Question

我是servlet概念的新手。我的要求就像将restful给定的URL转换为正文中的查询参数。

给定网址：

http://anydomain:8080/ServletBasics/HelloForm/India/Andhrapradesh

必填输出网址：

http://anydomain:8080/ServletBasics/HelloForm?Country=India&State=Andhrapradesh

使用给定的servlet代码完成了URL提取。任何人都可以帮助我将给定的URL转换为基于查询的URL。感谢

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

    response.setContentType("text/html");
    PrintWriter out = response.getWriter();
    out.println("<html><body>");
    String vid = request.getRequestURI();
    out.println("</body></html>");
    out.close();

}


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

修改后的代码：sdfd.java

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

    response.setContentType("text/html");

    String url = request.getRequestURI();
    PrintWriter out = response.getWriter();
    out.println("<html>");
    out.println("<body>");

if(url.equals("/servletTest/v1/code")) {    

    String[] words = url.split("/");
    String newURI = url.replace(url, "/ws/simple/Apicode?"+"first_name="+words[2]+"&"+"last_name="+words[3]);

    RequestDispatcher rd = request.getRequestDispatcher(newURI);
    rd.forward(request, response);

    out.println(newURI);


    }

    else 
    {
        out.println("bad");
    }



    out.println("</html>");
    out.println("</body>");

    out.close();
}

的web.xml

<servlet>
<servlet-name>sdfd</servlet-name>
<servlet-class>sdfd</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>sdfd</servlet-name>
<url-pattern>/v1/code</url-pattern>
</servlet-mapping>

我正在尝试转换

http://localhost:8080/servletTest/v1/code

到

http://localhost:8080/servletTest/ws/simple/Apicode?first_name=v1&last_name=code

但我得到的错误。

HTTP Status 404 - /servletTest/ws/simple/Apicode
type Status report
message /servletTest/ws/simple/Apicode
description The requested resource is not available.
Apache Tomcat/7.0.42

请帮助我哪里出错了？感谢

Answer 1

使用URLRewrite

您可以在以下网址中找到该文档：http://urlrewritefilter.googlecode.com/svn/trunk/src/doc/manual/4.0/index.html

例如：

 <rule>
    <from>^/HelloForm/([a-z]+)/([a-z]+)$</from>
    <to>/HelloForm?Country=$1&State=$2</to>
</rule>

要配置UrlRewrite，请阅读手册http://urlrewritefilter.googlecode.com/svn/trunk/src/doc/manual/4.0/index.html

如何使用Servlet将以下URL转换为查询参数

1 个答案: