[Solved] What is wrong with this bash snippet?input="" if [ -z input ]; then echo "Empty" fi - Ans: Variable input is not quoted inside -z test | Bash Scripting

Bash Scripting - Conditionals

What is wrong with this bash snippet?

input=""
if [ -z input ]; then
  echo "Empty"
fi

AVariable input is not quoted inside -z test

BUsing -z instead of -n

CMissing else branch

DUsing single brackets instead of double brackets

Step-by-Step Solution

Solution:

Step 1: Understand -z operator
-z checks if a string is empty; it requires the variable to be expanded properly.
Step 2: Importance of quoting variables
Without quotes, if input is empty, test sees no argument, causing syntax errors.
Final Answer:
Variable input is not quoted inside -z test -> Option A
Quick Check:
Change to [ -z "$input" ] fixes the issue [OK]

Quick Trick: Always quote variables in tests like -z "$var" [OK]

Common Mistakes:

MISTAKES

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