Question

我正在尝试在Websphere Liberty服务器上部署单个maven webapp。

我的代码（XCCQuery.java）：

...
@Transactional
public void run(String... arg0)
    ...
    bookdetailRepository.save(detail)
    book1.setBookdetail(detail);
    ...

我的web.xml如下所示：

package TTAQuery;

import java.io.IOException;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import javax.ws.rs.core.Response;
import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.PathParam;
import javax.ws.rs.core.Response;

@WebServlet(name="XCCQuery", urlPatterns={"/hello"})
public class XCCQuery extends HttpServlet {


    protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
  PrintWriter out = response.getWriter();
  String simpleParam = getServletConfig().getInitParameter("simpleParam");
      out.println("Hello World "+simpleParam);
          out.close();

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

}

我尝试访问： http://localhost:9087/XCCQueryServlet/rest/hello

但总是得到：

<!DOCTYPE web-app PUBLIC
 "-//Sun Microsystems, Inc.//DTD Web Application 2.3//EN"
 "http://java.sun.com/dtd/web-app_2_3.dtd" >

<web-app>
  <display-name>Archetype Created Web Application</display-name>

  <servlet>
    <servlet-name>XCCQuery</servlet-name>
    <display-name>XCCQuery</display-name>
    <description></description>
    <servlet-class>XCCQuery</servlet-class>
  </servlet>

  <servlet-mapping>
    <servlet-name>XCCQuery</servlet-name>
    <url-pattern>/rest/*</url-pattern>
  </servlet-mapping>
</web-app>

对我来说似乎是一个部署错误。

这里有什么问题？

javax.servlet.UnavailableException：SRVE0200E - Maven webapp

0 个答案: