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我在 Android 资源中有 5 个文件(test1、test2、test3、test4、test5)。所有资源都应该分配给变量parser。应该如何正确完成?

公共类 MainActivity 扩展 Activity {

@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);

    setContentView(R.layout.activity_main);

        XmlResourceParser parser = getResources().getXml(R.xml.test1);
        databaseHelper myDbHelper  = new databaseHelper(this);

    try {
        //myDbHelper = new databaseHelper(this);


        while (parser.next() != XmlPullParser.END_TAG) {
            if (parser.getEventType() != XmlPullParser.START_TAG) {
                continue;
            }
            String name = parser.getName();
            if (name.equals("test") || name.equals("test")) {
                String id = null, gname = null, pob = null, pob1 = null, memb = null, seq = null, site = null;
                while (parser.next() != XmlPullParser.END_TAG) {
                    if (parser.getEventType() != XmlPullParser.START_TAG) {
                        continue;
                    }
                    name = parser.getName();                            
                    if (name.equals("test")) {
                        id = readText(parser);
                    } else if (name.equals("test")) {
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