总的来说,我正在尝试编写一个脚本来捕获服务器对使用 java 的 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());
}
}
最佳答案
当你对 url 进行编码时,你的 url 会变成如下所示
http%3A%2F%2Fyahoo.com
不要编码,直到你有一些特别的东西。
你的程序也抛出类转换异常
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());
}
}
关于java - 尝试 HTTP POST 但收到 MalformedURLException : no protocol: yahoo. com,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/30253225/