ajax JSP 文件路径问题

againlove 2008-06-24 09:40:55

初学Ajax技术，遇到了一点问题请高手指教：

我要实现这样的功能：从JSP中获取一个值，传给Servlet， Servlet处理之后，返回一个字符串给JSP。
postingXML.jsp：

<%@ page language="java" contentType="text/html; charset=UTF-8"

	pageEncoding="UTF-8"%>

<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">

<html>

	<head>

		<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">

		<title>Sending an XML Request</title>

		<script type="text/javascript" language="javascript">

		var xmlHttp;

	

		function createXMLHttpRequest(){

				

			if(window.XMLHttpRequest){ //Mozilla

				

				xmlHttp = new XMLHttpRequest();

			}

			else if(window.ActiveXObject){

					

				try{

					

					xmlHttp = new ActiveXObject("Msxml2.XMLHTTP");

				} 

				catch(e){

					try{

							

						xmlHttp = new ActiveXObject("Microsoft.XMLHTTP");

					}

					catch(e){

							

						alert("创建XMLHttpRequest对象失败");

					}

				}

			}

		}

		

		function createXML() {

		

			createXMLHttpRequest()

		

			var xml = "<pets>";

			

			var options = document.getElementById("petTypes").childNodes;

			

			var option = null;

			

			for(var i = 0; i < options.length; i++) {

			

				option = options[i];

				if(option.selected) {

					

					xml = xml + "<type>" + option.value + "<\/type>";

				}			

			}

			xml = xml + "<\/pets>";

			

			return xml;

		}

		

		function sendPetTypes() {

		

			var xml = createXML();

			

			var url = "servlet/PostingXMLExample?timeStamp=" + new Date().getTime();

			

			xmlHttp.open("POST", url, true);

			

			xmlHttp.onreadystatechange = processResponse;//指定响应函数

			

			xmlHttp.setRequestHeader("Content-Type","application/x-www-form-urlencoded");

			

			xmlHttp.send(xml);

		}

		

		function processResponse(){

		

			if(xmlHttp.readyState == 4){

				

				if(xmlHttp.Status == 200){

					parseResults();

					//document.all.name.value = xmlHttp.responseText;

				}

				else{

	        	

	        		window.alert("你所请求的页面有异常");

	           	 	alert(xmlHttp.status);

	       		}

			}

		}

		

		function parseResults(){

		

			var responseDiv = document.getElementById("serverResponse");

			

			 if(responseDiv.hasChildNodes()) {

			 	

			 	 responseDiv.removeChild(responseDiv.childNodes[0]);

			 }

			 

			 var responseText = document.createTextNode(xmlHttp.responseText);

			 

			 responseDiv.appendChild(responseText);

		}

	</script>

	</head>

	<body>

		<h1>

			Select the types of pets in your home:

		</h1>







		<form action="#">



			<select id="petTypes" size="6" multiple="true">



				<option value="cats">

					Cats

				</option>



				<option value="dogs">

					Dogs

				</option>



				<option value="fish">

					Fish

				</option>



				<option value="birds">

					Birds

				</option>



				<option value="hamsters">

					Hamsters

				</option>



				<option value="rabbits">

					Rabbits

				</option>



			</select>



			<br />

			<br />



			<input type="button" value="Submit Pets" onclick="sendPetTypes()" />

			



		</form>



		<h2>

			Server Response:

		</h2>







		<div id="serverResponse"></div>

	</body>

</html>

Servlet:

PostingXMLExample.java

package org.ajax.servlet;



import java.io.BufferedReader;

import java.io.ByteArrayInputStream;

import java.io.IOException;

import java.io.PrintWriter;



import javax.servlet.ServletException;

import javax.servlet.http.HttpServlet;

import javax.servlet.http.HttpServletRequest;

import javax.servlet.http.HttpServletResponse;

import javax.xml.parsers.DocumentBuilderFactory;

import javax.xml.parsers.ParserConfigurationException;



import org.xml.sax.SAXException;



import org.w3c.dom.*;



public class PostingXMLExample extends HttpServlet {



	public void doGet(HttpServletRequest request, HttpServletResponse response)

			throws ServletException, IOException {



		doPost(request, response);

	}



	public void doPost(HttpServletRequest request, HttpServletResponse response)

			throws ServletException, IOException {

		 

		 String xml = readXMLFromRequestBody(request);

		 Document xmlDoc = null;

		 

		 try {

			xmlDoc = DocumentBuilderFactory.newInstance().newDocumentBuilder().

						parse(new ByteArrayInputStream(xml.getBytes()));

		} catch (SAXException e) {

			

			 System.out.println("SAXException: " + e);

		} catch (ParserConfigurationException e) {

			

			System.out.println("ParserConfigurationException: " + e);

		}

		

		NodeList selectedPetTypes = xmlDoc.getElementsByTagName("type");

		

		String type = null;

		String responseText = "Selected Pets: ";

		for(int i = 0; i < selectedPetTypes.getLength(); i++) 

		{

			type = selectedPetTypes.item(i).getFirstChild().getNodeValue();

			responseText = responseText + " " + type;

		}

//		System.out.println(type);

		response.setContentType("text/html");

		PrintWriter out = response.getWriter();

		out.print(responseText);

	}



	private String readXMLFromRequestBody(HttpServletRequest request) {

		

		StringBuffer xml = new StringBuffer();

		String line = null;

		

		try {

			BufferedReader reader = request.getReader();

			while((line = reader.readLine())!=null)

			{

				xml.append(line);

			}

		} catch (IOException e) {

			

			 System.out.println("Error reading XML: " + e.toString());

		}

		//System.out.println(xml.toString());

		return xml.toString();

	}





}

问题是这样的，
当我把postingXML.jsp文件放到项目的WebRoot根目录下时，也就是WebRoot、postingXML.jsp这样时，程序正常执行！
但是，我在WebRoot\ 下建一个文件夹（文件夹名为sendParam），将postingXML.jsp放到sendParam文件夹下，也就是WebRootsendParam\postingXML.jsp 时，运行报错，会弹出响应函数function processResponse()中的window.alert("你所请求的页面有异常");对话框，这是为什么啊？请高手帮忙！

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neo_yoho 2008-06-24

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../表示上一级目录
这里目录相对简单所以就用../了

但是当你不知道当前的JSP与SERVLET的是怎么相对的时候你可以用
/servlet/PostingXMLExample……
因为单独一个/表示根目录
此时/servlet/PostingXMLExample就表示到根目录下去找servlet文件夹下的PostingXMLExample文件

我说不清楚~ 如果要深入了解你可以底呕埃网上搜 “HTML相对路径”就可以了
再次对自己的表达能力汗一次

againlove 2008-06-24

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[Quote=引用 1 楼 neo_yoho 的回复:]
改
var url = "../servlet/PostingXMLExample?timeStamp=" + new Date().getTime();
试试
应该是路径问题
[/Quote]
按这样好使了，但我还不是不明白，为什么要在前面加“../”
而放在WebRoot\ 目录下就不需要加呢？还请指点一下！！

neo_yoho 2008-06-24