如何在android中阅读和使用网站内容

我需要一些帮助； 实际上，我必须在Android应用程序中阅读和使用某些网站的内容。我学习了一些教程，但没有成功。有人可以帮我

更新：

我实际上使用了两种不同的代码来获取网站的内容，但它们对我不起作用

public static String connect(String url)
{
    String result = "bubububu" ;

    HttpClient httpclient = new DefaultHttpClient();

    // Prepare a request object
    HttpGet httpget = new HttpGet(url); 

    // Execute the request
    HttpResponse response;
    try {
        response = httpclient.execute(httpget);
        // Examine the response status
        Log.i("Praeda",response.getStatusLine().toString());

        // Get hold of the response entity
        HttpEntity entity = response.getEntity();
        // If the response does not enclose an entity, there is no need
        // to worry about connection release

        if (entity != null) {

            // A Simple JSON Response Read
            InputStream instream = entity.getContent();
            result= convertStreamToString(instream);
            // now you have the string representation of the HTML request
            instream.close();
            return result ;
        }


    } catch (Exception e) {
        e.getMessage() ;
    }

    return result ;
}

    private static String convertStreamToString(InputStream is) {
    /*
     * To convert the InputStream to String we use the BufferedReader.readLine()
     * method. We iterate until the BufferedReader return null which means
     * there's no more data to read. Each line will appended to a StringBuilder
     * and returned as String.
     */
    BufferedReader reader = new BufferedReader(new InputStreamReader(is));
    StringBuilder sb = new StringBuilder();

    String line = null;
    try {
        while ((line = reader.readLine()) != null) {
            sb.append(line + "\n");
        }
    } catch (IOException e) {
        e.printStackTrace();
    } finally {
        try {
            is.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
    return sb.toString();
}


public static String connect(String url)
{
    String result = "bubububu" ;

    HttpClient httpclient = new DefaultHttpClient();

    // Prepare a request object
    HttpGet httpget = new HttpGet(url); 

    // Execute the request
    HttpResponse response;
    try {
        response = httpclient.execute(httpget);
        // Examine the response status
        Log.i("Praeda",response.getStatusLine().toString());

        // Get hold of the response entity
        HttpEntity entity = response.getEntity();
        // If the response does not enclose an entity, there is no need
        // to worry about connection release

        if (entity != null) {

            // A Simple JSON Response Read
            InputStream instream = entity.getContent();
            result= convertStreamToString(instream);
            // now you have the string representation of the HTML request
            instream.close();
            return result ;
        }


    } catch (Exception e) {
        e.getMessage() ;
    }

    return result ;
}

    private static String convertStreamToString(InputStream is) {
    /*
     * To convert the InputStream to String we use the BufferedReader.readLine()
     * method. We iterate until the BufferedReader return null which means
     * there's no more data to read. Each line will appended to a StringBuilder
     * and returned as String.
     */
    BufferedReader reader = new BufferedReader(new InputStreamReader(is));
    StringBuilder sb = new StringBuilder();

    String line = null;
    try {
        while ((line = reader.readLine()) != null) {
            sb.append(line + "\n");
        }
    } catch (IOException e) {
        e.printStackTrace();
    } finally {
        try {
            is.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
    return sb.toString();
}

及

这两个都给了我一个例外。第一个在

---

response=httpclient.execute（httpget）和exception.getMessage（）为“null”，而第二个在httpConn.setAllowUserInteraction（false）和exception.getMessage（）处出现异常。连接时出错。即使我在menifest文件中使用了Internet权限

看看这个问题的答案：

它有一个可以读取URL的代码

---

但是，最好更具体地说明StackOverflow，并解释您的问题到底是什么。

对这个问题不是100%确定，但您可以使用Apache HTTPClient（建议用于预姜饼）或HTTPURLConnection（姜饼和更高版本）并执行GET以获取网页。从这里，您可以浏览原始数据（通常是HTML，返回文本）。现在有很多关于HTTPClient和HTTPURLConnection的好教程，所以我不在这里解释

---

另一个选择通常是WebView，我承认它可能很混乱。WebView允许您登录并执行诸如从下一页提取结果URL之类的操作。问题是，在Android设备上，WebView的行为是不一样的。

你可能需要更具体地回答这个问题，但我会给出一个答案，如果你不允许从网站管理员那里获得Web服务，那不是一个好主意！谢谢朋友；你们在这里的意思更具体一些。我只有这些详细信息，即网站url，从这里获取一些特定内容，并在我的网站中使用这些内容app@Waza_Be... 我有网站管理员权限httpresponse response=httpclient.execute（httpget）给出异常，exception.getMessage（）给出消息“null”，即使我添加了

private String DownloadText(String URL)
{
    int BUFFER_SIZE = 2000;
    InputStream in = null;
    try {
        in = OpenHttpConnection(URL);
    } catch (IOException e1) {
        // TODO Auto-generated catch block
        e1.printStackTrace();
        return "";
    }

    InputStreamReader isr = new InputStreamReader(in);
    int charRead;
    String str = "";
    char[] inputBuffer = new char[BUFFER_SIZE];          
    try {
        while ((charRead = isr.read(inputBuffer))>0)
        {                    
            //---convert the chars to a String---
            String readString = String.copyValueOf(inputBuffer, 0, charRead);
            str += readString;
            inputBuffer = new char[BUFFER_SIZE];
        }
        in.close();
    } catch (IOException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
        return "";
    }    
    return str;        
}

private InputStream OpenHttpConnection(String urlString) 
        throws IOException
        {
    InputStream in = null;
    int response = -1;

    URL url = new URL(urlString); 
    URLConnection conn = url.openConnection();

    if (!(conn instanceof HttpURLConnection))                     
        throw new IOException("Not an HTTP connection");

    try{
        HttpURLConnection httpConn = (HttpURLConnection) conn;
        httpConn.setAllowUserInteraction(false);
        httpConn.setInstanceFollowRedirects(true);
        httpConn.setRequestMethod("GET");
        httpConn.connect(); 

        response = httpConn.getResponseCode();                 
        if (response == HttpURLConnection.HTTP_OK) {
            in = httpConn.getInputStream();                                 
        }                     
    }
    catch (Exception ex)
    {
        throw new IOException("Error connecting");            
    }
    return in;     
        }