Relations-Auflistung (DAO)

Gilt für: Access 2013, Office 2013

Eine Relations-Auflistung enthält gespeicherte Relation-Objekte für ein Database-Objekt (gilt nur für Microsoft Access-Datenbanken).

Hinweise

Sie können das Relation-Objekt verwenden, um neue Beziehungen zu erstellen und vorhandene Beziehungen in der Datenbank zu untersuchen. Wenn Sie der Relations-Auflistung ein Relation-Objekt hinzufügen möchten, erstellen Sie es zunächst mit der CreateRelation-Methode und fügen es dann an die Relations-Auflistung an, indem Sie die Append-Methode verwenden. Dadurch wird das Relation-Objekt gespeichert, wenn Sie das Database-Objekt schließen. Wenn Sie ein Relation-Objekt aus der Auflistung entfernen möchten, verwenden Sie die Delete-Methode.

Der Verweis auf ein Relation-Objekt in einer Auflistung erfolgt über dessen Ordnungszahl oder den Wert der Name-Eigenschaft, wobei Sie die folgenden Syntaxformen verwenden können:

Relations(0)

Beziehungen("Name")

Beziehungen! [Name]

Beispiel

This example shows how an existing Relation object can control data entry. The procedure attempts to add a record with a deliberately incorrect CategoryID; this triggers the error-handling routine.

    Sub RelationX() 
     
     Dim dbsNorthwind As Database 
     Dim rstProducts As Recordset 
     Dim prpLoop As Property 
     Dim fldLoop As Field 
     Dim errLoop As Error 
     
     Set dbsNorthwind = OpenDatabase("Northwind.mdb") 
     Set rstProducts = dbsNorthwind.OpenRecordset("Products") 
     
     ' Print a report showing all the different parts of 
     ' the relation and where each part is stored. 
     With dbsNorthwind.Relations!CategoriesProducts 
     Debug.Print "Properties of " & .Name & " Relation" 
     Debug.Print " Table = " & .Table 
     Debug.Print " ForeignTable = " & .ForeignTable 
     Debug.Print "Fields of " & .Name & " Relation" 
     With .Fields!CategoryID 
     Debug.Print " " & .Name 
     Debug.Print " Name = " & .Name 
     Debug.Print " ForeignName = " & .ForeignName 
     End With 
     End With 
     
     ' Attempt to add a record that violates the relation. 
     With rstProducts 
     .AddNew 
     !ProductName = "Trygve's Lutefisk" 
     !CategoryID = 10 
     On Error GoTo Err_Relation 
     .Update 
     On Error GoTo 0 
     .Close 
     End With 
     
     dbsNorthwind.Close 
     
     Exit Sub 
     
    Err_Relation: 
     
     ' Notify user of any errors that result from 
     ' the invalid data. 
     If DBEngine.Errors.Count > 0 Then 
     For Each errLoop In DBEngine.Errors 
     MsgBox "Error number: " & errLoop.Number & _ 
     vbCr & errLoop.Description 
     Next errLoop 
     End If 
     
     Resume Next 
     
    End Sub 

This example uses the CreateRelation method to create a Relation between the Employees TableDef and a new TableDef called Departments. Außerdem wird veranschaulicht, wie beim Erstellen einer neuen Beziehung auch alle erforderlichen Indizes in der Fremdtabelle (der DepartmentsEmployees-Index in der Tabelle Employees) erstellt werden.

    Sub CreateRelationX() 
     
     Dim dbsNorthwind As Database 
     Dim tdfEmployees As TableDef 
     Dim tdfNew As TableDef 
     Dim idxNew As Index 
     Dim relNew As Relation 
     Dim idxLoop As Index 
     
     Set dbsNorthwind = OpenDatabase("Northwind.mdb") 
     
     With dbsNorthwind 
     ' Add new field to Employees table. 
     Set tdfEmployees = .TableDefs!Employees 
     tdfEmployees.Fields.Append _ 
     tdfEmployees.CreateField("DeptID", dbInteger, 2) 
     
     ' Create new Departments table. 
     Set tdfNew = .CreateTableDef("Departments") 
     
     With tdfNew 
     ' Create and append Field objects to Fields 
     ' collection of the new TableDef object. 
     .Fields.Append .CreateField("DeptID", dbInteger, 2) 
     .Fields.Append .CreateField("DeptName", dbText, 20) 
     
     ' Create Index object for Departments table. 
     Set idxNew = .CreateIndex("DeptIDIndex") 
     ' Create and append Field object to Fields 
     ' collection of the new Index object. 
     idxNew.Fields.Append idxNew.CreateField("DeptID") 
     ' The index in the primary table must be Unique in 
     ' order to be part of a Relation. 
     idxNew.Unique = True 
     .Indexes.Append idxNew 
     End With 
     
     .TableDefs.Append tdfNew 
     
     ' Create EmployeesDepartments Relation object, using 
     ' the names of the two tables in the relation. 
     Set relNew = .CreateRelation("EmployeesDepartments", _ 
     tdfNew.Name, tdfEmployees.Name, _ 
     dbRelationUpdateCascade) 
     
     ' Create Field object for the Fields collection of the 
     ' new Relation object. Set the Name and ForeignName 
     ' properties based on the fields to be used for the 
     ' relation. 
     relNew.Fields.Append relNew.CreateField("DeptID") 
     relNew.Fields!DeptID.ForeignName = "DeptID" 
     .Relations.Append relNew 
     
     ' Print report. 
     Debug.Print "Properties of " & relNew.Name & _ 
     " Relation" 
     Debug.Print " Table = " & relNew.Table 
     Debug.Print " ForeignTable = " & _ 
     relNew.ForeignTable 
     Debug.Print "Fields of " & relNew.Name & " Relation" 
     
     With relNew.Fields!DeptID 
     Debug.Print " " & .Name 
     Debug.Print " Name = " & .Name 
     Debug.Print " ForeignName = " & .ForeignName 
     End With 
     
     Debug.Print "Indexes in " & tdfEmployees.Name & _ 
     " TableDef" 
     For Each idxLoop In tdfEmployees.Indexes 
     Debug.Print " " & idxLoop.Name & _ 
     ", Foreign = " & idxLoop.Foreign 
     Next idxLoop 
     
     ' Delete new objects because this is a demonstration. 
     .Relations.Delete relNew.Name 
     .TableDefs.Delete tdfNew.Name 
     tdfEmployees.Fields.Delete "DeptID" 
     .Close 
     End With 
     
    End Sub