Problems Trying Insert Data in Database using $this->input->post() - Codeigniter - php

Question

Im trying use "$this->input->post();" of Codeigniter to do not need specify each field in my form. But im getting troubles when try insert into Database.
Look my controller:
public function cadastrar(){
$var = $this->input->post(null, TRUE);
$this->load->model('m_clientes');
$this->m_clientes->inserir($var);
}
My controller is simplified here because i know how to handle database in codeigniter.
The result of this post was:
Array ( [nome] => Raphael [sobrenome] => Schubert [cpf] => 893.528.432-89 [rg] => 4529875231908472 [telefone] => (53) 2980-5792 [celular] => (53) 9 2180-7529 [rua] => Israel de Almeida [numero] => 859 [cep] => 88.312-000 [bairro] => São Vicente [cidade] => ITAJAÍ [estado] => Santa Catarina [email] => rfswdp#gmail.com [tipo] => pf [cnpj] => 34.827.481/2834-78 [inscricaoestadual] => 34120489032814930128 [razaosocial] => Teste [nomefantasia] => Disney [dataaberturaempresa] => 10/21/15 [proprietario] => Marcos Aurelio )
I normaly use this way to insert:
$data = array(
'user_name' => $this->input->post('user_name',TRUE);
'user_phone' => $this->input->post('user_phone',TRUE);
'user_role' => $this->input->post('user_role',TRUE);
);
$this->name_of_model->inserir($data);
And works...
But i was trying to use just $this->input->post(); to get all fields from form. Because my actualy application will have hundreds of fields and i was trying to do not write each line.
So my model actually was:
public function inserir($var){
if($var!=NULL):
print_r($var);
$this->db->insert('tb_usuarios',$var);
endif;
}
But i`m getting and error saying:
Message: Undefined property: Clientes::$db
and
Message: Call to a member function insert() on null
My table name is: "tb_usuarios"
I changed all fields in database to accept NULL to see if i`m get some field name wrong... but not work...
Any tips??

Answer 1

There is no need to catch the POST var inside $var. You can see POST variable inside the model very well. So all you need to do in the controller is:
public function cadastrar(){
$this->load->model('m_clientes');
$this->m_clientes->inserir();
}
,and inside your model:
public function inserir(){
if( count($this->input->post()) > 0):
$this->db->insert('tb_usuarios',$this->input->post());
endif;
}
The fields names in your form must correspond to the column names inside your table.
The message Message: Call to a member function insert() on null means you forgot to load the database library, just like remiheens said.
But my advice is, to use form validation for your fields, so you may be sure all necessary fields are completed using the required data format for each one of them. Although this may require allot of coding, there is no other safe way from errors on database operations. You cannot trust the user to insert the data correctly, that's why you need form validation.

Answer 2

In here $var = $this->input->post(null, TRUE); you use null. null is not valid input name. name = ''
and this will works $this->input->post('user_name',TRUE);
cz of it has input tag name (name = 'user_name').
We use ,TRUE) next to input post field to allows XSS Protection
$var = $this->input->post(null, TRUE); Is Wrong
while you trying this, it shows
Message: Undefined property: Clientes::$db and Message: Call to a member function insert() on null.
will not work
public function cadastrar(){
$var = $this->input->post(null, TRUE);//will never give valid response
$this->load->model('m_clientes');
$this->m_clientes->inserir($var);
}
Works well
public function cadastrar(){
$this->load->model('m_clientes');
$data = array(
'user_name' => $this->input->post('user_name',TRUE);
'user_phone' => $this->input->post('user_phone',TRUE);
'user_role' => $this->input->post('user_role',TRUE);
);
$this->name_of_model->inserir($data);
$this->load->model('m_clientes');
}

Answer 3

This is because your database isn't loaded into codeigniter instance. ($this->db)
Just try to autoload "database" library (config/autoload.php) or load/connect your database in your model with :
$this->load->database();
Don't forget to edit your config/database.php ;)

Answer 4

I handle hundreds of fields everytime but I also validate basically each one. I always do this:
$customer['name'] = $this->input->post('customer_name');
$customer['age'] = $this->input->post('customer_age');
$customer['country'] = $this->input->post('customer_country');
$customer['city'] = $this->input->post('customer_city');
// validations
if(isAgeValid($customer['age']) == FALSE)
{
echo 'Hold on, your age...hmm check it out!';
return;
}
$this->customers_model->add($customer);
The function that handles the insertion only has this:
public function add($data)
{
$data = $this->security->xss_clean($data);
$this->db->insert('customers', $data);
return $this->db->insert_id();
}
Pretty clean and simple. Now, if you don't want to validate the fields or just want to validate some of them and want to insert the others without validation this is what I purpose based on the previous code:
// validations..
if(isAgeValid());
$customer = array();
foreach($_POST as $key => $value)
$customer[$key] = $value;
$this->customers_model->add($customer);

Answer 5

$data = array(
'user_name' => $this->input->post('user_name');
'user_phone' => $this->input->post('user_phone');
'user_role' => $this->input->post('user_role');
);
$this->load->model('name_of_model');
$this->name_of_model->inserir($data);
Model:
public function inserir($data)
{
$this->db->insert('***', $data);
if ($this->db->affected_rows() > 0) {
return true;
}
}