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我开发了简单的应用程序，我从数据库中动态生成了网格中的复选框，但我的问题是当用户从网格中选择复选框和其他必填字段并按下提交按钮时，它会添加重复值，所以我想知道如何检查将数据提交到数据库时，复选框值和具有数据库值的其他字段值。

以下代码我用来生成所有选定的项目，然后也保存 db

    foreach ($this->addattendee->results as $key=>$value)
{
//print_r($value);
$id = $this->Attendee_model->save($value);
}

我正在使用codeigniter ....任何人都可以用示例代码给出这个想法吗？

 {
    $person = $this->Person_model->get_by_id($id)->row();
    $this->form_data->id = $person->tab_classid;
    $this->form_data->classtitle = $person->tab_classtitle;
    $this->form_data->classdate = $person->tab_classtime;
    $this->form_data->createddate = $person->tab_crtdate;
    $this->form_data->peremail = $person->tab_pemail;
    $this->form_data->duration = $person->tab_classduration;

    //Show User Grid - Attendee>>>>>>>>>>>>>>>>>>>>>>>>
    $uri_segment = 0;
    $offset = $this->uri->segment($uri_segment);
    $users = $this->User_model->get_paged_list($this->limit, $offset)->result();
    // generate pagination
    $this->load->library('pagination');
    $config['base_url'] = site_url('person/index/');
    $config['total_rows'] = $this->User_model->count_all();
    $config['per_page'] = $this->limit;
    $config['uri_segment'] = $uri_segment;
    $this->pagination->initialize($config);
    $data['pagination'] = $this->pagination->create_links();
    // generate table data
    $this->load->library('table');
    $this->table->set_empty(" ");
    $this->table->set_heading('Check', 'User Id','User Name', 'Email', 'Language');
    $i = 0 + $offset;
    foreach ($users as $user)
    {
        $checkarray=array('name'=>'chkclsid[]','id'=>'chkclsid','value'=>$user->user_id);
        $this->table->add_row(form_checkbox($checkarray), $user->user_id, $user->user_name, $user->user_email,$user->user_language
        /*,anchor('person/view/'.$user->user_id,'view',array('class'=>'view')).' '.
        anchor('person/update/'.$user->user_id,'update',array('class'=>'update')).' '.
        anchor('person/showattendee/'.$user->user_id,'Attendee',array('class'=>'attendee')).' '.
        anchor('person/delete/'.$user->user_id,'delete',array('class'=>'delete','onclick'=>"return confirm('Are you sure want to delete this person?')"))*/ );
    }
    $data['table'] = $this->table->generate();

//结束网格代码

    // load view
    // set common properties
    $data['title'] = 'Assign Attendees';
    $msg = '';
    $data['message'] = $msg;
    $data['action'] = site_url('person/CreateAttendees');
    //$data['value'] = "sssssssssssssssssss";
    $session_data = $this->session->userdata('logged_in');
    $data['username'] = "<p>Welcome:"." ".$session_data['username']. " | " . anchor('home/logout', 'Logout')." | ". "Userid :"." ".$session_data['id']; "</p>";
    $data['link_back'] = anchor('person/index/','Back to list of Classes',array('class'=>'back'));
    $this->load->view('common/header',$data);
    $this->load->view('adminmenu');
    $this->load->view('addattendee_v', $data);

}
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2 回答 2

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也许我和你有同样的麻烦。这就是我所做的。

<?php
public function set_news(){
        $this->load->helper('url');

        $slug = url_title($this->input->post('title'), 'dash', TRUE);

        $query = $this->db->query("select slug from news where slug like '%$slug%'");
        if($query->num_rows()>=1){
            $jum = $query->num_rows() + 1;
            $slug = $slug.'-'.$jum;
        }

        $data = array(
            'title' => $this->input->post('title'),
            'slug' => $slug,
            'text' => $this->input->post('text')
        );

        return $this->db->insert('news', $data);        
    }
?>

然后它工作。

于 2013-12-06T09:57:11.620 回答
0

代码非常混乱，但我认为我在我的应用程序中解决了一个类似的问题，我不确定它是否是最好的方法，但它可以工作。

function save_vote($vote,$show_id, $stats){
    // Check if new vote
    $this->db->from('show_ratings')
            ->where('user_id', $user_id)
            ->where('show_id', $show_id);
    $rs = $this->db->get();
    $user_vote = $rs->row_array();
            // Here we are check if that entry exists
    if ($rs->num_rows() == '0' ){
        // Its a new vote so insert data

        $this->db->insert('show_ratings', $rate);

    }else{
        // Its a not new vote, so we update the DB. I also added a UNIQUE KEY to my database for the user_id and show_id fields in the show_ratings table. So There is that extra protection.
        $this->db->query('INSERT INTO `show_ratings`  (`user_id`,`show_id`,`score`) VALUES (?,?,?) ON DUPLICATE KEY UPDATE `score`=?;', array($user_id, $show_id, $vote, $vote));
        return $update;
    }
}

我希望这个代码片段能让你知道该怎么做。

于 2012-11-11T23:47:11.263 回答