Question

在开发Hello World背后的Android应用程序时，我是初学者！所以我真的需要有人告诉我我的应用程序有什么问题。我的应用程序非常简单，用户在editText中输入问题并按下按钮，然后应用程序从Yahoo Answers API检索信息。但当我按下按钮时，我得到的是TextView中的“错误”消息。我真的需要另一只眼睛来看我的代码并告诉我出了什么问题。我的模拟器正在运行android 4.2.2。我不知道它的url是否会导致显示“错误”消息，或者是否是eclipse中的某些内容或我的代码中的内容。
有人请帮助!!

主要活动

import java.net.URL;




import javax.xml.parsers.SAXParser;

import javax.xml.parsers.SAXParserFactory;




import org.xml.sax.InputSource;

import org.xml.sax.XMLReader;




import android.app.Activity;

import android.os.Bundle;

import android.view.View;

import android.view.View.OnClickListener;

import android.widget.Button;

import android.widget.EditText;

import android.widget.TextView;







public class WhatsYourQuestion extends Activity implements OnClickListener {



    static final String baseURL = "http://answer.yahooapis.com/AnswerService/V1/questionSearh?appid=example&query=";

    EditText et; 

    TextView answer; 





    @Override

    protected void onCreate(Bundle icicle) {

        super.onCreate(icicle);

        setContentView(R.layout.main);



        et = (EditText) findViewById(R.id.editText1);




        answer = (TextView) findViewById(R.id.stockAnswerOuput);



        Button myButton = (Button) findViewById(R.id.button1);

        myButton.setOnClickListener(new OnClickListener()  {







            @Override

            public void onClick(View v) {

            String a = et.getText().toString(); 

           StringBuilder URL = new StringBuilder(baseURL);

            URL.append(a + "");

            String fullUrl = URL.toString();

            try{

                URL website = new URL(fullUrl);

                SAXParserFactory spf = SAXParserFactory.newInstance();

                SAXParser sp = spf.newSAXParser();

                XMLReader xr = sp.getXMLReader();

                HandlingXMLProperties doingWork = new HandlingXMLProperties();

                xr.setContentHandler(doingWork);

                xr.parse(new InputSource(website.openStream()));

                String information = doingWork.getInformation();

                et.setText(information);

                           }catch (Exception e){

                et.setText("error");

XMLCollectedData类

public class XMLCollectedData {



    private int answer = 0;

    public void setAnswer(int a){
        answer = a;
    }

    public String dataToString(){
        return "Here is your answer" + answer;
    }

处理xml属性类

import org.xml.sax.Attributes;
import org.xml.sax.SAXException;
import org.xml.sax.helpers.DefaultHandler;

public class HandlingXMLProperties extends DefaultHandler{

    XMLCollectedData info = new XMLCollectedData();

    public String getInformation() {
        return info.dataToString();
    }
    @Override
    public void startElement(String uri, String localName, String qName,
            Attributes attributes) throws SAXException {
        // TODO Auto-generated method stub

        if(localName.equals("ChosenAnswer")){
            String a = attributes.getValue("type");
            int answer = Integer.parseInt(a);
            info.setAnswer(answer);
        }

我还需要知道我需要使用哪种santax，所以我可以定义qName。有人帮助

在Application中显示API信息

0 个答案: