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NEW QUESTION: 1
インターネット履歴レコードの調査中に、Unique Resource Locator（URL）内に次の文字列が表示されます。
http://www.companysite.com/products/products.asp?productid=123
or 1=1
これはどのような種類の攻撃を意味しますか？
A. クロスサイトスクリプティング（XSS）
B. シェルコードインジェクション
C. 構造化照会言語（SQL）インジェクション
D. ディレクトリトラバーサル
Answer: A

NEW QUESTION: 2
Sie haben eine lokale Microsoft SQL Server-Instanz.
Sie erstellen eine Datenbank mit dem Namen DB1.
Sie müssen DB1 für die Unterstützung von In-Memory-OLTP konfigurieren.
Welche beiden Aktionen sollten Sie ausführen? Jede richtige Antwort ist Teil der Lösung.
HINWEIS; Jede richtige Auswahl ist einen Punkt wert.
A. Konfigurieren Sie eine neue Dateigruppe für Dateistream-Daten.
B. Dateistream für die Instanz aktivieren.
C. In-Memory OLTP für die Instanz aktivieren.
D. Konfigurieren Sie eine neue Datei für eine Dateigruppe.
E. Konfigurieren Sie eine neue Dateigruppe für speicheroptimierte Daten.
Answer: C,E
Explanation:
References:
https://docs.microsoft.com/en-us/sql/relational-databases/in-memory-oltp/survey-of-initial-areas-in-in-memory-oltp?view=sql-server-ver15

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);
}
}