I create a library and function show
, outputs (results) the Library sent for control,
but there's in view page following error. What is in your opinion problem?
i put outputs to Controller as return $info; return $results; return $offset;
and they of Controller echo
in view as: $data['num_count'] = $info; $data['results'] = $results; $data['offset'] = $offset;
error:
A PHP Error was encountered
Severity: Notice Message: Undefined variable: info Filename: admin/a开发者_如何转开发ccommodation.php Line Number: 29A PHP Error was encountered
Severity: Notice Message: Undefined variable: results Filename: admin/accommodation.php Line Number: 30Fatal error: Call to a member function result() on a non-object in D:\xampp\htdocs\Siran-mehdi\system\core\Loader.php(679) : eval()'d code on line 46
When you call return, it quits the function at that point, and it doesn't return literally $info so you can keep using that name, but the data inside of that variable.
At the end of your library, change the three returns to something like:
return array('num_count' => $info, 'results' => $results, 'offset' => $offset);
This will return an associative array.
and in your controller:
$data = $this->siran->show($where, $table, $url_pag);
Where $data will become that array returned by the library.
Return it as array
return array('num_count' => $info, 'results' => $results, 'offset' => $offset);
And actually, in MVC pattern, that much like Model job instead library, since it related with your database abstraction. You create a Library for other task, which ussually is a common task you need.
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