[Solved] You have a file records.txt with lines like:John|Doe|30|Engineer Jane|Smith|25|Designer Bob|Brown|40|ManagerWrite an awk command to print only the names (first and last) of people older than 28. | Bash Scripting

Bash Scripting - Text Processing in Scripts

You have a file records.txt with lines like:

John|Doe|30|Engineer
Jane|Smith|25|Designer
Bob|Brown|40|Manager

Write an awk command to print only the names (first and last) of people older than 28.

Aawk -F '|' '$3 < 28 {print $1, $2}' records.txt

Bawk -F ',' '$3 > 28 {print $1, $2}' records.txt

Cawk '{if ($3 > 28) print $1, $2}' records.txt

Dawk -F '|' '$3 > 28 {print $1, $2}' records.txt

Step-by-Step Solution

Solution:

Step 1: Set correct field separator
The file uses '|' as separator, so use -F '|' to split fields correctly.
Step 2: Filter by age and print names
Use condition $3 > 28 to select lines where age is greater than 28, then print first and second fields (first and last names).
Final Answer:
awk -F '|' '$3 > 28 {print $1, $2}' records.txt -> Option D
Quick Check:
Filter $3 > 28 and print $1, $2 [OK]

Quick Trick: Use -F '|' and condition $3 > 28 to filter [OK]

Common Mistakes:

MISTAKES

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