Question

如何获取此请求的完整网址？我只需要HttpGet get = new HttpGet(authUrlTemplate); BasicHttpParams bhp = new BasicHttpParams(); bhp.setParameter("client_id", getClientId()); bhp.setParameter("client_secret", getClientSecret()); bhp.setParameter("redirect_uri", redirectURL.toString()); bhp.setParameter("grant_type", "authorization_code"); get.setParams(bhp); System.err.println(get.getURI()); System.err.println(get.getRequestLine()); System.err.println(get.getParams());。我如何构建URL？

没有任何print语句打印出我想要的内容，getURI（）不会添加参数......

21:18:49,967 ERROR [stderr] (default task-7) https://accounts.google.com/o/oauth2/v2/auth
21:18:49,970 ERROR [stderr] (default task-7) GET https://accounts.google.com/o/oauth2/v2/auth HTTP/1.1
21:18:49,971 ERROR [stderr] (default task-7) org.apache.http.params.BasicHttpParams@28310ff2

输出：

            LayoutParams params = holder.itemView.getLayoutParams();
            params.height = screenWidth/2;
            params.width = screenHeight*24/100;
            holder.itemView.setLayoutParams(params);

我知道有一个库可以做到这一点，但是想这样做。

Answer 1

这是一个通用的工作示例。代码很冗长，可以缩短，但这给你一个整体的想法。

import org.apache.http.NameValuePair;
import org.apache.http.client.methods.HttpGet;
import org.apache.http.client.utils.URLEncodedUtils;
import org.apache.http.message.BasicNameValuePair;
import org.apache.http.params.BasicHttpParams;
import sun.management.resources.agent;

import java.util.*;

/**
 * Created by aditya on 9/24/2016.
 */
public class SOMain {
    static public void main(String[] args){
        String baseUrl="http://google.co.in";
        HashMap<String,String> parameters = new HashMap<String,String>();
        parameters.put("key1","value1");
        parameters.put("key3","value3");
        parameters.put("key2","value2");
        System.out.println(addParameters(baseUrl,parameters));
    }

    static protected String addParameters(String url, HashMap<String,String> parameters){
        if(!url.endsWith("?"))
            url += "?";

        List<NameValuePair> params = new LinkedList<NameValuePair>();
        if(parameters!=null){
            Iterator entries = parameters.entrySet().iterator();

            while (entries.hasNext()) {
                Map.Entry entry = (Map.Entry) entries.next();
                String key = (String)entry.getKey();
                String value = (String)entry.getValue();
                params.add(new BasicNameValuePair(key, value));            }
        }

        String paramString = URLEncodedUtils.format(params, "utf-8");

        url += paramString;
        return url;
    }
}

打印

http://google.co.in?key1=value1&key2=value2&key3=value3

Answer 2

以下是我如何做到这一点的简单版本：

public String getAuthUrl() {
    URIBuilder b = null;
    // Removed exception handling...
    b = new URIBuilder(AUTH_URL); // http://whatever.com

    List<NameValuePair> nvps = new ArrayList<NameValuePair>();
    nvps.add(new BasicNameValuePair("client_id", getClientId()));
    nvps.add(new BasicNameValuePair("redirect_uri", redirectURL.toString()));
    nvps.add(new BasicNameValuePair("state", STATE));
    nvps.add(new BasicNameValuePair("response_type", "code"));
    nvps.add(new BasicNameValuePair("scope", "email"));
    b.setParameters(nvps);
    return b.toString();
  }

HTTPClient 4 - 使用params生成url

2 个答案: