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NEW QUESTION: 1
A system administrator needs to use wsadmin to execute commands interactively. The WebSphere cell has global security enabled. To avoid the security login dialog when executing wsadmin, the administrator should:
A. edit the sas.client.props file.
B. login in to the local operating system (local OS) with an administrator user id.
C. edit the wsadmin.properties file.
D. use the argument -user <current_user>.
Answer: A

NEW QUESTION: 2
A particular database threat utilizes a SQL injection technique to penetrate a target system. How would an attacker use this technique to compromise a database?
A. An attacker utilizes an incorrect configuration that leads to access with higher-thanexpected privilege of the database
B. An attacker gains control of system to flood the target system with requests,preventing legitimate users from gaining access
C. An attacker uses poorly designed input validation routines to create or alter SQL commands to gain access to unintended data or execute commands of the database
D. An attacker submits user input that executes an operating system command to compromise a target system
Answer: C
Explanation:
Using the poorly designed input validation to alter or steal data from a database is a SQL injection attack.

NEW QUESTION: 3
あなたはモバイルアプリケーションを設計する会社で働いています。彼らは、プレーヤーの記録が異なるゲームに割り当てられるサーバーを維持しています。追跡システムは新しく、開発中です。
アプリケーションは、EntityFrameworkを使用してAzureデータベースに接続します。データベースには、PlayerテーブルとGameテーブルがあります。
プレーヤーを追加するとき、コードは新しいプレーヤーレコードを挿入し、既存のゲームレコードと新しいプレーヤーレコードの間に関係を追加する必要があります。
アプリケーションは、正しいgameIdとplayerIdを指定してCreatePlayerWithGameを呼び出し、プロセスを開始します。 （行番号は参照用にのみ含まれています。）

次の各ステートメントについて、ステートメントがtrueの場合は、[はい]を選択します。それ以外の場合は、[いいえ]を選択します。
注：正しい選択はそれぞれ1ポイントの価値があります。

Answer:
Explanation:

Explanation

Many-to-many relationships without an entity class to represent the join table are not yet supported. However, you can represent a many-to-many relationship by including an entity class for the join table and mapping two separate one-to-many relationships.
protected override void OnModelCreating(ModelBuilder modelBuilder)
{
modelBuilder.Entity<PostTag>()
HasKey(t => new { t.PostId, t.TagId });
modelBuilder.Entity<PostTag>()
HasOne(pt => pt.Post)
WithMany(p => p.PostTags)
HasForeignKey(pt => pt.PostId);
modelBuilder.Entity<PostTag>()
HasOne(pt => pt.Tag)
WithMany(t => t.PostTags)
HasForeignKey(pt => pt.TagId);
}
}

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