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NEW QUESTION: 1
What is the first step suggested before implementing any single CIS Control?
A. Perform a gap analysis
B. Develop a roll-out schedule
C. Perform a vulnerability scan
D. Develop an effectiveness test
Answer: A

NEW QUESTION: 2
What does the Cisco ASR 5000 PDSN do when subscriber's authentication is not found in subscriber's domain alias in source/AAA context?
A. PDSN selects AAA subscriber default in the context
B. PDSN rejects the call with Admin Prohibited
C. PDSN rejects the call with Insufficient Resources
D. PDSN proceeds without authentication
Answer: A

NEW QUESTION: 3
Sie arbeiten für ein Unternehmen, das mobile Anwendungen entwickelt. Sie unterhalten einen Server, auf dem die Spieleraufzeichnungen ihren verschiedenen Spielen zugewiesen werden. Das Trackingsystem ist neu und in Entwicklung.
Die Anwendung verwendet Entity Framework, um eine Verbindung zu einer Azure-Datenbank herzustellen. Die Datenbank enthält eine Spielertabelle und eine Spieltabelle.
Beim Hinzufügen eines Spielers sollte der Code einen neuen Spielerdatensatz einfügen und eine Beziehung zwischen einem vorhandenen Spielerdatensatz und dem neuen Spielerdatensatz hinzufügen.
Die Anwendung ruft CreatePlayerWithGame mit der richtigen gameId und der playerId auf, um den Vorgang zu starten.
(Zeilennummern dienen nur als Referenz.)

Wählen Sie für jede der folgenden Anweisungen Ja aus, wenn die Anweisung wahr ist. Andernfalls wählen Sie Nein.
HINWEIS: Jede richtige Auswahl ist einen Punkt wert.

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);
}
}