Question

我正在使用PHP Codeigniter进行项目，我需要连接多个字段并创建一个名为key的新字段，该字段应存储在同一个表中。例如，如果我插入名称：jack country：USA代码：+691那么我的代码应该将JACK-USA-691动态存储到我的表的关键字段中。我有一个执行我的CRUD操作的基本模型。下面是我的控制器代码，以获取新用户（经销商）以及我获得新用户（经销商）的模型：

控制器：

public function edit ($id = NULL)
{
    // Fetch a user or set a new one
    if ($id) {
        $this->data['user'] = $this->reseller_m->get($id);
        count($this->data['user']) || $this->data['errors'][] = 'User could not be found';
    }
    else {
        $this->data['user'] = $this->reseller_m->get_new();
    }

    // Set up the form
    $rules = $this->reseller_m->rules_admin;
    $id || $rules['password']['rules'] .= '|required';
    $this->form_validation->set_rules($rules);

    // Process the form
    if ($this->form_validation->run() == TRUE) {



$data = $this->reseller_m->array_from_post(array('sip_username','sip_password','key','allocation_block','name','email','password','phone','balance','user_num','address','country','created','modified','status'));


        $data['password'] = $this->reseller_m->hash($data['password']);

        $values=array($this->input->post('name'),$this->input->post('country'),$this->input->post('name'));

        $key=implode('-',$values);

        $this->db->insert('reseller',$key);



        $key=$this->reseller_m->save($data, $id);

        for($i=1; $i<=$data['user_num'];$i++)
            {
            $userdata=array('key'=>$key);
        // here users is taken name of user table with retailer_id is field
            $this->user_m->save($userdata,$id);
             }



        redirect('admin/reseller');
    }

    // Load the view
    $this->data['subview'] = 'admin/reseller/edit';
    $this->load->view('admin/_layout_main', $this->data);
}

模特：

public function get_new(){
    $user = new stdClass();

//          $user->id = '';
    $user->sip_username='';
    $user->sip_password='';
    $user->key='';
    $user->allocation_block='';
    $user->name='';
    $user->email = '';      
    $user->password = '';
    $user->phone=''; 
    $user->user_num=''; 
    $user->address = '';
    $user->status = '';
    $user->country=''; 
    $user->created = '';
    $user->modified  = '';
    $user->balance = '';
    return $user;
    $this->session->set_userdata($data);

}

我的一个解决方案是创建一个用户获取其id并通过连接字段来更新它。我也发现它是正确的，但如何连接数组的元素？以及如何动态地这样做？请帮我处理我的代码！我可以动态地将连接的字符串存储在我的数据库中吗？

Answer 1

使用implode()连接如下

$values=array($this->input->post('name'),$this->input->post('country'),$this->input->post('name'));
$key=implode('-',$value);

//output will be name-country-code

如何连接数组元素并将它们存储在codeigniter php中的字段中

1 个答案: