Question

我想开发一个在Tomcat上运行的servlet文件之间进行通信的应用程序。

下面是我的应用程序中的代码，它试图连接到这个servlet，发送请求并获得响应。

 private URLConnection getServletConnection() {
    try {
        URL servletURL = new URL("http://localhost:8080/test/servlet");
        URLConnection conn = servletURL.openConnection();

        conn.setDoInput(true);
        conn.setDoOutput(true);
        conn.setUseCaches(false);
        return conn;
    } catch (MalformedURLException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    } catch (IOException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }
    return null;
}

以下是引发异常的代码：

URLConnection conn = getServletConnection();
OutputStream outputStream = conn.getOutputStream();
ObjectOutputStream oos = new ObjectOutputStream(outputStream);
ServletRequestMessage srm = new ServletRequestMessage(2, username, password);
oos.writeObject(srm);
oos.flush();
oos.close();
InputStream inputStream = conn.getInputStream();

例外是：

java.io.FileNotFoundException: http://localhost:8080/test/servlet

任何人都可以帮助我吗？感谢。

Answer 1

服务器端

public class ServletImpl extends HttpServlet implements Servlet {

    ....
    public ServletImpl() {
      super();      
    }

 public void init(ServletConfig config) throws ServletException{
    super.init(config);         


    /*
     Application scope 
     Shared between all servlets, JSP pages, and custom tags within a J2EE application 
     or within the whole container if no applications are defined.
     The programmatic interface to the application scope is the 'ServletContext' object.         
     */


    ServletContext context = config.getServletContext();
    context.setAttribute("base", config.getInitParameter("base"));
      /* where "base" is iniy param in web.xml
         <init-param>
    <param-name>base</param-name>
    <param-value>/ServlrtName/sys</param-value>
    </init-param>       


       */



....



}
    ....
protected void doGet(HttpServletRequest request,HttpServletResponse response) throws ServletException, IOException
{
    doPost(request, response);
}


    protected void doPost(HttpServletRequest request,HttpServletResponse response) throws ServletException, IOException
   {
    Enumeration<?> paramNames = request.getParameterNames();


    while(paramNames.hasMoreElements()) {

        String paramName = (String)paramNames.nextElement();

        String[] paramValues = request.getParameterValues(paramName);


         if("sub".equals(paramName)){

            paramValues = request.getParameterValues(paramName);

            if(paramValues.length > 0){

                String param = paramValues[0];
                // do something
                ....
            }               

        }           
    }
}


....
// prepear response 
    response.setContentType("text/html");

    PrintWriter out = response.getWriter();

    out.println(mMessageResponseStr);
    out.close();

我在这里使用了sub标记，请参阅：if("sub".equals(paramName)){。

<强>的web.xml

<?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>ServlrtName</display-name>
 <servlet>
    <description>
    </description>
    <display-name>ServlrtName</display-name>
    <servlet-name>ServlrtName</servlet-name>
    <servlet-class>com.demo.servlet.ServletImpl</servlet-class>
  <init-param>
    <param-name>base</param-name>
    <param-value>/ServlrtName/sys</param-value>
    </init-param>
  <init-param>

        ....

</servlet>

...

 <servlet-mapping>
    <servlet-name>ServlrtName</servlet-name>
    <url-pattern>/sys/*</url-pattern>
  </servlet-mapping>

客户端

我使用了DefaultHttpClient和HttpPost。我发送了sub代码。这是一个将数据发送到Servlet的方法：

 public boolean send(String data) {

    DefaultHttpClient httpclient = null;
    boolean success = false;

    try {           
        httpclient = new DefaultHttpClient();

        String url = "your URL";


        HttpPost httpost = new HttpPost(url);

        List <NameValuePair> nvps = new ArrayList <NameValuePair>();
        nvps.add(new BasicNameValuePair("sub", data));


        httpost.setEntity(new UrlEncodedFormEntity(nvps, HTTP.UTF_8));


        HttpResponse response = httpclient.execute(httpost);

        HttpEntity entity = response.getEntity();

        if (entity != null) {

            StatusLine statusLine = response.getStatusLine();
            int statusCode = statusLine.getStatusCode();

            if(statusCode != 200){
                mResErr.onErrorResponse(statusCode);                    
            }

            InputStream is = entity.getContent();
            InputStreamReader isr = new InputStreamReader(is);
            BufferedReader br = new BufferedReader(isr);
            String line = null;
            while ( (line = br.readLine()) != null) {

                // get data from line


            }
            is.close();


        } else {
            //response is null/
        }
        success = true;

        mRes.onHttpResponse(mArr);


    } catch (Exception e) {
        mResErr.onErrorResponse(e);
        e.getStackTrace();
    }

    if (httpclient != null) {
        // resource cleanup
        httpclient.getConnectionManager().shutdown();
    }

    return success;
}

**注释，在开始检查连接之前，从服务器端删除用户/密码。如果一切按预期工作，请将其切换回来并在客户端使用：

Credentials cred = new UsernamePasswordCredentials("user", "pswd");


         httpclient.getCredentialsProvider().setCredentials(
                    new AuthScope(AuthScope.ANY_HOST, AuthScope.ANY_PORT, AuthScope.ANY_REALM),
                    cred);

尝试在应用程序和servlet之间进行连接时，URLConnection.getInputStream（）会抛出FileNotFoundException

1 个答案: