[Solved] What will be the outcome of executing this Specification query?Specification spec = (root, query, builder) -> builder.like(root.get("email"), "%@example.com");List users = userRepository.... | Spring Boot

Spring Boot - Advanced Patterns

What will be the outcome of executing this Specification query?

Specification spec = (root, query, builder) -> builder.like(root.get("email"), "%@example.com");
List users = userRepository.findAll(spec);

AReturns all users whose email ends with '@example.com'

BReturns all users whose email starts with '@example.com'

CReturns all users with email exactly '@example.com'

DThrows an exception because 'like' is not supported

Step-by-Step Solution

Solution:

Step 1: Analyze the Specification
The Specification uses builder.like with pattern "%@example.com", which matches any string ending with '@example.com'.
Step 2: Understand the query execution
userRepository.findAll(spec) will return all User entities matching this condition.
Final Answer:
Returns all users whose email ends with '@example.com' -> Option A
Quick Check:
Pattern "%@example.com" matches suffix [OK]

Quick Trick: Use % for wildcard in like queries [OK]

Common Mistakes:

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What will be the outcome of executing this Specification query?