4

嗨，我正在使用 codeIgniter，我设法简单地将 ID 号码和电话号码发布到名为“offers”的表中，这两个字段都是 INT，但是当我尝试更新与特定 ID 对应的电话号码时，我看不到任何变化数据库。我在下面列出了我的控制器、模型和视图

新报价控制器

   <?php 
    if ( ! defined('BASEPATH')) exit('No direct script access allowed');
  //insert data into db using offer_model model
  // other option update submit 
  class newOffer extends CI_Controller {
       function addOffer() {
       //if the form is submitted           
       $this->load->view("check");
       $this->load->model("offer_model");
         if ($this->input->post('mysubmit')) {
             $this->offer_model->entry_insert();
          }
       }

        function updateOffer (){
           $this->load->view("check2");
           $this->load->model("offer_model");
           if ($this->input->post('mysubmit')) {
                $this->offer_model->upddata();
            }
         }
    }
  ?>

报价模型 

class offer_model extends CI_Model{

 public function entry_insert(){
      $data = array(
           'idNum' => $this->input->post('idNum'),
            'phneNum' => $this->input->post('phneNum'),

        );
      $this->db->insert('offers',$data);
  }

   public function upddata($data) {
    $this->db->where('idNum', $idNum);
    $this->db->update('data' ,$data);
    //extract($data); 
  //$data['OfferName']
  //$this->db->where('OfferName' ,  $data['OfferName']); 
    //$this->db->update($Offers, array('OfferName' => $OfferName)); 
   return true;
    }
}
 ?>

更新值的视图

<?form _open(base_url()."index.php/newOffer/updateOffer")?>
<div class="form">
 // <?php echo form_open('newOffer/addOffer'); ?>
<legend>Please enter details for your new offer</legend>
<label for="ID Number">ID Number:  <span class="required">*</span></label>
<input type="text" name="idNum" id="idNum" placeholder="Please enter ID Number/>
<label for="phone Number">Phone Number:</label>
<input type="text" name="phneNum" id="phneNum " placeholder="Please enter phone Number"/>

 <fieldset class="submit_field">
     <?php echo form_submit('mysubmit', 'Submit Form'); ?>
 </fieldset>
 </div><!-- end of form div -->
   ?>
4

4 回答 4

1

您没有将任何数据传递给您的模型

$this->offer_model->upddata();

你需要添加类似的东西

$this->offer_model->upddata($this->input->post());

同样在您的模型代码$idNum未定义中，您也需要提供它。

例如：

public function upddata($data) {
  $idNum = $data['idNum'];
  unset($data['idNum']);
  $this->db->where('idNum', $idNum);
  $this->db->update('offers' ,$data);
  return true;
}
//etc
于 2013-09-24T15:01:40.513 回答
0

使用构造在控制器中加载模型并始终将发布数据传递给模型函数，然后只有你会发现表单发布数据到模型函数中。

<?php 
  if ( ! defined('BASEPATH')) exit('No direct script access allowed');

  function __construct()
  {
        parent::__construct();
        $this->load->model("offer_model");
  }

  class newOffer extends CI_Controller {
       function addOffer() {
       //if the form is submitted           
       $this->load->view("check");
       $this->load->model("offer_model");
         if ($this->input->post('mysubmit')) {
             $CI_Post = $$this->input->post();
             $this->offer_model->entry_insert($CI_Post);
          }
       }

        function updateOffer (){
           $this->load->view("check2");

           if ($this->input->post('mysubmit')) {
              $CI_Post = $$this->input->post();
              $this->offer_model->upddata($CI_Post);
           }
         }
    }
?>

这是在数组中获取表单发布数据的模型。

<?php
class offer_model extends CI_Model{

     public function entry_insert($param=""){
          $data = array(
               'idNum' => $param['idNum'],
                'phneNum' => $param['phneNum'],

            );
          $this->db->insert('offers',$data);
          return $this->db->insert_id();
      }

     public function upddata($param="") {
      $data = array(
             'idNum' => $param['idNum'],
              'phneNum' => $param['phneNum'],

      );
      $this->db->where('idNum', $idNum);
      $this->db->update('data' ,$data);
      return $this->db->affected_rows();
     }
}
?>
于 2016-12-27T10:09:04.777 回答
0

确保您来自插入或更新数据的同一个表。

public function upload($id_akun, $data)
    {
        $table = "foto";

        $this->db->where("id_akun", $data['id_akun']);
        $count = $this->db->count_all_results($table);


        if ($count < 1) {
            $this->db->insert($table, $data);
        } else {
            $this->db->where('id_akun', $data['id_akun']);
            $this->db->update($table, $data);
        }
    }

祝你好运 ：）

于 2016-10-07T05:35:32.963 回答
-1

在你的模型中做类似的事情

public function upddata() {
    $data=array();
    $data=$this->input->post(); //get all post value to data array
    $idNum=$data['idNum'];
    unset($data['idNum']); // unset unnecessary values 
    $this->db->where('idNum', $idNum)->update('offers' ,$data);
    return true;
}

这是过程。如果您有 100 个输入字段，则不可能将每个值都添加到数组中。如果您遇到任何问题，请告诉我。

更新

改变

<?php echo form_open('newOffer/addOffer'); ?>

至

<?php echo form_open('newOffer/updateOffer'); ?>

在视图中更新值

于 2013-09-24T15:13:59.207 回答