选择和显示数据取决于下拉菜单的值

时间:2017-02-06 09:05:08

标签: php jquery codeigniter

我想做出选择取决于下拉菜单,这是我的代码

视图: -

  
    

<select name="class" id="class"> <option selected value='-1'>--Select class--</option> <?php foreach($classes as $class) { echo '<option value="'.$class['class_id'].'">'.$class['class_name'].'</option>'; } ?> </select>

  

控制器: -

public function select_student($class_id){ <br />$class_id=$this->uri->segment(3);<br /> $this->load->model('select_model');<br />    header('Content-Type: application/x-json; charset=utf-8');<br />  echo(json_encode($this->select_model->student_subject($class_id)));<br />

}

这是我的模型

function student_subject($class_id){<br /> $this->db->select('*');<br /> $this->db->from('student s');<br /> $this->db->join('student_class c','s.student_id=c.student_id', 'left') ;<br /> $this->db-where('s.status',0);<br /> $this->db->where('c.date_of_end !=','0000-00-00');<br /> $this->db->where('c.date_of_end !=', $class_id);<br /> $query = $this->db->get();<br /> if($query->result_array()){<br />

                echo     '<table cellspacing="10" cellpadding=""><th>Names</th>';         
             $data='';
             $n=0;
    foreach($query->result_array() as $name)        {           $n++;

                            echo '<tr><td>' .$n .'</td><td>'. $name['firstname'].'</td><td> '.$name['middlename'].'</td><td>'.$name['lastname'].'</td><td><input type="checkbox" name="selected_stds[]"  value="'.$name['student_id'].'"></td></tr>';

                                    }

                    echo '</table>';
                     return $query->result_array(); }

1 个答案:

答案 0 :(得分:0)

在视图页面

<select name="class" id="class">
    <option>Select class</option>
    <?php foreach($classes as $class){?>
    <option value="<?php echo $class['class_id'] ?>">
        <?php echo $class['class_name'] ?>
    </option>
    <?php } ?>
</select>