SqlError.Source Eigenschap

Definitie

Naamruimte:
System.Data.SqlClient
Assembly:
System.Data.dll

Hiermee haalt u de naam op van de provider die de fout heeft gegenereerd.

public:
 property System::String ^ Source { System::String ^ get(); };
public string Source { get; }
member this.Source : string
Public ReadOnly Property Source As String

Waarde van eigenschap

String

De naam van de provider die de fout heeft gegenereerd.

Voorbeelden

In het volgende voorbeeld wordt elk SqlError in de SqlErrorCollection verzameling weergegeven.

public void DisplaySqlErrors(SqlException exception)
{
    for (int i = 0; i < exception.Errors.Count; i++)
    {
        Console.WriteLine("Index #" + i + "\n" +
            "Source: " + exception.Errors[i].Source + "\n" +
            "Number: " + exception.Errors[i].Number.ToString() + "\n" +
            "State: " + exception.Errors[i].State.ToString() + "\n" +
            "Class: " + exception.Errors[i].Class.ToString() + "\n" +
            "Server: " + exception.Errors[i].Server + "\n" +
            "Message: " + exception.Errors[i].Message + "\n" +
            "Procedure: " + exception.Errors[i].Procedure + "\n" +
            "LineNumber: " + exception.Errors[i].LineNumber.ToString());
    }
    Console.ReadLine();
}

Public Sub DisplaySqlErrors(ByVal exception As SqlException)
    Dim i As Integer

    For i = 0 To exception.Errors.Count - 1
        Console.WriteLine(("Index #" & i & ControlChars.NewLine & _
            "Source: " & exception.Errors(i).Source & ControlChars.NewLine & _
            "Number: " & exception.Errors(i).Number.ToString() & ControlChars.NewLine & _
            "State: " & exception.Errors(i).State.ToString() & ControlChars.NewLine & _
            "Class: " & exception.Errors(i).Class.ToString() & ControlChars.NewLine & _
            "Server: " & exception.Errors(i).Server & ControlChars.NewLine & _
            "Message: " & exception.Errors(i).Message & ControlChars.NewLine & _
            "Procedure: " & exception.Errors(i).Procedure & ControlChars.NewLine & _
            "LineNumber: " & exception.Errors(i).LineNumber.ToString()))
    Next i
    Console.ReadLine()
End Sub

Van toepassing op

Zie ook