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NEW QUESTION: 1
A Developer is creating a Lambda function and will be using external libraries that are not included in the standard Lambda libraries.
What action would minimize the Lambda compute time consumed?
A. Copy the external libraries to Amazon S3, and reference the external libraries to the S3 location.
B. Install the external libraries in Lambda to be available to all Lambda functions.
C. Install the dependencies and external libraries at the beginning of the Lambda function.
D. Create a Lambda deployment package that includes the external libraries.
Answer: D
Explanation:
Explanation
AWS documentation - Lambda Best Practices
Control the dependencies in your function's deployment package. The AWS Lambda execution environment contains a number of libraries such as the AWS SDK for the Node.js and Python runtimes (a full list can be found here: Lambda Execution Environment and Available Libraries). To enable the latest set of features and security updates, Lambda will periodically update these libraries. These updates may introduce subtle changes to the behavior of your Lambda function. To have full control of the dependencies your function uses, we recommend packaging all your dependencies with your deployment package.

NEW QUESTION: 2
Select and Place:

Answer:
Explanation:

NEW QUESTION: 3
The developer wants to write a criteria query that will return the number of orders made by customer of each county.
Assume that customer is an entity with a unidirectional one-to-many relationship to the Order entity and that Address is an embeddable class, with an attribute country of type String.
Which one of the queries below correctly achieves this?
A. CriteriaBuilder cb> = ...
CriteriaQuery cq = cb.createQuery();
Root<Custower> c = cq.from(Customer.class);
Join<Customer, Order> o = c.join(Customer_.orders);
cq.select(cb.count(o));
cq.groupBy(c.qet(Customer__.address) - get(Address_.country)) ;
B. CriteriaBuilder cb = ...
CriteriaQuery cq = cb.createQueryO;
Root<Customer> c = cq.from(Customer.class);
Join<Customer, Order> o = c.join(Customer_.orders);
Join<Address, String> country = c.join(Customer,.address) .join(Address cq.multiselect(cq.count(o), country );
cq.groupBy(c.get(Customer.address) - get (Address_ . country) ) ;
C. CriteriaBuilder cb> = ...
CriteriaQuery cq = cb.createQuery();
Root<Customer> c = cq.from(Customer.class);
Join<Customer, Order> o = c.join(Customer_.orders);
cq.multiselect(cb.count(0), c,get(customer_.address.get(address_.country) cq.groupBy (c.get(customer_.address) .get(address_.country))
D. CriteriaBuilder cb> = ...
CriteriaQuery cq = cb.createQuery();
Root<Customer> c = cq.from(Customer.class); cq.select (cb.count(c.join
(customer_. Orders)) , c.get(customers(0), c.get(customer_.address) . get (Address_'country)); (c.get(Customer_.address). get(address_.country));
Answer: C
Explanation:
Explanation/Reference:
Incorrect: Not B, Not C: Use multiselect, not select. Not D: Use one Join, not two.
Use multiselect, count and Group By. multiselect() method is used because we are going to get compound result and not one entity type.
Example:
CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery<Object[]> query = cb.createQuery(Object[].class);
Root<Department> d = query.from(Department.class);
Join<Department,Teacher> teachers = d.join("teachers");
query.multiselect(d.get("name"),cb.count(teachers)).groupBy(d.get("name")); Reference: Criteria group by clause