如何解决.tableExport不是函数错误

Question

所以我有这张桌子：

<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
<script type="text/javascript" src="js2/libs/FileSaver/FileSaver.min.js">    </script>
<script type="text/javascript" src="js2/libs/jsPDF/jspdf.min.js"></script>
<script type="text/javascript" src="js2/libs/jsPDF-AutoTable/jspdf.plugin.autotable.js"></script>
<script type="text/javascript" src="js2/tableExport.min.js"></script>

<?php
include_once("config.php"); //config for database

if(isset($_POST['detail_event_id'])) {
    $id_event = $_POST['detail_event_id'];
    $query = "SELECT id_user, id_event FROM user_event WHERE id_event = '$id_event'";
    $result = mysqli_query($db, $query);
    $output = '             
             <div class="bg-white">
                 <div class="table-hover margin-top-10px">
                     <table id="tablePendaftar" class="table table-hover">
                            <thead>
                            <tr>
                                <th>Nama</th>
                                <th>Tanggal Lahir</th>
                                <th>E-Mail</th>
                                <th>Nomor Telepon</th>
                                <th>Alamat</th>
                                <th>Deskripsi</th>
                            </tr>
                            </thead>

                            <tbody>

                            ';

    if (mysqli_num_rows($result) != 0) {
    while ($get = mysqli_fetch_array($result)) {
        $id_user = $get['id_user'];
        $getuser = "SELECT nama_depan, nama_belakang, tgl_lahir, email, tlp, alamat, deskripsi FROM user WHERE id_user = '$id_user'";
        $result2 = mysqli_query($db, $getuser);

        while ($getuser = mysqli_fetch_array($result2)) {
            $nama_depan = $getuser['nama_depan'];
            $nama_belakang = $getuser['nama_belakang'];
            $tgl_lahir = date('d F Y', strtotime($getuser['tgl_lahir']));
            $email = $getuser['email'];
            $tlp = $getuser['tlp'];
            $alamat = $getuser['alamat'];
            $deskripsi = $getuser['deskripsi'];
            $output .= '
                <tr>
                    <td>' . $nama_depan.' '.$nama_belakang . '</td>
                    <td>' . $tgl_lahir . '</td>
                    <td>' . $email . '</td>
                    <td>' . $tlp . '</td>
                    <td>' . $alamat . '</td>
                    <td>' . $deskripsi . '</td>
                </tr>
            ';
        }

    }

    }

    else{
        $output .= '
        <tr>
        <td>Tidak ada pendaftar</td>
        <td>Tidak ada pendaftar</td>
        <td>Tidak ada pendaftar</td>
        <td>Tidak ada pendaftar</td>
        <td>Tidak ada pendaftar</td>
        <td>Tidak ada pendaftar</td>
        </tr>
        ';
    }



    $output .=             '</tbody>
                     </table>
                 </div>
             </div>
    ';

    echo $output;

?>
    <a href="#" id="printpdf" class="btn btn-primary">
        Download PDF
    </a>
<?php

}
?>

<script>
$("#printpdf").click(function(){
    $('#tablePendaftar').tableExport({type:'pdf', jspdf: {orientation: 'l',
        format: 'a3',
        margins: {left:10, right:10, top:20, bottom:20},
        autotable: {styles: {fillColor: 'inherit',
            textColor: 'inherit'},
            tableWidth: 'auto'}
    }});
});
</script>

该表出现在html上并显示coorect，但问题是，当我点击表的下载PDF按钮时，控制台显示错误：

TypeError: $(...).tableExport is not a function

并且不会显示下载选项。

我已经按照插件中的教程进行了操作：

https://github.com/hhurz/tableExport.jquery.plugin

你们可以帮我解决这个错误吗？因为我不知道如何解决它，谢谢。