Question

我需要使用commons-httpclient-3.1.jar库，即使它现在已经结束了。

我正在尝试通过代理服务器地址“10.100.1.44”与代理端口“8080”从URL http://www.apache.org/执行简单的http GET，这不需要任何凭据。

以下是我的示例代码。

import java.io.IOException;
import org.apache.commons.httpclient.DefaultHttpMethodRetryHandler;
import org.apache.commons.httpclient.HttpClient;
import org.apache.commons.httpclient.HttpException;
import org.apache.commons.httpclient.HttpStatus;
import org.apache.commons.httpclient.UsernamePasswordCredentials;
import org.apache.commons.httpclient.auth.AuthScope;
import org.apache.commons.httpclient.methods.GetMethod;
import org.apache.commons.httpclient.params.HttpMethodParams;

public class HttpClientTutorial {

    private static String url = "http://www.apache.org/";

    public static void main(String[] args) {
        // Create an instance of HttpClient.
        HttpClient client = new HttpClient();

        // Create a method instance.
        GetMethod method = new GetMethod(url);

        // Provide custom retry handler is necessary
        method.getParams().setParameter(HttpMethodParams.RETRY_HANDLER,
                new DefaultHttpMethodRetryHandler(3, false));

        try {
            // setting proxy
            client.getParams().setAuthenticationPreemptive(true);
            client.getState().setProxyCredentials(
                    new AuthScope("10.100.1.44", 8080),
                    new UsernamePasswordCredentials("", ""));

            // Execute the method.
            int statusCode = client.executeMethod(method);

            if (statusCode != HttpStatus.SC_OK) {
                System.err.println("Method failed: " + method.getStatusLine());
            }

            // Read the response body.
            byte[] responseBody = method.getResponseBody();

            // Deal with the response.
            // Use caution: ensure correct character encoding and is not binary
            // data
            System.out.println(new String(responseBody));

        } catch (HttpException e) {
            System.err.println("Fatal protocol violation: " + e.getMessage());
            e.printStackTrace();
        } catch (IOException e) {
            System.err.println("Fatal transport error: " + e.getMessage());
            e.printStackTrace();
        } finally {
            // Release the connection.
            method.releaseConnection();
        }
    }
}

但是当我运行代码时，我会抛出异常

14 ส.ค. 2556 13:11:28 org.apache.commons.httpclient.HttpMethodDirector authenticateHost
WARNING: Required credentials not available for BASIC <any realm>@www.apache.org:80
14 ส.ค. 2556 13:11:28 org.apache.commons.httpclient.HttpMethodDirector authenticateHost
WARNING: Preemptive authentication requested but no default credentials available
Fatal transport error: www.apache.org
java.net.UnknownHostException: www.apache.org
    at java.net.PlainSocketImpl.connect(Unknown Source)
    at java.net.SocksSocketImpl.connect(Unknown Source)
    at java.net.Socket.connect(Unknown Source)
    at java.net.Socket.connect(Unknown Source)
    at java.net.Socket.<init>(Unknown Source)
    at java.net.Socket.<init>(Unknown Source)
    at org.apache.commons.httpclient.protocol.DefaultProtocolSocketFactory.createSocket(DefaultProtocolSocketFactory.java:80)
    at org.apache.commons.httpclient.protocol.DefaultProtocolSocketFactory.createSocket(DefaultProtocolSocketFactory.java:122)
    at org.apache.commons.httpclient.HttpConnection.open(HttpConnection.java:707)
    at org.apache.commons.httpclient.HttpMethodDirector.executeWithRetry(HttpMethodDirector.java:387)
    at org.apache.commons.httpclient.HttpMethodDirector.executeMethod(HttpMethodDirector.java:171)
    at org.apache.commons.httpclient.HttpClient.executeMethod(HttpClient.java:397)
    at org.apache.commons.httpclient.HttpClient.executeMethod(HttpClient.java:323)
    at jenkins.plugin.assembla.api.HttpClientTutorial.main(HttpClientTutorial.java:38)

有人可以告诉我我做错了什么，因为我是 HttpClient 库的新手，但我需要使用它，因为我无意使用 HttpComponents 库吗？ / p>

现在我找到了解决这个问题的方法。请参考下面的代码。

import java.io.IOException;
import org.apache.commons.httpclient.Credentials;
import org.apache.commons.httpclient.DefaultHttpMethodRetryHandler;
import org.apache.commons.httpclient.HttpClient;
import org.apache.commons.httpclient.HttpException;
import org.apache.commons.httpclient.HttpStatus;
import org.apache.commons.httpclient.UsernamePasswordCredentials;
import org.apache.commons.httpclient.auth.AuthScope;
import org.apache.commons.httpclient.methods.GetMethod;
import org.apache.commons.httpclient.params.HttpMethodParams;

public class HttpClientTutorial {

    private static String url = "http://www.apache.org/";

    public static void main(String[] args) {
        // Create an instance of HttpClient.
        HttpClient client = new HttpClient();

        // Create a method instance.
        GetMethod method = new GetMethod(url);

        // Provide custom retry handler is necessary
        method.getParams().setParameter(HttpMethodParams.RETRY_HANDLER,
                new DefaultHttpMethodRetryHandler(3, false));

        try {
            // setting proxy
            client.getHostConfiguration().setProxy("10.100.1.44", 8080);

            // Execute the method.
            int statusCode = client.executeMethod(method);

            if (statusCode != HttpStatus.SC_OK) {
                System.err.println("Method failed: " + method.getStatusLine());
            }

            // Read the response body.
            byte[] responseBody = method.getResponseBody();

            // Deal with the response.
            // Use caution: ensure correct character encoding and is not binary
            // data
            System.out.println(new String(responseBody));

        } catch (HttpException e) {
            System.err.println("Fatal protocol violation: " + e.getMessage());
            e.printStackTrace();
        } catch (IOException e) {
            System.err.println("Fatal transport error: " + e.getMessage());
            e.printStackTrace();
        } finally {
            // Release the connection.
            method.releaseConnection();
        }
    }
}

Answer 1

答案中存在类似的问题。我认为这与你的情况相同：

Using HttpProxy to connect to a host with preemtive authentication

从commons-httpclient-3.1.jar抛出的java.net.UnknownHostException

1 个答案: