使用Codeigniter更新MySql中的數據不起作用

Question

MySql數據庫未使用在視圖中輸入的新數據進行更新。遵循MVC架構。

我的控制器：

public function saveEdit() { 
      $this->load->helper('form'); 

      $id = $this->uri->segment(3); 

      $save=array(
      'userstory' => $this->input->post('userstoryarea'), 
      'datetocomplete' => $this-> input->post('datecomplete'), 
      'name'=> $this->input->post('developer') 
      ); 

      $this->load_userStory->saveEdited($id,$save); 
      $this->viewUStory(); 
     } 

我的模型

public function saveEdited($id,$save){ 

    $this->db->where('devid', $id); 
    $this->db->update('developer', $save); 


} 

在一些地方我都要求一些JavaScript方法將返回一些值顯示。

我查看

<?php foreach ($story as $UserStory): ?>     
     <form method="post" action="<?php echo base_url() ."addstories/saveEdit"?>" > 
     <p> Select Devoloper to add : 
     <select class="form-control" id="developer" name ="developer" style="width:200px;"> </p> 

     <?php 
      $val2 = $row['name']; 
      foreach($developers as $row) 
      { 
      echo '<option value="'.$row->full_name.'" >'.$row->full_name.'</option>'; 
      } 
      ?> 
      </select> 
      <p> 

      <p> 
      Select a date to complete the project : <i> 

      <input type="date" id="datecomplete" name ="datecomplete" value="<?php echo $UserStory->datetocomplete;?>" onchange="calculate()"> 
      </i> </p> 


      <p>    
      <input type="textarea" name="userStory" id="userStory" value=" <?php echo $UserStory->userstory; ?> "> 
      </p> 

      <input type="submit" name ="dsubmit" value="Save" id="submit"> 
      </form> 
      <?php endforeach; ?> 

Answer 1

通過第三值id在行動像我傳遞的$ id

<form method="post" action="<?php echo base_url() ."addstories/saveEdit/".$id?>" >