Question

我试图在post中将数据从servlet获取到ajax，但它总是会出现错误部分，就像“未定义”这样对ajax kinldy这个新手来说是新手。插入部分工作正常，但我试图获取值去错误部分??

的servlet

package Servlet;

import java.io.IOException;
import java.sql.Connection;
import java.sql.PreparedStatement;
import java.sql.ResultSet;
import java.sql.SQLException;
import java.util.ArrayList;

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 Databases.databases;

/**
 * Servlet implementation class Form
 */
@WebServlet(asyncSupported = true, urlPatterns = { "/Form" })
public class Form extends HttpServlet {
    private static final long serialVersionUID = 1L;

    /**
     * @see HttpServlet#HttpServlet()
     */
    public Form() {
        super();
        // TODO Auto-generated constructor stub
    }

    /**
     * @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response)
     */
    protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        // TODO Auto-generated method stub
    }

    /**
     * @see HttpServlet#doPost(HttpServletRequest request, HttpServletResponse response)
     */
    protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        // TODO Auto-generated method stub


        System.out.println("sssssssss");

        //System.out.println("input are====>"+input);

        //response.setContentType("text/plain;charset=UTF-8");

        response.setContentType("text/html");
        String name=request.getParameter("name");
        String job=request.getParameter("job");

        System.out.println("name===>"+name);
        System.out.println("job===>"+job);

        ArrayList al=new ArrayList();

        String a="";
        String b="";
        //databases d=new databases();
        try {
            Connection con=databases.getconnection();
            String sql="insert into senthiil(name,job)values(?,?)";
            PreparedStatement ps=con.prepareStatement(sql);
            ps.setString(1, name);
            ps.setString(2, job);

            System.err.println("sql===>"+ps);
        //  ps.execute();
            //con.commit();



            String select="select name,job from senthiil";
            PreparedStatement psselect=con.prepareStatement(select);
            System.out.println("select======>"+psselect);
            ResultSet rs=psselect.executeQuery();

            while(rs.next())
            {
                al.add(rs.getString(1));
                al.add(rs.getString(2));

                System.out.println("al====>"+al);

                request.setAttribute("data", al);

                /*a=rs.getString(1);
                b=rs.getString(2);
                */

            }

            //response.getWriter().write(a);
            response.getWriter().write("{issue:true}");

            response.getWriter().print(al);

            response.getWriter().close();

        } 
        catch (ClassNotFoundException | SQLException e) 
        {
            // TODO Auto-generated catch block
            e.printStackTrace();

            System.out.println("e");
        }


    }

}

JS：

$(document).ready(function()
{
    $("#submit").click(function(){

        alert("senthil");

        var name=$("#name").val();
        var job=$("#job").val();

    //  var datastring="name" +  + "&job=" + job; // option one         type:'post',

        //alert("datastring=======>"+datastring);
        $.ajax({            
            url: "Form",
            data: {"name": name, "job": job},
            dataType: 'text/html',
            type: 'post',
            success: function(data)
            {
                alert("ssssss");
                alert("data===>"+data);

                console.log(data);
            },
            error: function(e) {
                alert('Error: ' + e.message);
            }
        });     
    });
});

Answer 1

您应该将数据格式化为json类型，以便客户端可以在错误的情况下接收它。在您的服务器端，请更改您的响应内容输入：“application / json”

response.setContentType("application/json");

在你的ajax调用中，将dataType更改为json，并更改控制台日志以检查它是否有效：

dataType: 'json',
alert("data===>"+data.issue);

您应该将a1 List格式化为json字符串并将该字符串附加到response.getWriter().write("{issue:true}");的响应对象，此库可以帮助您将java对象解析为json字符串：

http://flexjson.sourceforge.net/#Serialization

如何从servlet获取数据到ajax成功

1 个答案: