我试图从这个URL获得简单的HTTP响应:http://realtorsipad.demowebsiteonline.net/eventsfeed.php
但令人惊讶的是,它没有返回预期的XML响应而是返回另一个HTML页面!
我无法理解什么是问题.
以下是示例活动:
public class MainActivity1 extends Activity {
String parsingWebURL = "http://realtorsipad.demowebsiteonline.net/eventsfeed.php";
String line = "";
Document docXML;
/** Called when the activity is first created. */
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.main);
line = getXML();
System.out.println(line);
}
// ------------------------------------------------
public String getXML() {
String strXML = "";
try {
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost(parsingWebURL);
HttpResponse httpResponse = httpClient.execute(httpPost);
HttpEntity httpEntity = httpResponse.getEntity();
strXML = EntityUtils.toString(httpEntity);
return strXML; …public class JavaPuzzler {
public static void main(String[] args) {
JavaPuzzler javaPuzzler = null;
System.out.println(javaPuzzler.get());
}
private static String get(){
return "i am a java puzzler";
}
}
您可能认为它应该抛出NullPointerException,因为main方法在局部变量上调用get()方法,该方法初始化为null,并且您不能在null上调用方法.
但是如果你运行这个程序,你会看到它打印出"我是一个java puzzler".
任何人都可以给我答案.提前致谢.
如果我试图检索第一个数组, (String) response.getProperty(0);但它返回一个完整的字符串.
**Here is the code of webservice calling**
public static Object getResponse(String methodName, String actionName, LinkedHashMap<String, String> valueStrings)
{
SoapObject soapObject = new SoapObject(LetUsClickAPIConstants.COMMON_NAMESPACE, methodName);
Object response = null;
for (Map.Entry<String, String> mapKeys : valueStrings.entrySet())
{
soapObject.addProperty(mapKeys.getKey(), mapKeys.getValue());
}
final SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);
envelope.dotNet = false;
envelope.setOutputSoapObject(soapObject);
final AndroidHttpTransport androidHttpTransport = new AndroidHttpTransport(LetUsClickAPIConstants.COMMON_URL);
try
{
androidHttpTransport.call(actionName, envelope);
response = envelope.getResponse();
Log.d("Response ", response.toString());
} catch (final IOException e)
{
Log.d("Exception", "" + e);
} catch …我正在使用编辑文本来支持粗体,斜体和下划线的属性.在选择文本并使其变为粗体后,我获得了成功.现在我想在点击Normal按钮后删除粗体.
Typeface.NORMAL在这里不起作用.任何人都可以建议其他选择.
Button btnBold = (Button) findViewById(R.id.btnBold);
btnBold.setOnClickListener(new OnClickListener() {
@Override
public void onClick(View v) {
startSelection = etx.getSelectionStart();
endSelection = etx.getSelectionEnd();
Spannable s = etx.getText();
s.setSpan(new StyleSpan(android.graphics.Typeface.BOLD), startSelection, endSelection, 0);
}
});
Button btnNormal = (Button) findViewById(R.id.btnNormal );
btnNormal .setOnClickListener(new OnClickListener() {
@Override
public void onClick(View v) {
**//What I have to do here.**
}
});
我正在开发一个类似记事本的应用程序,我想在其中动态更改所选的文本格式(颜色,更改字体样式,粗体,斜体,下划线等).如何格式化特定单词？
我有应用程序,我需要onTouch滑动向上和向下事件.我已经在应用程序中实现了onTouch事件,但我没有任何关于Swipe Up和Swipe Down的想法是否有人有一个代码,我该怎么做？
final ImageView imageView = listOfImageView.get(i);
imageView.setOnTouchListener(new OnTouchListener () {
private float oldPosition;
public boolean onTouch(View view, MotionEvent event) {
if (event.getAction() == android.view.MotionEvent.ACTION_DOWN) {
this.oldPosition = event.getY();
}
}
});
我有应用程序,用户可以从SD卡搜索图像,音频,视频,doc文件,并选择1个文件上传它在服务器上.使用下面的代码我可以打开图库并选择图像,音频,视频但我不知道如何从库中搜索文档.
这是我的代码.
Intent intent = new Intent();
intent.setAction(Intent.ACTION_GET_CONTENT);
//intent.setType("video/*");
//intent.setType("audio/*");
//intent.setType("image/*");
//**What I have to do for view document[.pdf/text/doc] file**
startActivityForResult(Intent.createChooser(intent, "Complete action using"), REQUEST_CODE);
有谁知道如何实现这一目标？任何帮助是极大的赞赏.
这是一个程序
public class MovieTitle {
public static void main(String[] args) {
Movie one = new Movie();
one.title = "I am title";
one.playIt();
System.out.println(one.title);
}
}
class Movie {
String title;
void playIt() {
this.title = "I am title of movie";
}
}
输出是"我是电影的标题" 我试图理解它,但直到现在我还是不理解它.我想知道:为什么不打印"我是头衔"