我使用 jQuery DataTables 并收到此警告消息:
DataTables warning: table=userTable - Invalid JSON response
servlet 从 MySQL 获取用户，我想将其显示在 jQuery 数据表中，但 Ajax 无法解析 JSON 或 servlet 中生成的 JSON 错误？
Servlet:
List<UserDTO> users = this.service.getAllUser();
Gson gson = new Gson();
request.setAttribute("users", gson.toJson(users));
request.getRequestDispatcher("listAllUser.jsp").forward(request, response);
JSP:
<%@ page language="java" contentType="text/html; charset=ISO-8859-1"
pageEncoding="ISO-8859-1"%>
<!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=ISO-8859-1">
<title>Registered Users</title>
<script src="http://code.jquery.com/jquery-1.10.2.js"></script>
<script src="http://code.jquery.com/ui/1.11.4/jquery-ui.js"></script>
<script
src="https://cdn.datatables.net/1.10.9/js/jquery.dataTables.min.js"></script>
<link rel="stylesheet"
href="http://cdn.datatables.net/1.10.9/css/jquery.dataTables.min.css" />
<link rel="stylesheet"
href="http://code.jquery.com/ui/1.11.4/themes/flick/jquery-ui.css">
<script>
$(document).ready(function() {
$('#userTable').dataTable({
"processing" : true,
"serverSide" : true,
"ajax" : {
"url" : "ListAllUserServlet",
"type" : "POST"
},
"columns" : [ {
"data" : "id"
}, {
"data" : "userName"
}, {
"data" : "firstName"
}, {
"data" : "lastName"
}, {
"data" : "email"
}, {
"data" : "phone"
}, {
"data" : "location"
}, {
"data" : "password"
}, {
"data" : "gender"
}, {
"data" : "birthday"
} ]
});
});
</script>
</head>
<body>
<table id="userTable" class="display">
<thead>
<tr>
<th colspan="10" id="userList">Users</th>
</tr>
<tr>
<th>User id</th>
<th>User name</th>
<th>First Name</th>
<th>Last Name</th>
<th>Email</th>
<th>Phone</th>
<th>Location</th>
<th>Password</th>
<th>Gender</th>
<th>Birth date</th>
</tr>
</thead>
<tfoot>
<tr>
<td colspan="10"><a href="index.jsp" id="toIndex">Back</a></td>
</tr>
</tfoot>
</table>
</body>
</html>
servlet 生成的 JSON:
[
{
"id": 1,
"userName": "userName1",
"firstName": "firstName1",
"lastName": "lastName1",
"email": "email1@gmail.com",
"phone": "36202080085",
"location": "location1",
"password": "password1",
"gender": "m",
"birthday": "1-02-2015"
}
]
最佳答案
您的代码存在几个问题:
您已使用
"serverSide": true启用服务器端处理模式，但您的数据格式改为客户端处理模式。删除"serverSide": true以使用客户端处理模式。您需要使用如下所示的
dataSrc: ""ash 来匹配您的 JSON 数据格式，请参阅dataSrc了解更多信息。"ajax" : { "url" : "ListAllUserServlet", "type" : "POST", "dataSrc": "" },
关于jquery - 数据表警告 : table=userTable - Invalid JSON response?，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/33582203/