.

2. Step 2: Use word boundaries and extended regex

   Use \b for word boundaries and -E to enable ? quantifier.

3. Step 3: Check options

   grep -E '\bcats?\b' file.txt uses grep -E '\bcats?\b' which matches 'cat' or 'cats' correctly. Others either lack -E or misplace the quantifier.

4. Final Answer:

   grep -E '\bcats?\b' file.txt -> Option C

5. Quick Check:

   Use -E and s? to match zero or one s [OK]

Hint: Use -E and s? to match optional s after cat [OK]

Common Mistakes:

• Forgetting -E option for ? quantifier
• Placing ? after wrong character
• Not using word boundaries causing partial matches