org.apache.http.ProtocolException:未指定目标主机

问题描述 投票:0回答:3

我编写了一个简单的http请求/响应代码,但出现以下错误。我在类路径中引用了 httpclient、httpcore、common-codecs 和 common-logging。我对java很陌生,不知道这里发生了什么。请帮我。

代码:

import org.apache.http.client.HttpClient;
import org.apache.http.client.methods.HttpGet;
import org.apache.http.HttpResponse;
import org.apache.http.impl.client.HttpClientBuilder;
import org.apache.http.Header;
import org.apache.http.HttpHeaders;

public class UnshorteningUrl {

    public static void main(String[] args) throws Exception
    {           
        HttpGet request=null;
        HttpClient client = HttpClientBuilder.create().build();         

        try {
            request = new HttpGet("trib.me/1lBFzSi");
            HttpResponse httpResponse=client.execute(request);

            Header[] headers = httpResponse.getHeaders(HttpHeaders.LOCATION);
           // Preconditions.checkState(headers.length == 1);
            String newUrl = headers[0].getValue();          
            System.out.println("new url" + newUrl);         
        } catch (IllegalArgumentException e) {
            // TODO: handle exception
        }finally {
            if (request != null) {
                request.releaseConnection();
            }           
        }
    }}

错误:

Exception in thread "main" org.apache.http.client.ClientProtocolException
    at org.apache.http.impl.client.InternalHttpClient.doExecute(InternalHttpClient.java:186)
    at org.apache.http.impl.client.CloseableHttpClient.execute(CloseableHttpClient.java:82)
    at org.apache.http.impl.client.CloseableHttpClient.execute(CloseableHttpClient.java:106)
    at org.apache.http.impl.client.CloseableHttpClient.execute(CloseableHttpClient.java:57)
    at UnshorteningUrl.main(UnshorteningUrl.java:26)
Caused by: org.apache.http.ProtocolException: Target host is not specified
    at org.apache.http.impl.conn.DefaultRoutePlanner.determineRoute(DefaultRoutePlanner.java:69)
    at org.apache.http.impl.client.InternalHttpClient.determineRoute(InternalHttpClient.java:124)
    at org.apache.http.impl.client.InternalHttpClient.doExecute(InternalHttpClient.java:183)
    ... 4 more
java httpclient httpresponse apache-httpclient-4.x
3个回答
90
投票

错误消息有点误导。您提供的值不代表完整的 URI

request = new HttpGet("trib.me/1lBFzSi");

缺少协议。

只需提供完整的 URI

request = new HttpGet("https://trib.me/1lBFzSi");

0
投票

此错误可能是由错误的 url 引起的。 检查网址:

  1. 协议缺失;
  2. url参数编码;

0
投票

就我而言,我解决了在开头添加 / 的问题: 请求 = new HttpGet("/trib.me/1lBFzSi");

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