Java 将servlet类扩展到自定义类以获得更自定义的响应
在我的项目中,用于插入、更新和删除我向servlet类请求的数据,然后它处理并返回响应 这一切都是通过jqueryajax实现的 现在,它只回应成功或失败如下Java 将servlet类扩展到自定义类以获得更自定义的响应,java,jakarta-ee,inheritance,servlets,Java,Jakarta Ee,Inheritance,Servlets,在我的项目中,用于插入、更新和删除我向servlet类请求的数据,然后它处理并返回响应 这一切都是通过jqueryajax实现的 现在,它只回应成功或失败如下 PrintWriter out = response.getWriter(); out.println("<custom message>"); import javax.servlet.ServletException; import javax.servlet.http.HttpServlet; import jav
PrintWriter out = response.getWriter();
out.println("<custom message>");
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
public class myservlet extends HttpServlet {
/**
*
*/
private static final long serialVersionUID = 1L;
public void doPost(HttpServletRequest req,HttpServletResponse resp) throws ServletException,java.io.IOException
{
String format = req.getParameter("format");
if(format == null)
{
format = "json";
}
else
{
if(format.equals("json"))
{
resp.setContentType("application/json");
}
else if(format.equals("xml"))
{
resp.setContentType("application/rss+xml");
}
else
{
//error
}
}
}
public void doGet(HttpServletRequest req,HttpServletResponse resp) throws ServletException,java.io.IOException
{
String format = req.getParameter("format");
if(format == null)
{
format = "json";
}
else
{
if(format.equals("json"))
{
resp.setContentType("application/json");
}
else if(format.equals("xml"))
{
resp.setContentType("application/rss+xml");
}
else
{
//error
}
}
}
}
上面的这个类扩展如下
导入java.io.PrintWriter
import javax.servlet.ServletException;
public class abc extends myservlet
{
private static final long serialVersionUID = 1L;
public void doPost(HttpServletRequest req,HttpServletResponse resp) throws ServletException,java.io.IOException
{
PrintWriter out = resp.getWriter();
out.println("{/"id/": /"file/"}");
//response must be converted to either json or to xml
}
}
有可能吗
如何将响应动态转换为
xml
或json
?不要在servlet中这样做。这是一个创建过滤器的好机会,它将完成转换工作