Question

我正在考虑从我的数据库事件表中获取member_id，这是与另一个表成员相关的外键。我想得到的是Incident表中与Members表相关的所有member_id，因此我想显示member_id的Name字段。这是我的代码

// My Model 
function get_mid(){
    $data = array();
    $query = $this->db->get('Incident');
    foreach($query->resul_array as $row) {
        $data[] = $row['member_id'];
    }
    return $data;
}

在我的控制器中我有

public function index() {
    $member_id = $this->incident_model->get_mid();
    if(count($member_id)){
        foreach($member_id as $id){
            $f_name =$this->incident_model->get_fname($id);
            $m_name =$this->incident_model->get_mname($id);
            $l_name =$this->incident_model->get_lname($id);
            $data = array(
                'first_name' => $f_name,
                'middle_name' => $m_name,
                'last_name' => $l_name
            );
        }
        $this->load->view('incident_list', $data);
    }
}

这给了我视图中所有具有相同名称的记录。

Answer 1

即时使用代码

 public function index(){
 $member_id = $this->incident_model->get_mid();
 if(count($member_id)){
    foreach($member_id as $id){
        $f_name =$this->incident_model->get_fname($id);
        $m_name =$this->incident_model->get_mname($id);
        $l_name =$this->incident_model->get_lname($id);
   $data = array(
        'first_name' => $f_name,
        'middle_name' => $m_name,
        'last_name' => $l_name
       );
     }
 $this->load->view('incident_list', $data);
   }

替换为

public function index(){
 $member_id = $this->incident_model->get_mid();
$temp = array();
$data = array();
 if(count($member_id)){
    foreach($member_id as $id){
        $f_name =$this->incident_model->get_fname($id);
        $m_name =$this->incident_model->get_mname($id);
        $l_name =$this->incident_model->get_lname($id);
   $temp = array(
        'first_name' => $f_name,
        'middle_name' => $m_name,
        'last_name' => $l_name
       );
    $data[] = $temp;
     }
 $this->load->view('incident_list', $data);
   }

尝试代码。

从codeigniter中的数据库中获取列记录

1 个答案: