Codeigniter If else 语句条件

Question

我正在尝试在我的 codeigniter 中包含 if else 语句。当用户搜索一个地点时，它可以从数据库中查看该地点的信息。我正在尝试添加 if else 语句。如果评论超过 5 个用户将该地方评为“嘈杂”，则系统将自动将其评为“该地方嘈杂”。 （噪音评论由用户在编写评论时使用单选按钮输入并将其保存到数据库中）。下面是我正在处理的代码。

//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

如果我理解你的问题是正确的：

型号：

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();
}

我会将它包含在控制器中：

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

然后在视野中：

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>

希望这会有所帮助。