Question

我正在尝试按照本教程创建简单的唯一用户名检查。

http://javaandj2eetutor.blogspot.com/2013/12/check-username-availability-using-java.html

但我对用户名的ajax调用失败了。我是ajax的新手，对于jsp来说有点新鲜。

这是我的index.jsp

<%@ page contentType="text/html;charset=UTF-8" language="java" %>
<html>
<head>
    <title>Username Availability</title>
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.0/jquery.min.js"></script>
</head>
<body>
<script type="text/javascript">

    $(document).ready(function () {
        $(".username").change(function () {
            var username = $(this).val();
            if (username.length >= 3) {
                $(".status").html("<font color=gray> Checking availability...</font>");
                $.ajax({
                    type: "POST",
                    url: "CheckAvalability",
                    data: "uname="+ username,
                    success: function (msg) {

                        $(".status").ajaxComplete(function (event, request, settings) {
                            $(".status").html(msg);

                        });
                    }
                });
            }
            else {

                $(".status").html("<font color=red>Username should be <b>3</b> character long.</font>");
            }

        });
    });
</script>   
<div>
    <label class="flable">User Name :</label> <input class="username" type="text" name="username"> <span class="status"></span>
</div>  
</body>
</html>

这是我的Servlet

package hsenid;

import org.apache.logging.log4j.LogManager;
import org.apache.logging.log4j.Logger;

import java.io.*;
import java.sql.*;
import javax.servlet.ServletException;
import javax.servlet.http.*;

public class CheckAvailability extends HttpServlet {
    private static final Logger logger = LogManager.getLogger(CheckAvailability.class);
    private static final long serialVersionUID = -734503860925086969L;

    protected void doPost(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {
        response.setContentType("text/html;charset=UTF-8");
        PrintWriter out = response.getWriter();
        try {
            logger.info("Check availability called");
            DBConnector dbPool = (DBConnector)getServletContext().getAttribute("DBConnection");
            Connection myConn = dbPool.getConn();

            String uname = request.getParameter("username");

            PreparedStatement ps = myConn.prepareStatement("select username from userdetails where username=?");
            ps.setString(1,uname);
            ResultSet rs = ps.executeQuery();

            if (!rs.next()) {
                out.println("<font color=green><b>"+uname+"</b> is avaliable</font>");
                logger.info("Username detected!!!");
            }
            else{
                out.println("<font color=red><b>"+uname+"</b> is already in use</font>");
            }
            out.println();

        } catch (Exception ex) {
            out.println("Error ->" + ex.getMessage());
        } finally {
            out.close();
        }
    }

    protected void doGet(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {
        doPost(request, response);
    }
}

这是我在web.xml中的servlet映射。

<servlet>
        <servlet-name>CheckAvailability</servlet-name>
        <servlet-class>hsenid.CheckAvailability</servlet-class>
    </servlet>
    <servlet-mapping>
        <servlet-name>CheckAvailability</servlet-name>
        <url-pattern>/CheckAvailability</url-pattern>
    </servlet-mapping>

这会检测是否添加了最小字符，并在输入时提供按摩检查可用性，因此我认为添加了jquery。我也在Eclipse Mars中运行servlet。它会检查用户名是否在表中。所以我认为调用servlet的问题是因为我无法看到log4j控制台输出。我无法找到我的代码有什么问题。提前致谢

Ajax唯一用户名搜索失败JSP

0 个答案: