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Exam Code: JN0-223
Exam Name: Automation and DevOps, Associate (JNCIA-DevOps)
Vendor: Juniper

60 Questions & Answers
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NEW QUESTION: 1
AWSクラウドアーキテクチャの設計原則は次のうちどれですか?
A. 単一障害点を実装します。
B. モノリシック設計を実装します。
C. 垂直スケーリングを実装します。
D. 疎結合を実装します。
Answer: D
Explanation:
Explanation
Loose coupling between services can also be done through asynchronous integration. It involves one component that generates events and another that consumes them. The two components do not integrate through direct point-to-point interaction, but usually through an intermediate durable storage layer. This approach decouples the two components and introduces additional resiliency. So, for example, if a process that is reading messages from the queue fails, messages can still be added to the queue to be processed when the system recovers.

NEW QUESTION: 2
Which of the following benefits are accurate with respect to using Terraform? (Number of correct answers: 4) Score 2
A. It can achieve exactly the same Dev, Staging, and Production environment by deploying a new environment from an existing template
B. It can achieve "immutable infrastructure"
C. It can make it easy to achieve CI (continuous integration) of application code
D. It is a declarative but not procedural language
E. It is an infrastructure orchestration tool: it's not only for resource configuration
Answer: A,B,E

NEW QUESTION: 3
You are developing an ASP.NET MVC application that uses forms authentication. The application uses SQL queries that display customer order data.
You need to prevent all SQL injection attacks against the application.
How should you secure the queries?
A. Implement parameterization.
B. Pattern check the input.
C. Filter out prohibited words in the input.
D. Escape single quotes on string-based input parameters.
Answer: A
Explanation:
With most development platforms, parameterized statements that work with parameters can be used (sometimes called placeholders or bind variables) instead of embedding user input in the statement. A placeholder can only store a value of the given type and not an arbitrary SQL fragment. Hence the SQL injection would simply be treated as a strange (and probably invalid) parameter value.
Reference: https://en.wikipedia.org/wiki/SQL_injection#Parameterized_statements


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