URLConnection不允许我访问有关Http错误的数据（404,500等）

Question

我正在制作搜寻器，无论是否为200，都需要从流中获取数据。CURL以及任何标准浏览器都在这样做。

以下内容实际上不会获得请求的内容，即使有一些内容，http错误状态代码也会引发异常。我想要输出，有办法吗？我更喜欢使用此库，因为它实际上会进行持久连接，这对于我正在执行的爬网类型非常理想。

package test;

import java.net.*;
import java.io.*;

public class Test {

    public static void main(String[] args) {

         try {

            URL url = new URL("http://github.com/XXXXXXXXXXXXXX");
            URLConnection connection = url.openConnection();

            DataInputStream inStream = new DataInputStream(connection.getInputStream());
            String inputLine;

            while ((inputLine = inStream.readLine()) != null) {
                System.out.println(inputLine);
            }
            inStream.close();
        } catch (MalformedURLException me) {
            System.err.println("MalformedURLException: " + me);
        } catch (IOException ioe) {
            System.err.println("IOException: " + ioe);
        }
    }
}

辛苦了，谢谢：这是我想出的-只是概念的粗略证明：

import java.net.*;
import java.io.*;

public class Test {

    public static void main(String[] args) {
//InputStream error = ((HttpURLConnection) connection).getErrorStream();

        URL url = null;
        URLConnection connection = null;
        String inputLine = "";

        try {

            url = new URL("http://verelo.com/asdfrwdfgdg");
            connection = url.openConnection();

            DataInputStream inStream = new DataInputStream(connection.getInputStream());

            while ((inputLine = inStream.readLine()) != null) {
                System.out.println(inputLine);
            }
            inStream.close();
        } catch (MalformedURLException me) {
            System.err.println("MalformedURLException: " + me);
        } catch (IOException ioe) {
            System.err.println("IOException: " + ioe);

            InputStream error = ((HttpURLConnection) connection).getErrorStream();

            try {
                int data = error.read();
                while (data != -1) {
                    //do something with data...
                    //System.out.println(data);
                    inputLine = inputLine + (char)data;
                    data = error.read();
                    //inputLine = inputLine + (char)data;
                }
                error.close();
            } catch (Exception ex) {
                try {
                    if (error != null) {
                        error.close();
                    }
                } catch (Exception e) {

                }
            }
        }

        System.out.println(inputLine);
    }
}

Answer 1

简单：

URLConnection connection = url.openConnection();
InputStream is = connection.getInputStream();
if (connection instanceof HttpURLConnection) {
   HttpURLConnection httpConn = (HttpURLConnection) connection;
   int statusCode = httpConn.getResponseCode();
   if (statusCode != 200 /* or statusCode >= 200 && statusCode < 300 */) {
     is = httpConn.getErrorStream();
   }
}

您可以参考Javadoc进行解释。我将处理此问题的最佳方法如下：

URLConnection connection = url.openConnection();
InputStream is = null;
try {
    is = connection.getInputStream();
} catch (IOException ioe) {
    if (connection instanceof HttpURLConnection) {
        HttpURLConnection httpConn = (HttpURLConnection) connection;
        int statusCode = httpConn.getResponseCode();
        if (statusCode != 200) {
            is = httpConn.getErrorStream();
        }
    }
}