我是delphi语言的新手,在这里我有一个疑问,我有一个名为vehicle.xml的xml文件。
看起来像这样
<data>
<vehicle>
<type>Car</type>
<model>2005</model>
<number>1568</number>
</vehicle>
<vehicle>
<type>Car</type>
<model>2009</model>
<number>1598</number>
</vehicle>
</data>
我的Delphi表单包含三个文本框:
在加载页面时,我想在文本框中显示vehicle.xml的内容,如:
答案 0 :(得分:11)
查看Delphi自己的TXMLDocument
组件,例如:
procedure TForm1.FormCreate(Sender: TObject);
var
Vehicle: IXMLNode;
begin
XMLDocument1.FileName :='vehicle.xml';
XMLDocument1.Active := True;
try
Vehicle := XMLDocument.DocumentElement;
txtType.Text := Vehicle.ChildNodes['type'].Text;
txtModel.Text := Vehicle.ChildNodes['model'].Text;
txtnumber.Text := Vehicle.ChildNodes['number'].Text;
finally
XMLDocument1.Active := False;
end;
end;
或者,直接使用IXMLDocument
接口(TXMLDocument
包装):
procedure TForm1.FormCreate(Sender: TObject);
var
Doc: IXMLDocument;
Vehicle: IXMLNode;
begin
Doc := LoadXMLDocument('vehicle.xml');
Vehicle := Doc.DocumentElement;
txtType.Text := Vehicle.ChildNodes['type'].Text;
txtModel.Text := Vehicle.ChildNodes['model'].Text;
txtnumber.Text := Vehicle.ChildNodes['number'].Text;
end;
更新:问题中的XML已更改为现在将vehicle
元素包含在data
元素内,并且具有多个vehicle
元素。所以上面的代码必须相应调整,例如:
procedure TForm1.FormCreate(Sender: TObject);
var
Doc: IXMLDocument;
Data: IXMLNode;
Node: IXMLNode;
I: Integer;
begin
Doc := LoadXMLDocument('vehicle.xml');
Data := Doc.DocumentElement;
for I := 0 to Data.ChildNodes.Count-1 do
begin
Node := Data.ChildNodes[I];
// if all of the child nodes will always be 'vehicle' only
// then this check can be removed...
if Node.LocalName = 'vehicle' then
begin
// use Node.ChildNodes['type'], Node.ChildNodes['model'],
// and Node.ChildNodes['number'] as needed...
end;
end;
end;
答案 1 :(得分:4)
您可以使用单位MSXML(或任何其他XML解析器)读取XML文件。
它为您提供表示XML文件的树结构。其中车辆是顶部节点而其他三个是子节点。
每个节点都有一个可用于获取值的text属性。您可以将其分配给表单上的文本框。
代码示例:
uses
ActiveX,
MSXML;
procedure TForm1.ReadFromXML(const AFilename: string);
var
doc : IXMLDOMDocument;
node : IXMLDomNode;
begin
CoInitialize; // Needs to be called once before using CoDOMDocument.Create;
if not FileExists(AFileName) then
Exit; // Add proper Error handling.
doc := CoDOMDocument.Create;
doc.load(AFileName);
if (doc.documentElement = nil) or (doc.documentElement.nodeName <> 'vehicle') then
Exit; // Add proper Error handling.
node := doc.documentElement.firstChild;
while node<>nil do begin
if node.nodeName = 'model' then
txtModel.Text := node.text;
if node.nodeName = 'number' then
txtNumber.Text := node.text;
if node.nodeName = 'type' then
txtType.Text := node.text;
node := node.nextSibling;
end;
end;
答案 2 :(得分:-1)
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