himXML (gesprochen himix ML)

[himitsu]

entweder du probierst es mal mit der Serialisierung (Beispiele siehe Projekt Demo_Serialize).
> dieses speichert aber bei Objekten nur published Properties (für "alles" Andere bietet mir Delphi keine oder nur unzureichende Informationen)

z.B. so in dieser Art:

Delphi-Quellcode:

XML := TXMLFile.Create;
Try
  XML.RootNode.AddNode('Computers').Serialize(Computers, ....);
  XML.SaveToFile('Computers.xml');
Finally
  XML.Free;
End;

XML := TXMLFile.Create;
Try
  XML.LoadFromFile('Computers.xml');
  XML.RootNode.AddNode('Computers').Deserialize(Computers, ....);
Finally
  XML.Free;
End;

oder du machst es selber, was aber auch nicht sooooo schwer ist

dieses ergibt dein XML-Beispiel (nur noch mit dem Count-Node):

XML-Code:

<?xml version="1.0" encoding="UTF-8" standalone="yes" ?>
<xml>
  <Computers>
    <Count>1</Count>
    <Computer>
      <Name>Hallo</Name>
      <LastIPByte>101</LastIPByte>
      <Position>
        <Left>100</Left>
        <Top>25</Top>
      </Position>
    </Computer>
  </Computers>
</xml>

Delphi-Quellcode:

// Speichern
XML := TXMLFile.Create;
Try
  XML.RootNode.AddNode('Computers\Count').Data := Computers.Count;
  For i := 0 to Computers.Count - 1 do
    With XML.RootNode.AddNode('Computers\Computer') do Begin
      AddNode('Name').Data          := Computers[i].Name;
      AddNode('LastIPByte').Data    := Computers[i].LastIPByte;
      AddNode('Position\Left').Data := Computers[i].Left;
      AddNode('Position\Top').Data  := Computers[i].Top;
    End;
  XML.SaveToFile('Computers.xml');
Finally
  XML.Free;
End;

// laden
XML := TXMLFile.Create;
Try
  XML.LoadFromFile('Computers.xml');
  Computers.Count := XML.RootNode.Node['Computers\Count'].Data;
  For i := 0 to Computers.Count - 1 do
    With XML.RootNode.NodeList['Computers\Computer'][i] do Begin
      Computers[i].Name       := Node['Name'].Data;
      Computers[i].LastIPByte := Node['LastIPByte'].Data;
      Computers[i].Left       := Node['Position\Left'].Data;
      Computers[i].Top        := Node['Position\Top'].Data;
    End;
Finally
  XML.Free;
End;

und mit Attributen:

XML-Code:

<?xml version="1.0" encoding="UTF-8" standalone="yes" ?>
<xml>
  <Computers Count="1">
    <Computer Name="Hallo" LastIPByte="101" Left="100" Top="25">
  </Computers>
</xml>

Delphi-Quellcode:

// Speichern
XML := TXMLFile.Create;
Try
  XML.RootNode.AddNode('Computers').Attributes['Count'] := Computers.Count;
  For i := 0 to Computers.Count - 1 do
    With XML.RootNode.AddNode('Computers\Computer') do Begin
      Attribute['Name']       := Computers[i].Name;
      Attribute['LastIPByte'] := Computers[i].LastIPByte;
      Attribute['Left']       := Computers[i].Left;
      Attribute['Top']        := Computers[i].Top;
    End;
  XML.SaveToFile('Computers.xml');
Finally
  XML.Free;
End;

// laden
XML := TXMLFile.Create;
Try
  XML.LoadFromFile('Computers.xml');
  Computers.Count := XML.RootNode.Node['Computers'].Attribute['Count'];
  For i := 0 to Computers.Count - 1 do
    With XML.RootNode.NodeList['Computers\Computer'][i] do Begin
      Computers[i].Name       := Attribute['Name'];
      Computers[i].LastIPByte := Attribute['LastIPByte'];
      Computers[i].Left       := Attribute['Left'];
      Computers[i].Top        := Attribute['Top'];
    End;
Finally
  XML.Free;
End;

man könnte auch Count weglassen und zählen:

Delphi-Quellcode:

// Speichern
XML := TXMLFile.Create;
Try
  For i := 0 to Computers.Count - 1 do
    With XML.RootNode.AddNode('Computers\Computer') do Begin
      ...
    End;
  XML.SaveToFile('Computers.xml');
Finally
  XML.Free;
End;

// laden
XML := TXMLFile.Create;
Try
  XML.LoadFromFile('Computers.xml');
  Computers.Count := Length(XML.RootNode.Node['Computers'].NodeList['Computer']);
  For i := 0 to Computers.Count - 1 do
    With XML.RootNode.NodeList['Computers\Computer'][i] do Begin
      ...
    End;
Finally
  XML.Free;
End;

// laden 2 (wenn es eh keine anderen Subnodes im Node "Computers" gibt)
XML := TXMLFile.Create;
Try
  XML.LoadFromFile('Computers.xml');
  Computers.Count := XML.RootNode.Node['Computers'].Nodes.Count;
  For i := 0 to Computers.Count - 1 do
    With XML.RootNode.Node['Computers'].Node[i] do Begin
      ...
    End;
Finally
  XML.Free;
End;

ich hoff mal, es ist jetzt nicht zu schwer?