Question

我有一个包含近14000条记录的位置表，我需要对服务器端数据进行ajax分页。我使用了下面的代码，但是没有用。

    <table class="table table-bordered table-striped table-hover dataTable js-exportable" id="htmltableID">
    <thead>
    <tr>

        <th>SNO</th>
        <th>Location</th>
        <th>City</th>
        <th>State</th>


    </tr>
    </thead>
    <tbody>

    <?php
    $i = 1;
    foreach ($data as $row) {
        echo "<tr>";
        echo "<td>" . $i . "</td>";
        echo "<td>" . $row->location . "</td>";
        echo "<td>" . $row->city . "</td>";
        echo "<td>" . $row->state . "</td>";
        echo "</tr>";
        $i++;
    }
    ?>
    </tbody>
</table>

<script>
    var oTable = "";
    $(document).ready(function () {
        oTable = $('#htmltableID').dataTable({
            "sPaginationType": "full_numbers",
            "bServerSide": true,
            "sAjaxSource": "location",
            "sServerMethod": "POST",
            "iDisplayLength": 5
        });
    });
</script>

通过使用此代码，我收到错误消息“ DataTables警告：表id = htmltableID-无法重新初始化DataTable。有关此错误的更多信息，请参见http://datatables.net/tn/3”和“ DataTables警告：表id = htmltableID-无效的JSON响应。有关此错误的更多信息，请参见http://datatables.net/tn/1“

Answer 1

您应该在主文件中具有以下代码类型。假设Maindata.php

<table class="table table-bordered table-striped table-hover dataTable js-exportable" id="htmltableID">
    <thead>
        <tr>
            <th>SNO</th>
            <th>Location</th>
            <th>City</th>
            <th>State</th>
        </tr>
    </thead>
</table>

<script>
    var oTable = "";
    $(document).ready(function () {
        oTable = $('#htmltableID').dataTable({
            "sPaginationType": "full_numbers",
            "bServerSide": true,
            "sAjaxSource": "location",
            "sServerMethod": "POST",
            "iDisplayLength": 5
        });
    });
</script>

需要从其他文件中获取数据（json），例如loadrecords.php（在您的情况下为location/loadRecord）

<?php
// $data is the list of records fetched from database

// No need to print the data since we need to provide json data to dataTable, so Below code not required
/*
$i = 1;
foreach ($data as $row) {
    echo "<tr>";
    echo "<td>" . $i . "</td>";
    echo "<td>" . $row->location . "</td>";
    echo "<td>" . $row->city . "</td>";
    echo "<td>" . $row->state . "</td>";
    echo "</tr>";
    $i++;
}
*/

// Data which you needs to send in 'location/loadRecord' should be like this (In my case it was `loadrecords.php`)
/*{
    "data": [
        ["1","Location 1","City 1","State 1"],
        ["2","Location 2","City 2","State 2"],
        .....
        .....
        ["N","Location N","City N","State N"]
    ]
}*/

// Loop the data records fetched from database and prepare array in below format
$json_arr = array(
    "data" => array(
        array("1","Location 1","City 1","State 1"),
        array("2","Location 2","City 2","State 2"),
        ...............
        ...............
        array("N","Location N","City N","State N"),
    )
);
echo json_encode($json_arr);

数据将具有4列的N个行数，因为您要在表中显示4列，因此需要在json中提供相关数据

数据表服务器端Ajax分页

1 个答案: