如何从servlet获取数据到ajax成功
我试图从servlet中获取数据到ajax但是它总是会出现错误部分，就像“未定义”这样对ajax kinldy这是一个新手。 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); } }); }); });
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键入：“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response.getWriter().write("{issue:true}"); ，这个库可以帮助你解析java对象到json字符串：