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.
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>
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