如何将参数添加到宁静的请求？

Question

我正在使用以下代码发送REST请求，请求失败（401错误），因为它需要用户名和密码。

如何将它们添加到网址？即使当我在浏览器中复制url并在其末尾添加用户名和密码时，浏览器弹出登录页面，所以我想在代码中我应该添加用户名和密码参数，但是如何？

URL url = new URL("www.example.com/user?ID=1234"); 
     HttpURLConnection conn = (HttpURLConnection) url.openConnection(); 
     conn.setDoOutput(true); 
     conn.setRequestMethod("POST"); 
     conn.setRequestProperty("Content-Type", "application/json"); 
       if (conn.getResponseCode() != HttpURLConnection.HTTP_CREATED) { 
      throw new RuntimeException("Failed : HTTP error code : " 
       + conn.getResponseCode()); 
     } 

我在代码中添加了以下内容，但仍遇到401错误。

   conn.setRequestProperty("username", "myusername"); 
      conn.setRequestProperty("password", "mypassword"); 

Answer 1

试试这个：

String params="username=myusername&password=mypassword"; 
conn.getOutputStream().write(params.getBytes()); 
conn.getOutputStream().flush(); 
conn.getOutputStream().close(); 

---

或者，也许你需要编码PARAMS：

DataOutputStream dos = new DataOutputStream(conn.getOutputStream()); 

String postContent = URLEncoder.encode("username", "UTF-8") + "=" + 
        URLEncoder.encode(myusername, "UTF-8") + "&" + 
        URLEncoder.encode("password", "UTF-8") + "=" + 
        URLEncoder.encode(mypassword, "UTF-8") ; 

dos.write(postContent.getBytes()); 
dos.flush(); 
dos.close(); 

---

关于Basic Authorization

String userPassword = username + ":" + password; 
String encoding = new sun.misc.BASE64Encoder().encode(userPassword.getBytes()); 
URLConnection uc = url.openConnection(); 
uc.setRequestProperty("Authorization", "Basic " + encoding); 
uc.connect();