Question

我正在尝试在我的codeigniter中包含if else语句。当用户搜索某个地点时，它可以从数据库中查看该地点的信息。我试图添加if else语句。如果评论超过5个用户将该地点评为＆＃34; Noisy＆＃34;，则系统会自动将其评为＆＃34;该地点为Noisy＆＃34;。（在撰写评论时，用户使用单选按钮输入噪声评论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();
}

我会把它包含在Controller中：

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

然后在视图中：

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>

希望这有帮助。

Codeigniter If else语句条件

1 个答案: