A PHP Error was encountered Severity: Notice Message: Undefined variable: company_n |
controller code:
Code: $vendor=$this->vendormodel->vendor_details($login_email); view code: Code: <div class="form-group"> error:: A PHP Error was encountered Severity: Notice Message: Undefined variable: company_name Filename: views/vendor_dashboard.php print_r($vendor); results to:: Array ( [0] => stdClass Object ( [company_name] => P.A ) ) plz help
controller
PHP Code: $vendor=$this->vendormodel->vendor_details($login_email); view PHP Code: <?php echo $vendor->company_name ?>
I'd suggest using row() at the end of your model query instead of result_array() because it's not returning a single object, it's returning an array of objects.
so, you've two options... 1: in your model: PHP Code: $query = $this->db->where('email', $login_email)->get('mytable')->limit(1)->row(); PHP Code: // Controller It's then converted to $company_name, $user_name, $IP_Address, etc in your view. ----- Option 2 (while this will work, it's ugly and will cause confusion later) Pass $vendor[0] in your view call. PHP Code: $this->load->view('vendor_dashboard',$vendor[0]); then access as $company_name, $user_name, etc in your view...
(07-03-2016, 06:32 PM)pdthinh Wrote: controller Not quite... PHP Code: // view file |
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