Android，如何在不知道文件路徑的情況下將文件加載為“文件”對象

Question

在我的應用程序中，我需要從xml文件中讀取元素的內容。

因此，我以這種方式在LocalRead.java類中編寫了方法“ getValueOfElement”：

[...]
    public String getValueOfElement (String filename, String what){

    try{
        File xmlDocument = new File ("/Unknow_Path/"+filename);

        DocumentBuilderFactory documentBuilderFactory = DocumentBuilderFactory.newInstance();

        DocumentBuilder documentBuilder = documentBuilderFactory.newDocumentBuilder();

        Document document = documentBuilder.parse(xmlDocument);

        String lookingFor = document.getElementsByTagName(what).item(0).getTextContent();

        return lookingFor;


    } catch (FileNotFoundException e) {
        System.err.println("----------------- File not found -----------------");
        e.printStackTrace();
        return "";
    } catch (ParserConfigurationException e) {
        System.err.println("----------------- Error creating DocumentBuilder -----------------");
        e.printStackTrace();
        return "";
    } catch (SAXException e) {
        System.err.println("----------------- Error creating the document(Sax) -----------------");
        e.printStackTrace();
        return "";
    } catch (IOException e) {
        System.err.println("----------------- Error creating the document(IO) -----------------");
        e.printStackTrace();
        return "";
    }
}

 [...]

如您所見，當我創建文件“ xmlDocument”時，我不知道我的xml文件所在的路徑。 我使用此類創建文件。

import android.content.Context;

import java.io.FileOutputStream;
import java.io.IOException;

public class FileBuilder {

    private String xmlContent;
    private String filename;
    private Context context;
    private FileOutputStream outputStream;

    public FileBuilder(Context context){

        this.context = context;

    }

    public boolean createUserFile(String username, String password){

        this.xmlContent = "<?xml version='1.0' encoding='UTF-8'?>\n" +
                "<giocatore>\n" +
                "<username>"+username+"</username>\n" +
                "<password>"+password+"</password>\n" +
                "</giocatore>";

        this.filename = "[D&D]User.xml";

        try{
            outputStream = context.openFileOutput(filename, context.MODE_PRIVATE);
            outputStream.write(xmlContent.getBytes());
            outputStream.close();
            System.out.println("----------------- File created -----------------");
            return true;
        } catch (IOException e) {
            e.printStackTrace();
            return false;
        }

    }

}

我如何找出路徑？

Answer 1

就像CommonsWare在評論中所說， Document可以解析InputStream 。 因此，您可以使用openFileInput()將文件作為Document加載。 這里是完整的代碼：

[...]    
public String getValueOfElement (String filename, String what){

            try{

                DocumentBuilderFactory documentBuilderFactory = DocumentBuilderFactory.newInstance();

                DocumentBuilder documentBuilder = documentBuilderFactory.newDocumentBuilder();

                Document document = documentBuilder.parse(context.openFileInput(filename));

                String lookingFor = document.getElementsByTagName(what).item(0).getTextContent();

                return lookingFor;


            } catch (FileNotFoundException e) {
                System.err.println("----------------- File not found -----------------");
                e.printStackTrace();
                return "";
            } catch (ParserConfigurationException e) {
                System.err.println("----------------- Error creating DocumentBuilder -----------------");
                e.printStackTrace();
                return "";
            } catch (SAXException e) {
                System.err.println("----------------- Error creating the document(Sax) -----------------");
                e.printStackTrace();
                return "";
            } catch (IOException e) {
                System.err.println("----------------- Error creating the document(IO) -----------------");
                e.printStackTrace();
                return "";
            }
        }
[...]

請記住， openFileInput()需要一個Context 。

感謝@CommonsWare的答案