方法Post和Get In CodeIgniter

Question

I am using CodeIgniter.

if(isset($query1))
{
    foreach($query1 as $row)
    {
        echo '<tr>';
        echo '<td><a href="'.base_url().'site/companyDetail">'.$row->companyName.'</a></td>';
        echo '<td>'.$row->address.'</td>';
        echo '<td>'.$row->contactPerson.'</td>';
        echo '<td>'.$row->contactnum.'</td>';
        echo '</tr>';
    }
}

I want to pass $row->companyName in the URL to be a part of site/companyDetail?name=CompanyName where companyDetail is a file. The values are from a SQL database. I want to load companyDetail of CompanyName. How do I do it? Thanks.

Answer 1

1) Pass $row->companyName as parameter,

if(isset($query1))
{
    foreach($query1 as $row)
    {
        echo '<tr>';
        echo '<td><a href="'.base_url().'site/companyDetail/"'.$row->companyName.'>'.$row->companyName.'</a></td>';
        echo '<td>'.$row->address.'</td>';
        echo '<td>'.$row->contactPerson.'</td>';
        echo '<td>'.$row->contactnum.'</td>';
        echo '</tr>';
    }
}

2) After clicking on the link it'll reach your Controller (applications/controllers/site.php),

class Site extends CI_Controller
{
    public function __construct()
    {
        parent::__construct();
    }

    public function companyDetail($companyName)
    {
       // Uncomment below to check whether you are getting company name
       // echo $companyName; exit;

       $data['company'] = $this->abc_model->get_company_details($companyName);

      // Uncomment below to check the data
      // echo '<pre>'; print_r($data); exit;

       $this->load->view('views/company_detail.php', $data);
    }
}

3) Your view ("applications/views/company_detail.php")

<table>
<?php foreach($company as $c) { ?>
   <tr>
       <td><?php echo $c['name']; ?></td>
       <td><?php echo $c['founder']; ?></td>
       <td><?php echo $c['assets']; ?></td>
   </tr>
<?php } ?>
</table>