DataTables警告:table = userTable - 无效的JSON响应？

Question

DataTables警告:table = userTable - 无效的JSON响应？

use*_*360 5 jquery jsp servlets datatables gson

我使用jQuery DataTables并收到此警告消息:

DataTables警告:table = userTable - 无效的JSON响应

servlet从MySQL中获取用户,我希望在jQuery Datatable中显示,但是Ajax无法解析JSON或者在servlet中生成错误的JSON？

Servlet的:

    List<UserDTO> users = this.service.getAllUser();
                Gson gson = new Gson();
                request.setAttribute("users", gson.toJson(users));
                request.getRequestDispatcher("listAllUser.jsp").forward(request, response);

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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>

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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"
    }
]

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Answer 1

Gyr*_*com 8

您的代码有几个问题:

您已启用服务器端处理模式,"serverSide": true但您的数据已格式化为客户端处理模式.删除"serverSide": true以使用客户端处理模式.
您需要使用dataSrc: ""下面显示的灰匹配您的JSON数据格式,dataSrc有关详细信息,请参阅.
```
"ajax" : {
    "url" : "ListAllUserServlet",
    "type" : "POST",
    "dataSrc": ""
},
```
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Answer 2

Cod*_*rld 5

即使所有数据都正确显示，并且当您收到此警告时，也可以将其处理为，

$.fn.dataTable.ext.errMode = function(obj,param,err){
                var tableId = obj.sTableId;
                console.log('Handling DataTable issue of Table '+tableId);
        };

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它之所以有效，是因为有时点击 ajax 不会返回任何内容。确实很有帮助 (2认同)

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