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

Question

我正在制作一个爬虫,并且需要从流中获取数据,无论它是否为200.CURL正在这样做,以及任何标准浏览器.

以下实际上不会获取请求的内容,即使有一些,也会引发http错误状态代码的异常.我想要输出,是否有办法？我更喜欢使用这个库,因为它实际上会执行持久连接,这对于我正在进行的爬行类型来说是完美的.

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);
        }
    }
}

工作,谢谢:这是我想出的 - 就像一个粗略的概念证明:

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.我要处理的最好方法如下:

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();
        }
    }
}

Answer 2

您需要在致电后执行以下操作openConnection.

1. 将URLConnection转换为HttpURLConnection

2. 调用getResponseCode

3. 如果响应成功,请使用getInputStream,否则使用getErrorStream

(成功的测试应该是200 <= code < 300因为除了200之外还有有效的HTTP成功代码.)

我正在制作一个爬虫,并且需要从流中获取数据,无论它是否为200.

请注意,如果代码是4xx或5xx,那么"数据"很可能是某种错误页面.

最后一点应该是您应该始终尊重"robots.txt"文件...并在抓取/抓取其所有者可能关心的网站内容之前阅读服务条款.简单地吹嘘GET请求可能会让网站所有者感到烦恼......除非你已经与他们达成某种"安排".