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| Controller validation works, Config/Validation - Syntax error |
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Posted by: BilltheCat - 05-19-2020, 08:59 AM - Forum: CodeIgniter 4 Support
- Replies (3)
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Maybe this will be an obvious question, but I can't see what's wrong with it.
If I use this validation rule in my Controller, it works fine. Doing basically the same thing in Config/Validation, and my IDE gives me a syntax error message
Controller:
PHP Code: // validate form input
$this->validation->setRules([
'identity' => [
'label' => str_replace(':', '', lang('Auth.login_identity_label')),
'rules' => 'required'
],
'password' => [
'label' => str_replace(':', '', lang('Auth.login_password_label')),
'rules' => 'required'
]
]);
Config/Validation:
PHP Code: public $login = [
'identity' => [
'label' => str_replace(':', '', lang('Auth.login_identity_label')),
'rules' => 'required'
],
'password' => [
'label' => str_replace(':', '', lang('Auth.login_password_label')),
'rules' => 'required'
]
];
Syntax error:
![[Image: kGeO1B.jpg]](https://snipboard.io/kGeO1B.jpg)
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| curl lib response getBody return binary string |
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Posted by: sandip002 - 05-19-2020, 05:52 AM - Forum: CodeIgniter 4 Discussion
- Replies (1)
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Today I'm trying to fetch some testing data from "https://jsonplaceholder.typicode.com/todos/1",
but when I get a response and try to get data it will give binary string.
Code: $this->client = \Config\Services::curlrequest([
'base_uri' => 'https://jsonplaceholder.typicode.com'
]);
$data = $this->client->get('/todos/1');
d( $data->getBody() ); // also try getJSON()
output:
Code: ���RPP*-N-�LQ�R0�q3̒̒�T%+��Ԝ���b���N�U+H��-�I-IiIK�)N�7��S
is anything I'm doing wrong.
EDIT:
I'm trying to get data in json or array format from api.
ALREADY TRIED THINGS:
Code: $data = $this->client->get('/todos/1');
dd( $data->getBody() ); // give binary string,
dd( json_decode( $data ) ); // error: json_decode() expects parameter 1 to be string, object given
dd( $data->getJSON() ); // error: Failed to parse json string, error: "Malformed UTF-8 characters, possibly incorrectly encoded".
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| prepared statements |
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Posted by: joseCarlos - 05-19-2020, 03:57 AM - Forum: CodeIgniter 4 Support
- Replies (4)
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Hi again.
So I was trying to improve safety in my project by using prepared statements, and it just doesn't seems to work.
would you be so kind as to take a look and see if you find anything out of place?
this is the model function
PHP Code: public static function insert_contactos($data_array)
{
$db = db_connect();
$pQuery = $db->prepare(function ($db) {
return $db->table('contactos')
->insert([
'ativo_contacto' => '',
'nome_contacto' => '',
'cod_cliente_contacto' => ''
]);
});
$result = $pQuery->execute(
$data_array['ativo_contacto'],
$data_array['nome_contacto'],
$data_array['cod_cliente_contacto']
);
return $result;
}
This is the controller function
PHP Code: public function insert()
{
if ($this->request->isAJAX()) {
$data_array['ativo_contacto'] = $_POST['ativo_contacto'];
$data_array['nome_contacto'] = $_POST['nome_contacto'];
$data_array['cod_cliente_contacto'] = $_POST['cod_cliente_contacto'];
if (Contactos_model::insert_contactos($data_array)) {
echo 'success';
} else {
echo 'fail';
}
}
}
after running in debug mode, i get this exception from
Exception has occurred.
ErrorException: Uninitialized string offset: 1
"C:\xampp\htdocs\pgp4\system\Database\Query.php"
As always, thank you for your time.
Messed with it again by trying a manually built statement, but still hangs on that exception, the thing is it is inserting the data correctly, but it stops afterwards.
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| [SOLVED] Model Update Fails |
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Posted by: 68thorby68 - 05-19-2020, 03:51 AM - Forum: CodeIgniter 4 Support
- No Replies
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Hi,
I'm trying to update 1 record in my table. But $this->update($id, $data) just fails.
I've tried every which way but ... well my frustration ...
My db (has aprox 50 records and player_id 12 does exist)
TABLE `player_master` (
`player_id` int(10) unsigned NOT NULL,
`first_name` varchar(50) NOT NULL,
`last_name` varchar(50) NOT NULL,
`image_path` varchar(200) NOT NULL,
`telephone` varchar(15) DEFAULT NULL,
`email` varchar(50) DEFAULT NULL,
`handicap` int(3) NOT NULL,
PRIMARY KEY (`player_id`) USING BTREE
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
My model
PHP Code: namespace App\Models\Admin;
use CodeIgniter\Model;
class Players extends Model {
protected $table = 'player_master';
protected $primaryKey = 'player_id';
protected $returnType = 'array';
function player ($id) {
return $this->where('player_id', $id)
->find();
}
function player_update($id, $data) {
return $this->update($id, $data);
}
}
PHP Code: $players_func = new \App\Models\Admin\Players;
$this->$data['player_name'] = $players_func->player($id);
//Player Data returned OK
$id=12;
$data=[
'image_path'=>'path_to_my_image'
];
$this-$data['update']= $players_func->player_update($id, $data);
//This just fails - no error, just fails
Note: the function player_update does not receive the var $id and $data.
Please help I am at a total loss ....
OK, so after thinking I'm going mad (more mad). I think I've sussed this out.
By defining allowed fields in the class, the update seems to have worked OK.
My model now reads
PHP Code: namespace App\Models\Admin;
use CodeIgniter\Model;
class Players extends Model {
protected $table = 'player_master';
protected $primaryKey = 'player_id';
protected $returnType = 'array';
protected $allowedFields = ['image_path'];
function player ($id) {
return $this->where('player_id', $id)
->find();
}
}
ie.
PHP Code: $players_func = new \App\Models\Admin\Players;
$this->$data['player_name'] = $players_func->player($id);
//Player Data returned OK
$id=12;
$data=[
'image_path'=>'path_to_my_image'
];
$this-$data['update'] = $players_func->update($id, $data);
Hope this help others
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| SQL injection question with codeigniter |
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Posted by: ashilnayak2 - 05-18-2020, 09:51 PM - Forum: General Help
- Replies (1)
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I'm working on a project using codeigniter where I've got a form that writes to a database.
I'm using CI's Active Record Class with form validation rules for the input and escaping the inputs as well.
The db driver being used is mysql. Am I doing enough to prevent sql injection attacks or missing the point on how to keep this site safe?
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| Problems in views when reinstall appserv |
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Posted by: Edel - 05-18-2020, 05:58 PM - Forum: CodeIgniter 4 Support
- Replies (2)
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Hello everyone. I am developing in CI 4. I changed the appserv version recently.
I saved my project from the old www folder and copied it into the new www in my new installation. but when I send variables to the view it doesn't recognize them.
PHP Code: $result->adverts = $adverts; //ad list
$dat['result'] = $result;
$view = \ Config\Services::renderer();
$view->setData($dat);
$data['previews'] = $view->render('tools/previews');
In my view : APPPATH/Views\tools\previews.php
Code: <?
print_r($result); //line 2
print_r($result->adverts);//line 3
foreach($result->adverts as $advert){
$time = Time::parse($advert->date);
?>
<div class=" preview col-lg-2 col-md-2 col-sm-3 col-xs-12">
<div class="box">
<a href="<?=$advert->getHref(NULL, $m_category)?>"> //line 9
the error that return is "ErrorException, Undefined variable: advert" in APPPATH/Views\tools\previews.php at line 9
the lines 2 and 3 don't print anything.
why if lines two an three does not print anything code is executed inside the foreach
I have not changed anything within my project. I suppose the problem is related to some php or apache configuration.
php 7.3.10
Thanks in advance.
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| Where to place Services? |
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Posted by: Zark - 05-18-2020, 01:59 PM - Forum: CodeIgniter 4 Support
- Replies (3)
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I am wondering what's the best practice or convention for a location wherein to store custom 'Services'?
E.g. Authentication service.
If I may suggest a change to docs, adding in this info under Services would be helpful.
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| Change output in post_system Event |
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Posted by: mickrus - 05-18-2020, 04:33 AM - Forum: CodeIgniter 4 Support
- Replies (1)
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In CI3 i used display_override - hook like this:
PHP Code: function placeholders()
{
$CI =&get_instance();
$buffer = $CI->output->get_output();
$buffer = str_replace(array('{field1}','{field2}'), array('data1','data2'), $buffer);
$CI->output->set_output($buffer);
$CI->output->_display();
}
?>
How can I access from CI4 Event's (post_system) controller to output?
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