我正在用java创建一个Web应用程序。在客户端,我有一个条形图,显示存储在由java服务器页面创建的tsv文件中的一些数据。通过单击按钮,服务器将更新文件中的这些数据。现在我想读取刷新的数据,但是我得到了更旧的数据。似乎浏览器缓存了该文件,因此无法获取更改的文件。
这是我的servlet代码:
public class GetDataServlet extends HttpServlet
{
private static final long serialVersionUID = 1L;
private User user;
Utility utility;
public void init() throws ServletException {
reset();
}
public void doGet (HttpServletRequest request,HttpServletResponse response) throws ServletException, IOException {
response.setHeader("Cache-Control", "no-cache");
response.setHeader("Pragma", "no-cache");
PrintWriter out = response.getWriter();
user.getProfile().get(0).setWeigth(user.getProfile().get(0).getWeigth()+0.03);
user.getProfile().get(1).setWeigth(user.getProfile().get(1).getWeigth()+0.02);
user.getProfile().get(5).setWeigth(user.getProfile().get(5).getWeigth()+0.01);
utility.createTsvFile(user, "/usr/local/apache-tomcat-7.0.50/webapps/Visualizer/data.tsv");
String message = String.format("data.tsv");
i++;
out.print(message);
}
public void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
if(request.getParameter("reset").compareTo("yes")==0)
reset();
}
private void reset(){
List<Concept> children = new ArrayList<Concept>();
Concept food = new Concept();
food.setWeigth(0.10);
food.setLabel("food");
food.setColor("#98abc5");
Concept dish = new Concept();
dish.setWeigth(0.08);
dish.setLabel("dish");
dish.setParent(food);
dish.setColor("#8a89a6");
Concept cuisine = new Concept();
cuisine.setWeigth(0.06);
cuisine.setLabel("cuisine");
cuisine.setParent(food);
cuisine.setColor("#8a89a6");
children.add(dish);
children.add(cuisine);
food.setChildren(children);
children.clear();
Concept pizza = new Concept();
pizza.setWeigth(0.05);
pizza.setLabel("pizza");
pizza.setParent(dish);
pizza.setColor("#6b486b");
Concept spaghetti = new Concept();
spaghetti.setWeigth(0.05);
spaghetti.setLabel("spaghetti");
spaghetti.setParent(dish);
spaghetti.setColor("#6b486b");
Concept sushi = new Concept();
sushi.setWeigth(0.06);
sushi.setLabel("sushi");
sushi.setParent(dish);
sushi.setColor("#6b486b");
children.add(pizza);
children.add(spaghetti);
children.add(sushi);
dish.setChildren(children);
List<Concept> profile = new ArrayList<Concept>();
profile.add(food);
profile.add(dish);
profile.add(cuisine);
profile.add(pizza);
profile.add(spaghetti);
profile.add(sushi);
user = new User("mario", profile);
utility = new Utility("");
}
}
这是调用servlet的javascript代码:
function ajaxSyncRequest(reqURL) {
//Creating a new XMLHttpRequest object
var xmlhttp;
if (window.XMLHttpRequest) {
xmlhttp = new XMLHttpRequest(); //for IE7+, Firefox, Chrome, Opera, Safari
} else {
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP"); //for IE6, IE5
}
//Create a asynchronous GET request
xmlhttp.open("GET", reqURL, false);
xmlhttp.send(null);
//Execution blocked till server send the response
if (xmlhttp.readyState == 4) {
if (xmlhttp.status == 200) {
document.getElementById("message").innerHTML = xmlhttp.responseText;
update(xmlhttp.responseText);
//alert(xmlhttp.responseText);
} else {
alert('Something is wrong !!');
}
}
}
function update(file){
d3.tsv(file, function(error, data) {
......
在html页面中我也提到了这个:
<meta http-equiv="Cache-control" content="no-cache">
但它不起作用。
使用这段代码,我第一次调用servlet时可以显示正确的数据,即使文件tsv中的数据发生变化,我也会得到第一个。
哪种方法可以读取文件中刷新的数据?
答案 0 :(得分:0)
好的,我正面临着像这样的浏览器缓存问题,所以可以完成两个小时
1)您可以创建一个过滤器并指示过滤器不要缓存某些类似Prevent user from seeing previously visited secured page after logout
的内容2)每次你点击tsv文件的url时,最后添加一个随机变量。(这通常是互联网浏览器的情况)