我试图在视图中添加分页链接，但不能使用“$data->links()”，因为我在ajax中传递了数据，在表tbody中传递了响应。 此外，如果你能帮助我完成我的分页根据我目前的结构，我将不胜感激。

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function getAllUsers() {
  $.ajax({
    url: base_url + '/users/listing',
    method: 'GET',
    success: function(result) {
      var html = '';
      $.each(result.data, function(i, row) {
        html += '<tr>'
        html += '<td class="text-center">' + ++i + '</td>'
        html += '<td>' + row.name + '</td>'
        html += '<td>' + row.email + '</td>'
        html += '<td>' + row.phone + '</td>'
        html += '<td>' + row.created_at + '</td>'
        html += '<td class="text-center"><ul class="table-controls">'
        html += '<li><a href="" id="userEdit" data-id="' + row.id + '" data-toggle="modal" data-target="#userModal" data-placement="top" title="Edit"></a></li>'
        html += '<li><a href="" id="userDel" data-id="' + row.id + '" data-toggle="tooltip" data-placement="top" title="Delete"></a></li>'
        html += '</ul></td>'
        html += '</tr>';
      })

      $('table tbody').html(html)
    }
  });
}

// Routes
Route::get('/users', 'Backend\UserController@index')->name('users');
Route::get('/users/listing', 'Backend\UserController@listing');


//Controller
public function index() {
  return view('backend/users/listing');
}

public function listing() {
  $users=User::paginate(5);
  return $users;
}

<div class="table-responsive">
  <table class="table table-bordered table-hover table-striped table-checkable table-highlight-head mb-4">
    <thead>
      <tr>
        <th class="text-center">No.
        </th>
        <th class="">Name</th>
        <th class="">Email</th>
        <th class="">Phone</th>
        <th class="">Register at</th>
        <th class="text-center">Actions</th>
      </tr>
    </thead>
    <tbody>

    </tbody>
  </table>
</div>

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