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Php 如何使用点燃的数据表从复选框中的数据库传递字段id。?_Php_Codeigniter_Datatables - Fatal编程技术网

Php 如何使用点燃的数据表从复选框中的数据库传递字段id。?

Php 如何使用点燃的数据表从复选框中的数据库传递字段id。?,php,codeigniter,datatables,Php,Codeigniter,Datatables,我试图在生成数据表时在复选框中传递一个id,我的代码如下 控制器 public function list_view() { $this->datatables->select('c.id,,c.image,c.first_name,c.email_primary,c.contact_mobile,c.contact_home,cn.country_name,c.unique_id,c.title') ->from('co

我试图在生成数据表时在复选框中传递一个id,我的代码如下

控制器

  public function list_view() {

            $this->datatables->select('c.id,,c.image,c.first_name,c.email_primary,c.contact_mobile,c.contact_home,cn.country_name,c.unique_id,c.title')
             ->from('contact c')  
             ->join('country cn','cn.id=c.country_id','left');
            $this->datatables->add_column('action', '<input id="list-chk_" class="top-label" type="checkbox" value=c.id />', 'id');
            $this->datatables->add_column('sync', '<span class="list-sync" ></span>', 'c.id');
            $this->datatables->add_column('email', '<span class="list-mail" ></span>', 'c.id');
            echo $this->datatables->generate('json');

        }
我只想将c.id与复选框一起传递,因为我需要在datatables中使用该id进行进一步的操作


多谢各位

您可以将变量作为{{c.id}传递

例如:

$result = DB::table('users')
        ->select('users.id as id','users.first_name', 'users.last_name', 'users.email', 'users.username', 'users.activated');

        return Datatables::of($result)
        ->add_column('actions', '<a aid="{{ $id }}" style="cursor: pointer;" class="edit_admin" data-toggle="modal" data-target="#edit_admin_modal" title="Edit Admin" ><i class="fa fa-pencil"></i></a>    <a style="cursor: pointer;" aid="{{ $id }}" class="delete_admin" data-toggle="modal" data-target="#delete_admin_modal" title="Remove Admin" ><i class="fa fa-times-circle"></i></a>    <a aid="{{ $id }}" style="cursor: pointer;" class="enable_disable_admin" title="Enable/Disable Admin" ><i class="fa fa-check-circle-o"></i></a>')                        
        ->remove_column('id')
        ->make();