我在學習Jquery,偶然發現了我無法繞過的問題。 在下面的例子中,我從互聯網上獲得了JSP表單(簡單加法計算器)和Servlet,它們將兩個數字相加在一起,並在按下「計算」按鈕時返回結果。JQuery + Ajax - 如何清除重置按鈕的結果部分?
我的問題,我無法找到一個方法後按「重置」按鈕如何清除JSP結果部分。我正在談論的部分是這樣的:在JSP文件末尾的<div id="result"></div>
。
JSP:
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
<title>Ajax Example in JSP And Servlet</title>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function() {
$('#btn').click(function() {
var data = $('#xForm').serializeArray();
$.ajax({
type:"POST",
url:"CalcAjax",
data:data,
cache: false,
success:function(msg)
{
$("#result").hide();
$("#result").html("Addition Value is: " + msg + " ").fadeIn("slow");
}
});
});
});
</script>
</head>
<body>
<form id="xForm" method="post" ACTION="/CalcAjax">
Enter number1:
<input id="number" type="text" name="number" />
Enter number2:
<input id="number1" type="text" name="number1" />
<input id="btn" type="button" value="Calculate" name="btn"/>
<input id="reset" type="reset" name="reset" value="Reset"/>
</form>
<div id="result"></div>
</body>
</html>
的Servlet:
package com.test;
import java.io.IOException;
import java.io.PrintWriter;
import javax.servlet.ServletException;
import javax.servlet.RequestDispatcher;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
public class CalcAjax extends HttpServlet {
private static final long serialVersionUID = 1L;
String redir="";
String urlencode="";
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
try {
redir=("/ajax.jsp");
urlencode = response.encodeRedirectURL(redir.toString());
response.sendRedirect(urlencode);
} catch (Throwable t) {
RequestDispatcher dispatcher4 = request.getRequestDispatcher("/ajax.jsp");
dispatcher4.forward(request, response);
}
}
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
// TODO Auto-generated method stub
int number = 0,number1=0;
number = Integer.parseInt(request.getParameter("number"));
number1 = Integer.parseInt(request.getParameter("number1"));
PrintWriter out=response.getWriter();
out.println(number+number1+" <br>");
}
}
任何幫助將不勝感激!
完美,謝謝!我現在明白了:-) –