Question

我想在Ajax上使用Java创建一个简单的Web应用程序。我有一个包含一个表的本地数据库 - 两列：ID和Surname。

我的应用程序应该向用户显示正确的surname（例如，他在网站id = 1上写道，而servlet应该从db返回姓氏。

这是我的servlet：

package db.connection;

import java.io.IOException;
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 java.io.IOException;
import java.sql.Connection;
import java.sql.ResultSet;
import java.sql.SQLException;
import java.sql.Statement;

/**
 * Servlet implementation class ConnectionServlet
 */
@WebServlet("/ConnectionServlet")
public class ConnectionServlet extends HttpServlet {
    private static final long serialVersionUID = 1L;

    public void doGet(HttpServletRequest req, HttpServletResponse res)
            throws ServletException, IOException {

        String key = req.getParameter("key");
        if (key != null) {

            String surname = "";

            ResultSet result = null;
            Connection con = null;
            try{
            con = DatabaseConnector.getConnection();
            Statement select = con.createStatement();
            result = select
                    .executeQuery("SELECT surname from person where ID = '"
                            + key + "';");

            //surname=result.getString(0);

            while (result.next()) {
                surname = result.getString("surname");
               // System.out.println(surname);

            }
            res.setContentType("text/xml");
            res.setHeader("Cache-Control", "no-cache");
            // write out the response string
            res.getWriter().write(surname);
            }
            catch (SQLException e)
            {
                {
                    System.out.println(e.getMessage());
                } }

            finally
                {
                    if (con != null)
                    {
                        try
                        {
                            con.close();
                        } catch (SQLException e) {
                        }
                    }
                }
            }
    }
    }

Ajax.js

 var req; 
  function getSurDB(){ 
    var key = document.getElementById("key"); 
    var keypressed = document.getElementById("keypressed"); 
    keypressed.value = key.value; 
    var url = "/DBcon/Ajax?key="+ escape(key.value); 
    if (window.XMLHttpRequest){ 
      req = new XMLHttpRequest(); 
    } 
    else if (window.ActiveXObject){ 
      req = new ActiveXObject("Microsoft.XMLHTTP"); 
    } 
    req.open("Get",url,true); 
    req.onreadystatechange = callback; 
    req.send(null);
  } 

  function callback(){ 
    if (req.readyState==4){ 
      if (req.status == 200){ 
        var surname = document.getElementById('surname'); 
        surname.value = req.responseText; 
      } 
    }
    clear();
  } 
  function clear(){ 
    var key = document.getElementById("key"); 
    key.value=""; 
  }
  function focusIn(){ 
    document.getElementById("key").focus();
  }

Answer 1

很难找到您的问题，因为我们不知道您的DatabaseConnector类是否正常工作。同样在评论中你说tomcat显示错误。您可以在IDE或tomcat日志文件中检查tomcat控制台并分析错误（可能是google）。

但我发现了一个问题。您将响应的内容类型设置为＆＃34; text / xml＆＃34;但是你返回一个String。尝试将其设置为纯文本：

res.setContentType("text/plain");

或将姓氏返回为xml。

不工作的servlet从数据库

1 个答案: