Question

我正在尝试使用DataTables实现this，并且在数据显示正确时，每当我尝试使用搜索栏，过滤器时，都会收到以下错误：

DataTables警告：table id = checkin - 无效的JSON响应。有关此错误的详细信息，请参阅http://datatables.net/tn/1

当我点击该链接并按照说明操作时，这是来自response.php的json：

{"draw":1,"recordsTotal":3,"recordsFiltered":3,"data":[["28","708089113","Nicole Foster","","","2016-04-15 14:27:36"],["29","708089113","Larry Quaglia","","","2016-04-15 14:28:38"],["30","708089113","Nicole Foster","Test","test@syr.edu","2016-04-21 09:18:59"]]}

我仍然是JSON和AJAX的新手，但当我检查该页面上的JSON Lint和JSON Parser链接输出的内容时，它表示它们是有效的。

这是我的代码：

Reports.php（只显示调用DataTables的脚本）

<script>
    $( document ).ready(function() {
        $('#checkin').DataTable({
            "bProcessing": true,
            "serverSide": true,
            "dom": 'lBfrtip',
            "ajax":{
                url :"response.php", // json datasource
                type: "post"  // type of method  ,GET/POST/DELETE
                },
            "buttons":[
                {
                    extend: 'collection',
                    text: 'Export',
                    buttons: [
                            'copy',
                            'excel',
                            'csv',
                            'pdf',
                            'print'
                    ]
                }
            ]
            });   
    });
</script>

Response.php

<?php
//include connection file 
include_once("../connection.php");

// initilize all variable
$params = $columns = $totalRecords = $data = array();

$params = $_REQUEST;

//define index of column
$columns = array( 
    0 => 'id',
    1 => 'suid', 
    2 => 'staffMember',
    3 => 'studentName',
    4 => 'studentEmail',
    5 => 'checkinDateTime'
);

$where = $sqlTot = $sqlRec = "";

// check search value exist
if( !empty($params['search']['value']) ) {   
    $where .=" WHERE ";
    $where .=" ( studentName LIKE '".$params['search']['value']."%' ";    
    $where .=" OR staffMember LIKE '".$params['search']['value']."%' ";
    $where .=" OR studentEmail LIKE '".$params['search']['value']."%' ";
    $where .=" OR suid LIKE '".$params['search']['value']."%' ";
    $where .=" OR checkinDate LIKE '".$params['search']['value']."%' )";
}

// getting total number records without any search
$sql = "SELECT * FROM `checkin` ";
$sqlTot .= $sql;
$sqlRec .= $sql;
//concatenate search sql if value exist
if(isset($where) && $where != '') {

    $sqlTot .= $where;
    $sqlRec .= $where;
}


$sqlRec .=  " ORDER BY ". $columns[$params['order'][0]['column']]."   ".$params['order'][0]['dir']."  LIMIT ".$params['start']." ,".$params['length']." ";

$queryTot = mysqli_query($VisitorManagement, $sqlTot) or die("database error:". mysqli_error($VisitorManagement));


$totalRecords = mysqli_num_rows($queryTot);

$queryRecords = mysqli_query($VisitorManagement, $sqlRec) or die("error to fetch check-in data");

//iterate on results row and create new index array of data
while( $row = mysqli_fetch_row($queryRecords) ) { 
    $data[] = $row;
}   

$json_data = array(
        "draw"            => intval( $params['draw'] ),   
        "recordsTotal"    => intval( $totalRecords ),  
        "recordsFiltered" => intval($totalRecords),
        "data"            => $data   // total data array
        );

echo json_encode($json_data);  // send data as json format
?>

Connection.php

$hostname_VisitorManagement = "localhost";
$database_VisitorManagement = "visitor-management";
$username_VisitorManagement = "***";
$password_VisitorManagement = "***";
$VisitorManagement = mysqli_connect($hostname_VisitorManagement, $username_VisitorManagement, $password_VisitorManagement, $database_VisitorManagement);

if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}

date_default_timezone_set('America/New_York');

有什么我想念的吗？我在他们的论坛上四处看看，但我没有看到类似我的问题的解决方案，就像我的代码一样。

Answer 1

尝试将选项serverSide更改为false。如果设置为true，则每次尝试过滤/排序/更改页面时，表都会查询ajax源。如果您打算先检索所有数据，然后在客户端工作，serverSide选项应为false

更新：所以我猜测问题是数据表在进行ajax调用时没有发送任何参数，然后查询构建错误。

尝试将这些参数添加到ajax.data选项中：

"ajax":{
     url :"response.php", // json datasource
     type: "post",  // type of method  ,GET/POST/DELETE
     data: {}//data to send
 },

DataTables - 使用搜索栏时显示无效的JSON消息

1 个答案: