Question

我是Jquery和ajax的新手，编写了一个简单的jsp页面，它将显示每个onkeyup事件的文本，ajax发出的请求将通过AjaxServlet。我没有使用.ajax（）方法获得输出，而如果我使用简单的ajax使用XhtmlRequest它给出输出。请作为sson尽可能

** * * JSP页面 * ** * ** *

<%@ page language="java" contentType="text/html; charset=ISO-8859-1"
    pageEncoding="ISO-8859-1"%>
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<script>
 function showName(str){



        $.ajax({
              url: '/AjaxServlet',
              type: 'GET',
               data: 'name='+name,
              success: function(response){
               $("#showName").html(response);

                  }

        });



}
</script>
<meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
<title>Insert title here</title>
</head>
<body>
<input type="text" onkeyup="showName(this.value)">Enter your name</input>
<p>Suggestions: <span id="showName"></span></p> 

</body>
</html>

的的 **** AjaxServlet 的 * ** < EM> * ** * *

protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        String name=request.getParameter("name");
        response.setContentType("text/xml");
        PrintWriter responseWriter=response.getWriter();
        responseWriter.write("Welcome "+name);

    }

Answer 1

你的jQuery库在哪里？您是否忘记下载并将其包含在HTML页面中？

Answer 2

是'name'实际上是正确使用的东西吗？函数是showName(str)？

尝试以下方法：

    $.ajax({
          url: '/AjaxServlet',
          type: 'GET',
          data: {name:str}, // use a object literal here
          success: function(response){
             $("#showName").html(response);
          },
          error: function(xhr, status, error){

             alert("help I died! [" + status + "] [" + error +"]" ); 
          }
    });


protected void doGet(HttpServletRequest request, HttpServletResponse response) throws  ServletException, IOException {
    String name=request.getParameter("name");
    // your output isn't xml! ... use text/plain in stead
    response.setContentType("text/plain");
    PrintWriter responseWriter=response.getWriter();
    responseWriter.write("Welcome "+name);

}

.ajax（）方法不起作用

2 个答案: