java 如何使用 HttpClient 将 XML 文件作为请求传递给我的 Servlet 之一？

How can I pass the XML file to one of my Servlet as the request using HttpClient?

提问by arsenal

Below is my code, in which I am trying to generate an XML file and then as soon as I generate an XML, I need to send this XML file to one of my own Servlet which is runnung locally on my box. I am able to generate an XML file but I am not sure how should I send that XML file to one of my servlet so that in the doGet method, I can parse that XML file.

下面是我的代码，我试图在其中生成一个 XML 文件，然后一旦我生成一个 XML，我就需要将此 XML 文件发送到我自己的一个 Servlet，它在我的机器上本地运行。我能够生成一个 XML 文件，但我不确定如何将该 XML 文件发送到我的 servlet 之一，以便在 doGet 方法中，我可以解析该 XML 文件。

public static void main(String[] args) throws SAXException, XPathExpressionException, ParserConfigurationException, IOException,
    TransformerException {

String xml = generateXML();
send("http://localhost:8080/ServletExample/SampleServlet", xml);
}


/**
 * A simple method to generate an XML file
 *
 */  
public static String generateXML(String conn, String funcAddr) throws ParserConfigurationException, SAXException, IOException,
    XPathExpressionException, TransformerException {

DocumentBuilderFactory docFactory = DocumentBuilderFactory.newInstance();
DocumentBuilder docBuilder = docFactory.newDocumentBuilder();

// Some code here to make an XML file

String xmlString = sw.toString();

// print xml
System.out.println("Here's the xml:\n" + xmlString);

return xmlString;
}


/**
 * A simple method to send the XML to servlet class
 *
 */
public static void send(String urladdress, String file) throws MalformedURLException, IOException {
String charset = "UTF-8";
String s = URLEncoder.encode(file, charset);

// I am not sure what should I do here so that I can pass the 
//  above XML file that I made to my servlet class.

}

My servlet is running locally on 8080. Below is the snippet from my servlet class-

我的 servlet 在 8080 上本地运行。下面是我的 servlet 类的片段-

protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {

    BufferedReader reader = request.getReader();

    //Parse the XML file here?

    System.out.println(reader.readLine());

}

Updated Code:-

更新代码：-

I have created a Servlet class named SampleServletin a new dynamic web project. I have started the server in debug mode. Below is the code in my Servlet-

我创建了一个以SampleServlet新的dynamic web project. 我已经在调试模式下启动了服务器。以下是我的 Servlet 中的代码-

protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {

    BufferedReader reader = request.getReader();
    System.out.println(reader.readLine());

}

protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {

    BufferedReader b = new BufferedReader(request.getReader());  
    System.out.println(reader.readLine());

}

And my web.xml file is like this-

而我的 web.xml 文件是这样的——

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" id="WebApp_ID" version="2.5">
  <display-name>ServletExample</display-name>
  <welcome-file-list>
    <welcome-file>index.html</welcome-file>
    <welcome-file>index.htm</welcome-file>
    <welcome-file>index.jsp</welcome-file>
    <welcome-file>default.html</welcome-file>
    <welcome-file>default.htm</welcome-file>
    <welcome-file>default.jsp</welcome-file>
  </welcome-file-list>
  <servlet>
    <description></description>
    <display-name>SampleServlet</display-name>
    <servlet-name>SampleServlet</servlet-name>
    <servlet-class>com.servlet.example.SampleServlet</servlet-class>
  </servlet>
  <servlet-mapping>
    <servlet-name>SampleServlet</servlet-name>
    <url-pattern>/SampleServlet</url-pattern>
  </servlet-mapping>
</web-app>

I have put the breakpoint in both the methods above. As soon as I hit this url from the browser-

我在上面的两种方法中都设置了断点。一旦我从浏览器点击这个网址 -

http://localhost:8080/ServletExample/SampleServlet

http://localhost:8080/ServletExample/SampleServlet

my breakpoint always gets hit in doGet method.

我的断点总是在 doGet 方法中被击中。

Now I have created a new Java Project in the eclipse which is my client and which will call the servlet doPost method as I need to pass an XML file to my servlet as a request.

现在我在 eclipse 中创建了一个新的 Java 项目，它是我的客户端，它将调用 servlet doPost 方法，因为我需要将 XML 文件作为请求传递给我的 servlet。

Below is my code-

下面是我的代码-

public static void main(String[] args) {

    HttpPost post = new HttpPost("http://localhost:8080/ServletExample/SampleServlet");
    post.setHeader("Content-Type", "application/xml");
    post.setEntity(new StringEntity(generateNewXML()));
    HttpClient client = new DefaultHttpClient();
    HttpResponse response = client.execute(post);
}

But somehow as soon as I run my above main program as a Java Application, it doesn't hit the breakpoint I have put in my servlet class. And I am not sure why it is happening and no exceptions is getting thrown. Any idea why it is happening?

但是不知何故，只要我将上面的主程序作为 Java 应用程序运行，它就不会到达我在 servlet 类中放置的断点。而且我不确定为什么会发生这种情况并且没有抛出异常。知道为什么会这样吗？

采纳答案by Septem

Everything seems ok, I just copied your code into a new project

一切正常，我只是将您的代码复制到一个新项目中

public class SampleServlet extends HttpServlet {

    @Override
    protected void doPost(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
        System.out.println("doPost is called");
    }
}

and run the client:

并运行客户端：

public class PostClient {

    public static void main(String[] args) throws ClientProtocolException, IOException {
        HttpPost post = new HttpPost("http://localhost:8080/ServletExample/SampleServlet");
        post.setHeader("Content-Type", "application/xml");
        post.setEntity(new StringEntity("<xml></xml>"));
        HttpClient client = new DefaultHttpClient();
        HttpResponse response = client.execute(post);
    }

}

the message "doPost is called" was printed in cosole, everything works as expected

消息“doPost被调用”打印在cosole中，一切都按预期工作

回答by Juned Ahsan

To post something to a servlet, using HTTP POST/doPost is a better option. GET/doGet is to get a resource. Here is the relevant code for the same:

要将某些内容发布到 servlet，使用 HTTP POST/doPost 是更好的选择。GET/doGet 是获取资源。这是相同的相关代码：

Servlet doPost

小服务程序

public void doPost(HttpServletRequest req,  
                    HttpServletResponse res)  
      throws ServletException, IOException {  
      try {  
            BufferedReader b = new BufferedReader(req.getReader());  
            StringBuffer xmlBuffer = new StringBuffer();    
            String xmlString = "";          
            while((xmlString = b.readLine()) != null) {  
                   xmlBuffer.append(xmlString);  
            }  
            xmlString = xmlBuffer.toString();  
            if (workBuffer.length() > 0) {  
              System.out.println("Got XML: " + workString);  
            }      
            else {  
                 System.out.println("No XML document received");  
            }  
      }   

Http POST Client code:

Http POST 客户端代码：

private void postMessage(TextMessage xmlMsg, String urlString)  
                throws Exception  
{  
  try  
  {  
    URL url = new URL(urlString);  
    URLConnection uc = url.openConnection();  
    HttpURLConnection conn = (HttpURLConnection) uc;  
    conn.setDoInput(true);  
    conn.setDoOutput(true);  
    conn.setRequestMethod("POST");  
    conn.setRequestProperty("Content-type", "text/xml");          
    PrintWriter pw = new PrintWriter(conn.getOutputStream());  
    pw.write(xmlMsg.getText());  
    pw.close();  
    BufferedInputStream bis = new BufferedInputStream(conn.getInputStream());  
    bis.close();  

  }  
  catch (Exception e)  
  {  
    e.printStackTrace();  
  }  
}   

回答by Septem

If it is not a "safe" method, you'd better use POST instead, then you can send xml in the body of post as below:

如果它不是一个“安全”的方法，你最好使用 POST 代替，然后你可以在 post 正文中发送 xml，如下所示：

HttpPost post = new HttpPost("http://localhost:8080/ServletExample/SampleServlet");
post.setHeader("Content-Type", "application/xml");
post.setEntity(new StringEntity(generateXML()));
HttpClient client = new DefaultHttpClient();
HttpResponse response = client.execute(post);

If you'd rather use GET, one way to do this is to encode the xml in query string:

如果您更愿意使用 GET，一种方法是在查询字符串中对 xml 进行编码：

String xml = generateXML();
HttpGet get = new HttpGet("http://localhost:8080/ServletExample/SampleServlet?xml=" + URLEncoder.encode(xml, "UTF-8"));
HttpClient client = new DefaultHttpClient();
HttpResponse response = client.execute(get);

parse xml in servlet:

在servlet中解析xml：

String xml = request.getParameter("xml");