Question

我正在使用CodeIgniter并且数据库存在一些问题。我使用表格＆＃39;客户＆＃39;中的客户名称填充我的下拉列表。在DB中。我需要从下拉列表中选择名称，输入项目名称并将其添加到数据库表项目中，但在此表中，我应该没有客户名称，但是ID。我该如何实现呢？当我想发送不是id，但是名字 - 我有一个错误：customerName有NULL值。

这是我的控制器

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

    class Add extends CI_Controller {

    public function __construct() {
        parent::__construct();
        $this->load->helper(array('url', 'form'));
        $this->load->model("addmodel");
        $this->load->library('session');
    }

    private function view($page, $data=false) {
        $this->load->view("header.php", $data);
        $this->load->view($page, $data);
        $this->load->view("footer.php", $data);
        }

     public function add_project() { 
         $this->load->database();  
         $data['customers']=$this->addmodel->select_customer();  
         $this->view('add/add_project.php', $data);  

      }   
       public function new_project() {
      $this->load->database();  
      $data['customers']=$this->addmodel->select_customer(); 

      $this->load->view("add/add_project.php", $data);

      $projectName = $this->input->post("projectName");
      $customerID = $this->input->post("customerName");

      $this->addmodel->add_project($projectName, customerID); 

      redirect("add_project", "refresh");
      }
    }

这是我的模特

<?php
 class AddModel extends CI_Model {

  public function __construct() {

      $this->load->database();

  }

   function add_project($projectName, $customerID) {

  $this->db->insert('projects', array(
      'projectName' => $projectName,
      'customerID' => $customerID
  ));

  }

 function select_customer() {  
     $query = $this->db->get('customers');  
     return $query; 
  }  
   }
   ?>

最后，这是我的观点

<?php echo form_open('new_project'); ?>


  <div id="form-main">
  <div id="form-div">
    <form class="form" id="form1">

     <select class="feedback-input" id="customer_selecter"> 
        <option name="customerName">Select customer</option>
        // <?php 
            foreach($customers->result() as $row){ 
              echo '<option value="'.$row->customerID.'">'.$row->customerName.'</option>';
            }
            ?>
      </select>

      <p class="projectName">
          <input type="text" name="projectName" placeholder="Project name" required class="feedback-input" id="projectName" />
      </p>


      <div class="submit">
          <button type="submit" id="button-red">Add</button>
      <div class="ease"></div>
          </form>   
      </div>
    </form>
  </div>

Answer 1

将选项值更改为$ row-＆gt; customerName，以便您可以获取客户名称

 <select class="feedback-input" id="customer_selecter" name="customerName> 
    // <?php 
        foreach($customers->result() as $row){ 
          echo '<option value="'.$row->customerID.'">'.$row->customerName.'</option>';
        }
        ?>
  </select>

Answer 2

更改您的Html代码选择框名称，如下所示：

<select class="feedback-input" id="customer_selecter" name="customerName"> 
        <option >Select customer</option>
        // <?php 
        foreach($customers->result() as $row){ 
          echo '<option value="'.$row->customerID.'">'.$row->customerName.'</option>';
        }
        ?>
</select>

并且需要使用$为customerId传递corrct变量，如下所示：

$this->addmodel->add_project($projectName, $customerID);

db codeigniter的下拉值

2 个答案: