URLConnection不允许我访问有关Http错误的数据(404,500等)
URLConnection不允许我访问有关Http错误的数据(404,500等)
提问于 2012-02-03 21:58:42
我正在做一个爬虫，并需要从流中获取数据，无论它是200或不是。CURL正在做这件事，就像任何标准的浏览器一样。
下面的代码不会实际获取请求的内容，即使有，也会抛出异常，并显示http错误状态代码。不管怎样，我都想要输出，有没有办法？我更喜欢使用这个库，因为它实际上会做持久连接，这对于我正在做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);
}
}
}成功了，谢谢:这是我想出来的--只是粗略的概念证明：
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);
}
}回答 1
Stack Overflow用户
回答已采纳
发布于 2012-02-03 22:15:21
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。我处理这个问题的最好方法是：
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();
}
}
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/9129766
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