如何访问的资产文件夹在我的Android应用程序？ [英] How to access assets folder in my android app?
问题描述
我正在写在给朋友的小项目。一个notecard的应用程序。 我的计划是把notecards XML格式，所以我可以导入并轻松导出。 我所在的XML文件的资产/ XML / mynotecard.xml文件夹，但我只是不能管理才能访问这个文件。
I'm writing at a little project for a friend. A notecard application. My plan is to put the notecards in an xml format so i can import and export them easily. I located the xml files at the assets/xml/mynotecard.xml folder but i just can't manage to get access to this file.
每当我尝试跨preT XML文件（将投入到它的阶级之后）我得到的异常：的test.xml不是一个文件。 这是我的code摘录：
Whenever i try to interpret the xml file (will be put to it's on class later) i get the exception with: test.xml is not a file. This is an extract of my code:
公共类NotecardProActivity延伸活动{
public class NotecardProActivity extends Activity {
List<String> xmlFiles;
public ArrayList<File> xmlFileList;
XMLInterpreter horst;
/** Called when the activity is first created. */
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.main);
final AssetManager mgr = getAssets();
displayFiles(mgr, "",0);
xmlFiles = displayFiles(mgr, "xml",0);
for (int e = 0; e<=xmlFiles.size()-1;e++)
{
Log.v("Inhalt List"+e+": ", xmlFiles.get(e));
}
xmlFileList = new ArrayList<File>();
for (int i = 0; i<=xmlFiles.size()-1;i++)
{
xmlFileList.add(new File("./assets/xml/"+xmlFiles.get(i)));
}
for (int i = 0; i<=xmlFileList.size()-1;i++)
{
Log.v("Name: ", xmlFileList.get(i).getName());
Log.v("Filelocation: ",xmlFileList.get(i).getPath());
Log.v("Filelocation: ",xmlFileList.get(i).getAbsolutePath());
}
Log.v("DEBUG","XML FILE LISTE erfolgreich geschrieben!");
//Alternative zum ausgelagerten XML INTERPRETER
try
{
if(xmlFileList.get(0) == null){
Log.v("Debug", "XMLFILE IS NULL");
}
else{
DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
Document doc = dBuilder.parse(xmlFileList.get(0));
doc.getDocumentElement().normalize();
Log.v("Root element :", doc.getDocumentElement().getNodeName());
NodeList nList = doc.getElementsByTagName("notecard");
for (int temp = 0; temp < nList.getLength(); temp++) {
Node nNode = nList.item(temp);
if (nNode.getNodeType() == Node.ELEMENT_NODE) {
Element eElement = (Element) nNode;
Log.v("GesetzesText: " , getTagValue("legal_text", eElement));
// System.out.println("Kommentar : " + getTagValue("comment", eElement));
}
}
}
}
catch (Exception e) {
e.printStackTrace();
}
mgr.close();
}
private List<String> displayFiles (AssetManager mgr, String path, int level) {
List<String> abc = new ArrayList<String>();
Log.v("Hello","enter displayFiles("+path+")");
try {
String list[] = mgr.list(path);
abc.addAll(Arrays.asList(list));
Log.v("Hello1","L"+level+": list:"+ Arrays.asList(list));
} catch (IOException e) {
Log.v("Fail","List error: can't list" + path);
}
return abc;
}
private String getTagValue(String sTag, Element eElement) {
NodeList nlList = eElement.getElementsByTagName(sTag).item(0).getChildNodes();
Node nValue = (Node) nlList.item(0);
return nValue.getNodeValue();
}
}
将是真正真棒，如果你们能帮助我：）
Would be really awesome if you guys could help me :)
推荐答案
邮编你的XML文件夹（xml.zip为例），并把它的资产。获得资产，AssetsManager很慢。此外，这样做可以保持你的目录结构，只需将它复制到SD卡并提取它。这将是比让每一个文件的目录快了很多。你可以使用这个职位的解压：如何以编程方式在Android的解压文件。希望这有助于。
Zip your xml folder (xml.zip for example) and put it in assets. Getting assets with AssetsManager is very slow. Also, doing so you can maintain your directory structure and just copy it to sdcard and extract it there. This will be a lot faster than getting each file in the directory. You can use this post for the unzipping: how to unzip files programmatically in android?. Hope this helps.
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