如何使用ajax在同一个servlet中实现doPost()和doGet()请求？
如何使用ajax在同一个servlet中实现doPost()和doGet()请求？
提问于 2018-08-13 01:27:30
我正在做一个关于jsp servlet的小项目，我想要一个示例来实现doGet()和doPost()方法，使用相同的servlet，并在jsp页面的表单中通过ajax调用发送和发布。
还要说明为什么不推荐在同一个servlet.How中使用doGet()和doPost()，以便在同一个jsp页面中使用两个不同的servlet来实现相同的功能。
提前感谢，任何帮助都是有益的。
我的jsp代码：-
<form action="/mamababu.do" method="POST">
<select name="command_no">
<c:forEach var="items" items="${scriptItems}">
<option value="${items.command}" name="command">${items.command}</option>
</c:forEach>
</select>
<input type="submit" value="submit"></input>
</form> 我的servlet类：
package com.project.mamabhagne;
import java.io.IOException;
import java.io.PrintWriter;
import java.sql.SQLException;
import java.util.List;
import javax.servlet.RequestDispatcher;
import javax.servlet.ServletException;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import com.google.gson.Gson;
@WebServlet("/mamababu.do")
public class mamababu extends HttpServlet {
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
// TODO Auto-generated method stub
//get the data from database ie the model class
try {
List<Script> scriptitems=modelDBUtil.getScriptList();
// String json = new Gson().toJson(scriptitems);
request.setAttribute("scriptItems", scriptitems);
} catch (ClassNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (SQLException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
/*String itemsfood[]={"biriyani","rice"};
request.setAttribute("itemsfood",itemsfood)*/;
//redirect to a different page
RequestDispatcher dispatcher =request.getRequestDispatcher("scriptviewer.jsp");
dispatcher.forward(request, response);
}
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
//TODO Auto-generated method stub
System.out.println("hi");
response.setContentType("text/html");
PrintWriter out = response.getWriter();
out.println("<HTML><HEAD><TITLE>Hello World!</TITLE>"+
"</HEAD><BODY>Hello World!</BODY></HTML>");
out.close();
RequestDispatcher dispatcher =request.getRequestDispatcher("scriptviewer.jsp");
dispatcher.forward(request, response);
}
}我在发布表单数据时遇到此错误：-
HTTP Status 404 - /mamababu.do
type Status report
message /mamababu.do
description The requested resource is not available.
Apache Tomcat/8.0.52回答 1
Stack Overflow用户
发布于 2018-08-13 02:37:40
我不清楚你的问题，顺便说一下，据我所知，你需要一个servlet并响应GET和POST请求(通过AJAX)。
@WebServlet(name = "MyServelet", urlPatterns = {"/myservletForm"})
public class MyServlet extends HttpServlet {
@Override
public void doGet(HttpServletRequest request, HttpServletResponse response)
throws IOException {
response.getWriter().println("Hello This is GET Response.");
}
@Override
public void doPost(HttpServletRequest request, HttpServletResponse response)
throws IOException{
response.getWriter().println("Hello This is POST Response.");
}
}如上所述，当你发送一个“get”请求到doGet，“post”请求到doPost。您可以通过声明特定的ajax请求类型来选择需要调用GET或POST。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/51811190
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