Codeigniter If else語句條件

Question

我想包含if else語句在我的codeigniter中。當用戶搜索一個地方時，它可以從數據庫中查看該地點的信息。我試圖添加if else語句。如果評論超過5位用戶評價爲「嘈雜」，那麼系統會自動評定爲「該地方很吵」。 （噪音評論是由用戶在寫評論時使用單選按鈕輸入並保存到數據庫中）。以下是我工作的代碼。

//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

如果我明白你的問題正確：

型號：

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(); 
} 

而且我會包括在控制器：

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

然後在視圖：

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> 

希望這有助於。