我是java servlets和dataTables的新手我无法让我的java servlet执行：
我使用netbeans 8.2 IDE与tomcat 8.0.27和DataTables 10.1.16
我的NetBeans结构如下所示：
战争看起来像这样： 我的index.jsp看起来像这样：
DataTable定义（在html头部分）
<script lang='javascript'>
$(document).ready(function () {
$('#memberList').dataTable( {
"processing": true,
"serverSide": true,
"ajax": {
"url": "${pageContext.request.contextPath}/SubSearch",
"type": "GET"
}
});
});
</script>
Html Body: <body>
<h1>Member TXN Display</h1>
<div>
(Enter Search Criteria)<br/>
<form action="${pageContext.request.contextPath}/SubSearch" method="post" >
<input type="text" id="SearchCritiera" style="width:322px">
<input type="submit" value="FIND">
<table id="memberList">
<thead>
<tr>
<th>Member #</th>
<th>Last Name</th>
<th>First Name</th>
</tr>
</thead>
</table>
</form>
</div>
</body>
</html>context.xml看起来像这样：
<?xml version="1.0" encoding="UTF-8"?>
<Context path="/MemberTXN"/>
单击Submit（Find）按钮调用servlet，但是'ajax'似乎根本没有命中。我在servlet中有断点，当我在'Debug'模式下运行时，由于document.ready代码而没有骰子，我希望它会被击中。
Servlet代码：
package member;
import java.io.IOException;
import java.io.PrintWriter;
import javax.servlet.ServletException;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import com.google.gson.*;
/**
*
* @author Ainsworth
*/
@WebServlet(name = "SubSearch", urlPatterns = {"/SubSearch"})
public class SubSearch extends HttpServlet {
/**
* Processes requests for both HTTP <code>GET</code> and <code>POST</code>
* methods.
*
* @param request servlet request
* @param response servlet response
* @throws ServletException if a servlet-specific error occurs
* @throws IOException if an I/O error occurs
*/
protected void processRequest(HttpServletRequest request, HttpServletResponse
response)
throws ServletException, IOException {
JsonObject jMembers = new JsonObject();
jMembers.addProperty("Echo","1");
jMembers.addProperty("TotalRecords", 7);
jMembers.addProperty("TotalDisplayRecords", 7);
JsonArray data = new JsonArray();
JsonArray row = new JsonArray();
row.add("123456789");
row.add("Trump");
row.add("Donald");
data.add(row);
row = new JsonArray();
row.add("123456799");
row.add("Clinton");
row.add("Hillary");
data.add(row);
row = new JsonArray();
row.add("123456809");
row.add("Shcumer");
row.add("Chuck");
data.add(row);
row = new JsonArray();
row.add("123456819");
row.add("Warren");
row.add("Elizabeth");
data.add(row);
row = new JsonArray();
row.add("123456829");
row.add("Sanders");
row.add("Bernie");
data.add(row);
row = new JsonArray();
row.add("123456839");
row.add("DeVoss");
row.add("Betsy");
data.add(row);
row = new JsonArray();
row.add("123456849");
row.add("Meyers");
row.add("Seth");
data.add(row);
jMembers.add("Data", data);
response.setContentType("application/Json");
response.getWriter().print(jMembers.toString());
}
/**
* Handles the HTTP <code>GET</code> method.
*
* @param request servlet request
* @param response servlet response
* @throws ServletException if a servlet-specific error occurs
* @throws IOException if an I/O error occurs
*/
@Override
protected void doGet(HttpServletRequest request, HttpServletResponse
response)
throws ServletException, IOException {
processRequest(request, response);
}
/**
* Handles the HTTP <code>POST</code> method.
*
* @param request servlet request
* @param response servlet response
* @throws ServletException if a servlet-specific error occurs
* @throws IOException if an I/O error occurs
*/
@Override
protected void doPost(HttpServletRequest request, HttpServletResponse
response)
throws ServletException, IOException {
processRequest(request, response);
}
/**
* Returns a short description of the servlet.
*
* @return a String containing servlet description
*/
@Override
public String getServletInfo() {
return "Short description";
}
}
这是提交按钮的输出：
如果有人可以指出我的方式的错误，我会非常感激
选项1：
您可以使用ajax.dataFilter选项来更改JSON结构，因为API默认情况下需要这样的结构：
{
"draw": 1,
"recordsTotal": 57,
"recordsFiltered": 57,
"data": [{...},{...}]
}
使用dataFilter来改变JSON结构：
$('#memberList').DataTable( {
"processing": true,
"serverSide": true,
"ajax": {
"url": "${pageContext.request.contextPath}/SubSearch",
"type": "GET",
dataFilter: function(data){
var json = jQuery.parseJSON( data );
json.recordsTotal = json.TotalRecords;
json.recordsFiltered = json.TotalDisplayRecords;
json.data = json.Data;
return JSON.stringify( json ); // return JSON string
}
}
方案2：
如果您有机会更改服务器API数据结构，请使用如下所示的适当密钥构建JSON。
jMembers.addProperty("echo","1"); //important case-sensitive
jMembers.addProperty("recordsTotal", 7);
jMembers.addProperty("recordsFiltered", 7);
jMembers.add("data", data);
还看到：
引用Firefox：嗯，这很尴尬：
我终于确定我的Javascript没有被执行。问题是我的<script type="text/javascript" src="js/jquery-3.2.1.min.js"></script>在>之前错过了</script>
现在再次感谢Gurkan，我正在执行我的服务器代码，你的帖子将帮助我确定为什么我没有填写我的表。
你应该在doGet或doPost中做你的算法