Javascript 使用ajax从数据库加载数据时向datatable添加复选框

使用ajax绑定数据库中的数据时，我无法将复选框添加到数据表中。如何添加服务器端数据加载复选框

我的jQuery：

var table = $('#example').DataTable({
        "ajax": "getBusperOrder.php",
        "bPaginate": true,
        "retrieve": true,
        "bProcessing": true,
        "pageLength": 10,
        "columns": [{
                mData: 'district'
            }, {
                mData: 'deponame'
            }, {
                mData: 'busname'
            }, {
                mData: 'bonnetnumber'
            }, {
                mData: 'routename'
            }, {
                mData: 'bustype'
            }, {
                mData: 'status'
            }
        ],
    });

HTML:

gerbusperoder.php

<?php
require('database/db.php');
$sql = "select * from bus as B left join depo as D on B.depoid=D.depoid left join district as DS on D.district=DS.id left join bustype as BS on B.bustypeid=BS.bustypeid left join route as R on B.routeid=R.routeid LEFT JOIN bustype as BT on B.bustypeid=BT.bustypeid WHERE B.busid IN(SELECT busid from bus where busid NOT IN (SELECT DISTINCT(bus_id) from advt_book_side AS ABS INNER JOIN booking as B on ABS.booking_number=B.bookingnumber WHERE B.todate>CURDATE() GROUP BY bus_id HAVING COUNT(sides_id)=4))";
$resultset = mysqli_query($db, $sql) or die("database error:" . mysqli_error($db));
$data = array();
while ($rows = mysqli_fetch_assoc($resultset)) {
    $data[] = $rows;
}
$results = array(
    "sEcho" => 1,
    "iTotalRecords" => count($data),
    "iTotalDisplayRecords" => count($data),
    "aaData" => $data
);
echo json_encode($results);
?>

我需要在id为的每个td的第一列添加一个复选框，您可以为此使用选项：

"columns": [{
        mData: 'district'
        render: (_,__,rowData) => `<input type="checkbox" value="${rowData.busid}">${rowData.busid}</input>`
    },
    ...
]

---

请找到这个答案。您可以从服务器端本身填充复选框

var table = $('#example').DataTable({
        "processing": false,
        "serverSide": true,
        "order": [],
        "ajax": {
            "url": "getBusperOrder.php",
            "type": "POST"
        }

在HTML中，您需要添加

<table id="example">
<thead>
    <tr>
        <th></th>
        <th>District</th>
        <th>Depo Name</th>
        <th>Bus Number</th>
        <th>Bonnet Number</th>
        <th>Route Name</th>
        <th>Bus Type</th>
        <th>Action</th>
    </tr>
</thead>

PHP的变化

<?php
require('database/db.php');
$sql = "select * from bus as B left join depo as D on B.depoid=D.depoid left join district as DS on D.district=DS.id left join bustype as BS on B.bustypeid=BS.bustypeid left join route as R on B.routeid=R.routeid LEFT JOIN bustype as BT on B.bustypeid=BT.bustypeid WHERE B.busid IN(SELECT busid from bus where busid NOT IN (SELECT DISTINCT(bus_id) from advt_book_side AS ABS INNER JOIN booking as B on ABS.booking_number=B.bookingnumber WHERE B.todate>CURDATE() GROUP BY bus_id HAVING COUNT(sides_id)=4))";
$resultset = mysqli_query($db, $sql) or die("database error:" . mysqli_error($db));
$data = array();
while ($rows = mysqli_fetch_assoc($resultset)) {
    $row = array();
    $row[] = '<div class="table-checkbox table-checkbox-data"><input type="checkbox" value="'. $rows['id'] .'"></div>';
   //insert other columns in $row array
    $data[] = $rows;
}
$output = array(
            "recordsTotal" => count($data),,
            "recordsFiltered" => count($data),,
            "data" => $data,
        );

    echo json_encode($output);
?>

---

您的表中没有id列。这也应该和checkbox一起动态生成吗？但是我需要设置checkbox的值，从这里开始，我尝试了{render:mData=>}和{render:mData=>+id}。否。我得到了另一个值，比如，{mData:'district'}，{mData:'deponame'}。我需要将{mData:'busid'}分配给checkbox@Linu：如果您需要将源数据对象的另一个属性的值放入单元格内容中，您可以引用整行数据，如我上面给出的示例中所示，但是，如何在datatable上显示此复选框？datatable将根据列数处理此问题。很高兴其他解决方案对你有效。

<?php
require('database/db.php');
$sql = "select * from bus as B left join depo as D on B.depoid=D.depoid left join district as DS on D.district=DS.id left join bustype as BS on B.bustypeid=BS.bustypeid left join route as R on B.routeid=R.routeid LEFT JOIN bustype as BT on B.bustypeid=BT.bustypeid WHERE B.busid IN(SELECT busid from bus where busid NOT IN (SELECT DISTINCT(bus_id) from advt_book_side AS ABS INNER JOIN booking as B on ABS.booking_number=B.bookingnumber WHERE B.todate>CURDATE() GROUP BY bus_id HAVING COUNT(sides_id)=4))";
$resultset = mysqli_query($db, $sql) or die("database error:" . mysqli_error($db));
$data = array();
while ($rows = mysqli_fetch_assoc($resultset)) {
    $row = array();
    $row[] = '<div class="table-checkbox table-checkbox-data"><input type="checkbox" value="'. $rows['id'] .'"></div>';
   //insert other columns in $row array
    $data[] = $rows;
}
$output = array(
            "recordsTotal" => count($data),,
            "recordsFiltered" => count($data),,
            "data" => $data,
        );

    echo json_encode($output);
?>