Codeigniter If else語句條件

Question

我正在嘗試在我的codeigniter中包含if else語句。 用戶搜索地點時，可以從數據庫中查看該地點的信息。 我正在嘗試添加if else語句。 如果評論多於5位用戶，則將該地點評價為“嘈雜”，然后系統會自動將其評價為“地點嘈雜”。 （噪聲檢查是在編寫檢查並將其保存到數據庫時由用戶使用單選按鈕輸入的）。 下面是我正在處理的代碼。

//view.php

<style>
#searchbutton{
position: absolute;
left:300px;
top:30px;
}
fieldset {
background-color:#EFEAEA;
margin: 0px 0px 10px 0px;
 padding: 20px;
border-radius: 1px;
width:900px;
margin-left:220px;
margin-top:-10px;
}
#user{
font-style:italic;
font-size: 12px;
text-align:right;
}
#titlereview {
font-style: italic;
font-size:20px;
}
#review {
font-size:16px;
}
</style>

<?=form_open_multipart('viewreview/view');?>

<?php $search = array('name'=>'search',);?>
<?php $noise = array('name'=>'noise',);?>

<div id = "searchbutton">
<?=form_input($search);?><input type=submit value="Search" /></p>
</div>
<?=form_close();?>

<div class = "tablestyle">

<fieldset>
<?php foreach ($query as $row): ?>

<div id = "user">User: <?php echo $row->name; ?><br>
Visited time: <?php echo $row->visitedtime; ?><br>
</div>

 <div id = "titlereview">"<?php echo $row->titlereview; ?>"<br></div>

 <div id = "noise"><?php echo $row->noise; ?><br></div>

 <div id = "review"><?php echo $row->yourreview; ?><br><hr><br></div>

<?php endforeach; ?>
</fieldset>

<!--$noise is the field form database!-->
<?php if ($noise='yes'>5){
echo 'The place is Noisy';
}
else {
echo 'The place is Not Noisy';
}
?>
</div>

//控制器

<?php
class viewreview extends CI_Controller {

public function view($page = 'viewreview') //writereview page folder name
{
    $this->load->model('viewreview_model');
    $data['query'] = $this->viewreview_model->get_data();
    $this->load->vars($data);
    if ( ! file_exists('application/views/viewreview/'.$page.'.php')) //link
    {
        // Whoops, we don't have a page for that!
        show_404();
    }

    $data['title'] = 'View Review'; 
    //$data['title'] = ucfirst($page); // Capitalize the first letter
    $this->load->helper('html');
    $this->load->helper('url');
    $this->load->helper('form');
    $this->load->view('templates/header', $data);
    $this->load->view('viewreview/'.$page, $data);
    $this->load->view('templates/footer', $data);
 }
}
?>

//模型

<?php
class viewreview_model extends CI_Model {

public function __construct()
{
    $this->load->database();
}
public function get_data()
{
    $match = $this->input->post('search');
    $this->db->like('sitename',$match);
    $this->db->or_like('titlereview',$match);
    $this->db->or_like('yourreview',$match);
    $this->db->or_like('suggestion',$match);

    $query = $this->db->get('review');      //pass data to query
    return $query->result();

  }
}
?>

Answer 1

如果我已正確理解您的問題：

模型：

public function number_of_noise_report() {
      $this->db->select('id'); // Change it to what column name you have for id
      $this->db->from('table');
      $this->db->where('noise', 'Yes') // 'Yes' or 'yes', depending on what you have in db
      $query = $this->db->get();
      return $query->num_rows();
}

我將其包含在Controller中：

// Code before
$data['query'] = $this->viewreview_model->get_data();
$data['noise_stat'] = $this->viewreview_model->number_of_noise_report(); // ** UPDATED
// Code after

然后來看：

if($noise_stat > 5){
    echo '<div id = "noise">Noisy<br></div>';
} else {
    echo '<div id = "noise">Do Something Here<br></div>';
}

// OR SIMPLY
<div id = "noise">
<?php if($noise_stat > 5){ echo 'Noisy<br>';
} else { echo 'Somrthing<br>'; } ?>
</div>

希望這可以幫助。