无法使用 CodeIgniter 将数据插入数据库

Question

我正在使用最新版本的 CI (3.1.6) 问题是我要插入的数据无法发布到控制器，所以如果我从视图中打印出数据，它就无法从表单中捕获任何数据。我认为问题出在表格中。但我不知道如何解决它。谁能帮我？ 这是我的代码

我的 form.php 代码

<div id="register" class="text-center">
   <div class="container">
    <div class="section-title center">
      <h2>Register</h2>
      <hr>
      <form action="" method="post">
      <link href="<?php echo base_url();?>assets/user/css/login.css" rel="stylesheet">
      <input type="text" name="firstname" placeholder="First Name">
      <input type="text" name="lastname" placeholder="Last Name">
      <input type="text" name="fullname" placeholder="Full Name">
      <input type="text" name="username" placeholder="Username">
      <input type="text" name="email" placeholder="Email">
      <input type="password" name="pass" placeholder="Password">
   <!--    <input type="password" name="pass1" placeholder="Retype Password"> -->
      <ul>
      <p1>Gender</p1>
      <select name="gender" id="" class="pilihan">
        <option value="men">Male</option>
        <option value="women">Female</option></select>
      </ul>
      <ul>
      <p1>Occupation</p1>
      <li><input type="radio" name="job" value="doctor" checked> Doctor<br></li>
      <li><input type="radio" name="job" value="nurse"> Nurse<br></li>
    </ul>
      <link rel="stylesheet" type="text/css" href="assets/user/login.css">
      <a href="<?php echo base_url('c_register/do_insert'); ?>" class="btn btn-primary btn-lg active" role="button">Primary link</a>
  </form>

我的控制器代码

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

class C_register extends CI_Controller {

    function __construct(){
        parent::__construct();
        $this->load->model('m_register');
    }


    function do_insert(){
        $this->load->helper('form');
        $this->load->library('form_validation');

        $this->form_validation->set_rules('firstname', 'firstname', 'required');
        $this->form_validation->set_rules('lastname', 'lastname', 'required');
        $this->form_validation->set_rules('fullname', 'fullname', 'required');
        $this->form_validation->set_rules('username', 'username', 'required');
        $this->form_validation->set_rules('email', 'email', 'required');
        $this->form_validation->set_rules('pass', 'pass', 'required');
        $this->form_validation->set_rules('gender', 'gender', 'required');
        $this->form_validation->set_rules('job', 'job', 'required');

        if ($this->form_validation->run() === FALSE) {
            echo "gagal diinsert";
            // $this->load->view('templates/header', $data);
            // $this->load->view('news/comment_form');
            // $this->load->view('templates/footer');
        } else {
            $this->m_register->InsertData();
            $this->load->view('dashboard');
        }
    }
}?>

我的模式代码

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

class M_register extends CI_Model {

    public function InsertData($tabelName, $data){
         $datac = array(
            'firstname' => $this->input->post('firstname'),
            'lastname' => $this->input->post('lastname'),
            'fullname' => $this->input->post('fullname'),
            'username' => $this->input->post('username'),
            'email' => $this->input->post('email'),
            'pass' => $this->input->post('pass'),
            'gender' => $this->input->post('gender'),
            'job' => $this->input->post('job')
        );

        $res =  $this->db->insert('member', $datac);
        return $res;

    }

    }
?

Answer 1

您没有像这样在控制器构造函数上加载数据库类。

  function __construct(){
    parent::__construct();
    $this->load->model('m_register'); 
    $this->load->database();
}

Answer 2

首先要解决的问题是模型的方法InsertData()。您已在模型中定义它需要两个参数 - $tableName 和 $data

public function InsertData($tabelName, $data){

但是在控制器中你调用方法而不提供任何参数。

    } else {
        $this->m_register->InsertData(); //No parameters???

因此，您要么需要传递参数，要么需要更改方法的定义。这是一个更简单的版本，可以完成插入工作，您无需更改控制器。

public function InsertData()
{
    $datac = $this->input->post();
    return $this->db->insert('member', $datac);
}

您不需要一次获取每个输入。在不传递字段名称的情况下调用 input->post() 将返回一个包含所有字段的数组。这意味着您不必手动创建$datac 数组。

此外，简单地返回 $this->db->insert 返回的内容而不是将其保存到 var 然后返回 var 的输入更少。

现在，为什么新插入的数据不显示？可能是因为在您调用$this->m_register->InsertData(); 之后，您调用了$this->load->view('dashboard');，但没有提供任何数据供“仪表板”视图显示。

我认为您真正想要做的是redirect 到显示仪表板的控制器/方法。假设控制器是Dashboard，方法是index()，do_insert() 的最后一部分应该是这样的。

if($this->form_validation->run() === FALSE)
{
    echo "gagal diinsert";
    // $this->load->view('templates/header', $data);
    // $this->load->view('news/comment_form');
    // $this->load->view('templates/footer');
}
else
{
    if($this->m_register->InsertData())
    {
        redirect('dashboard'); 
    }
    else
    {
        //show an error page or error message about the failed insert
    }
}

Answer 3

您不需要添加模型。您可以像这样使用简单的方法：

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

class C_register extends CI_Controller {

    function __construct(){
        parent::__construct();
        $this->load->model('m_register');
    }


    function do_insert(){
        $this->load->helper('form');
        $this->load->library('form_validation');

        $this->form_validation->set_rules('firstname', 'firstname', 'required');
        $this->form_validation->set_rules('lastname', 'lastname', 'required');
        $this->form_validation->set_rules('fullname', 'fullname', 'required');
        $this->form_validation->set_rules('username', 'username', 'required');
        $this->form_validation->set_rules('email', 'email', 'required');
        $this->form_validation->set_rules('pass', 'pass', 'required');
        $this->form_validation->set_rules('gender', 'gender', 'required');
        $this->form_validation->set_rules('job', 'job', 'required');

        if ($this->form_validation->run() === FALSE) {
            echo "gagal diinsert";
        } else {
            //check in the array that it is same as column name in table like 'firstname' will be present on the table        
            $datac = array(
                'firstname' => $this->input->post('firstname'),
                'lastname' => $this->input->post('lastname'),
                'fullname' => $this->input->post('fullname'),
                'username' => $this->input->post('username'),
                'email' => $this->input->post('email'),
                'pass' => $this->input->post('pass'),
                'gender' => $this->input->post('gender'),
                'job' => $this->input->post('job')
            );

            $this->db->insert('member', $datac);
            $insert_id = $this->db->insert_id();
            if(!empty($insert_id)){
                $this->load->view('dashboard');
            }else{
                echo "failed";           
            }

        }
    }
}?>