尝试 HTTP POST 但获取 MalformedURLException：无协议：yahoo.com

Question

总的来说，我正在尝试编写一个脚本，使用 java 捕获服务器对 HTTP POST 的响应。

不幸的是，我一直在对它的 URL 部分进行编码。虽然我遵循了几个关于编码 URL 的在线示例，但我仍然得到 MalformedURLException...

知道编码过程中可能出现什么问题吗？

错误：

$ java client_post

Sending Http POST request
Exception in thread "Main Thread" java.net.MalformedURLException: no 
protocol: http%3A%2F%2Fyahoo.com
     at java.net.URL.<init>(URL.java:567)
     at java.net.URL.<init>(URL.java:465)
     at java.net.URL.<init>(URL.java:414)
     at client_post.sendPost(client_post.java:30)
     at client_post.main(client_post.java:23)

代码：

//package client_post;

import java.io.BufferedReader;
import java.io.DataOutputStream;
import java.io.InputStreamReader;
import java.net.HttpURLConnection;
import java.net.URLEncoder;
import java.net.URL;
import javax.net.ssl.HttpsURLConnection;

public class client_post {

     private final String USER_AGENT = "Mozilla/5.0";

     public static void main(String[] args) throws Exception {

             client_post http = new client_post();

             System.out.println("\nSending Http POST request");
             http.sendPost();

     }
     // HTTP POST request
     private void sendPost() throws Exception {

             //String url =<host:port/create/service>
             String url = "http://yahoo.com";
             String EncoderUrl = URLEncoder.encode(url, "UTF-8");
             URL obj = new URL(EncoderUrl);
             HttpsURLConnection con = (HttpsURLConnection) obj.openConnection();

             //add reuqest header
             con.setRequestMethod("POST");
             con.setRequestProperty("User-Agent", USER_AGENT);
             con.setRequestProperty("Accept-Language","en-US,en;q=0.5");

             String urlParameters = "<string base64>";
// Send post request
             con.setDoOutput(true);
             DataOutputStream wr = new DataOutputStream(con.getOutputStream());
             wr.writeBytes(urlParameters);
             wr.flush();
             wr.close();

             int responseCode = con.getResponseCode();
             System.out.println("\nSending 'POST' request to URL : " + url);
             System.out.println("Post parameters : " + urlParameters);
             System.out.println("Response Code : " + responseCode);

             BufferedReader in = new BufferedReader(
                     new InputStreamReader(con.getInputStream()));
             String inputLine;
             StringBuffer response = new StringBuffer();

             while ((inputLine = in.readLine()) != null) {
                     response.append(inputLine);
             }
             in.close();

             //print result
             System.out.println(response.toString());
            System.out.println(response.toString());

       }

}

Answer 1

当您编码 url 时，您的 url 如下所示

http%3A%2F%2Fyahoo.com

在你有一些特别的东西之前不要编码。

你的程序也抛出 class 转换异常

 HttpsURLConnection con = (HttpsURLConnection) obj.openConnection();

上面应该像下面这样

 HttpURLConnection con = (HttpURLConnection) obj.openConnection();

下面是工作程序。

package com.ds.portlet.library;

import java.io.BufferedReader;
import java.io.DataOutputStream;
import java.io.InputStreamReader;
import java.net.HttpURLConnection;
import java.net.URL;

import javax.net.ssl.HttpsURLConnection;


public class client_post {

     private final String USER_AGENT = "Mozilla/5.0";

     public static void main(String[] args) throws Exception {

             client_post http = new client_post();

             System.out.println("\nSending Http POST request");
             http.sendPost();

     }
     // HTTP POST request
     private void sendPost() throws Exception {

             //String url =<host:port/create/service>
             String url = "http://yahoo.com";
//             String EncoderUrl = URLEncoder.encode(url, "UTF-8");
             URL obj = new URL(url);
             HttpURLConnection con = (HttpURLConnection) obj.openConnection();

             //add reuqest header
             con.setRequestMethod("POST");
             con.setRequestProperty("User-Agent", USER_AGENT);
             con.setRequestProperty("Accept-Language","en-US,en;q=0.5");

             String urlParameters = "<string base64>";
// Send post request
             con.setDoOutput(true);
             DataOutputStream wr = new DataOutputStream(con.getOutputStream());
             wr.writeBytes(urlParameters);
             wr.flush();
             wr.close();

             int responseCode = con.getResponseCode();
             System.out.println("\nSending 'POST' request to URL : " + url);
             System.out.println("Post parameters : " + urlParameters);
             System.out.println("Response Code : " + responseCode);

             BufferedReader in = new BufferedReader(
                     new InputStreamReader(con.getInputStream()));
             String inputLine;
             StringBuffer response = new StringBuffer();

             while ((inputLine = in.readLine()) != null) {
                     response.append(inputLine);
             }
             in.close();

             //print result
             System.out.println(response.toString());
            System.out.println(response.toString());

       }

}

Answer 2

您似乎正在尝试对 整个 URL 进行编码，包括 :// 和类似字符。 URL 编码的目的是隐藏路径或查询部分中的那些字符，它们不应该在主 URL 中编码。仅对参数或 application/x-www-form-urlencoded 内容使用 URLEncoder。

尝试 HTTP POST 但获取 MalformedURLException：无协议：yahoo.com

Trying to HTTP POST But Getting MalformedURLException: no protocol: yahoo.com

java

http-post