[英]My Servlet return a whole page with only one write or one forward (using AJAX call)
我有一个AJAX get函数,它正在调用Servlet并在警报中显示数据。
如果我使用write或forward函数,结果是相同的:我的所有数据(+20 000行)...
我的Servlet
package com.suptrip.servlets;
import java.io.IOException;
import javax.servlet.ServletException;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
@WebServlet("/Alert")
public class Alert extends HttpServlet {
/**
*
*/
private static final long serialVersionUID = 1L;
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
//request.getRequestDispatcher("OTHERS/alert.html").forward(request, response);
response.setContentType("text/html");
response.setCharacterEncoding("UTF-8");
response.getWriter().write("KO");
}
}
我的AJAX
// Load an alert to inform the use that the register form is done with success
//$('#place-for-alert').load('/Alert', function(responseTxt, statusTxt, xhr){});
$.get("Alert", function(responseText, statusTxt, xhr){
if(statusTxt == "success")
alert("External content loaded successfully!");
if(statusTxt == "error")
alert("Error: " + xhr.status + ": " + xhr.statusText);
alert(responseText);
});
有任何错误。
Servlet给出了错误的内容...
让我知道您是否需要更多详细信息/信息!
谢谢 !
如果我使用“很长的路要走”来编写ajax请求,那么它就可以工作...
像这样
$.ajax({
url: "Alert",
//type: "post",
data: objectArray,
cache: false,
success: function(data) {
$("#place-for-alert").html(data);
}
});
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