如何在数据库中获取数据并在View（Codeigniter）中显示特定数据

I am a newbie to Codeigniter framework, I have created a simple app that will fetch data from database and then display with condition in Views. I'm trying to display specific data from database to my View, So far, I have no luck using the code below.

我是Codeigniter框架的新手,我创建了一个简单的应用程序,它将从数据库中获取数据,然后在视图中显示条件。我正在尝试将数据库中的特定数据显示到我的View中,到目前为止,我没有运气使用下面的代码。

Model (Datamodel.php)

function getData() {
    $this->db->select("name, value");
    $this->db->from('settings');
    $query = $this->db->get();
    return $query->result_array();
}

Controller (maincontroller.php)

function index()
{
    $data['metadata'] = $this->Datamodel->getData();
    $this->load->view('viewdata', $data);
}

View

  <h3>Site Information</h3>
  <?php foreach($metadata as $data) {?>
      <h4>Site Title: <?php echo $data['site_title']; ?></h4>
      <h4>Site Description: <?php echo $data['site_description']; ?></h4>
      <h4>Site keyword: <?php echo $data['site_keyword']; ?></h4>
      <h4>Site Year: <?php echo $data['site_year']; ?></h4>
  <?php }?>

I want to display something like these on browser:

我想在浏览器上显示类似的内容:

Site Information

Site Title: My first codeigniter app
Site Description: This is sample ci application
Site keyword: simple ci app, first app
Site Year: 2015

网站标题:我的第一个codeigniter app网站描述:这是示例ci应用程序网站关键字:simple ci app,first app Site年份:2015

{ Any help & suggestion is accepted. thanks! pls see this image [ http://prntscr.com/722hsm ] for my database structure !}

{接受任何帮助和建议。谢谢!请参阅此图像[http://prntscr.com/722hsm]了解我的数据库结构!}

2 个解决方案

#1

Appreciate that you are taking initiative.

感谢您采取主动。

Just some modifications:

只是一些修改:

Model

<?php
function getData() {
  $siteData = array();
  $this->db->select("name, value");
  $this->db->from('settings');
  $query = $this->db->get();
  if ($query->num_rows()) {
    foreach ($query->result_array() as $row) {
    // Your data is coming from multiple rows, so need to loop on it.
      $siteData[$row['name']] = $row['value'];
    }
  }
  return $siteData;
}

View:

<h3>Site Information</h3>
<?php foreach($metadata as $dataName => $dataValue) {?>
    <!-- You are getting the names and values in loop, so, 
use it in loop. Also, use ucwords() and str_replace() to beautify your
 output. for more informtion on this, read php.net documentation.
       -->
  <h4><?php echo ucwords(str_replace('-', ' ', $dataName))?>: <?php echo $dataValue; ?></h4>
<?php }?>

SECOND APPROACH (WITHOUT FOREACH):

第二种方法(没有前提):

<h3>Site Information</h3>
<h4>Site Title: <?php echo $metadata['site_title']; ?></h4>
<h4>Site Description: <?php echo $metadata['site_description']; ?></h4>
<h4>Site keyword: <?php echo $metadata['site_keyword']; ?></h4>
<h4>Site Year: <?php echo $metadata['site_year']; ?></h4>

#2

Just load your model in controller:

只需在控制器中加载模型:

function index()
{
    $this->load->model('Datamodel');
    $data['metadata'] = $this->Datamodel->getData();
    $this->load->view('viewdata', $data);
}

try this on your views

试试看你的观点吧

<h3>Site Information</h3>

      <h4>Site Title: <?php echo $metadata['site_title']; ?></h4>
      <h4>Site Description: <?php echo $metadata['site_description']; ?></h4>
      <h4>Site keyword: <?php echo $metadata['site_keyword']; ?></h4>
      <h4>Site Year: <?php echo $metadata['site_year']; ?></h4>

#1