how to get values from model and display in controller - codeigniter

Question

I am making a site in codeigniter, here is my controller, model and view:
controller:
function checking()
{
$email=$this->input->post('email');
$this->load->model('login_model');
$data['dbemail']=$this->login_model->email();// assign your value to CI variable
$this->load->view('home', $data); //passing your value to view
}
model:
public function email()
{
$query = $this->db->query("SELECT email FROM change_password");
$result = $query->result_array();
return $result;
}
view:
foreach ( $dbemail as $new_dbemail )
{
echo $new_dbemail['database_field'];//in here you can get your table header.
//Ex if your table has name field and you need to sho it you can use $new_dbemail['name']
}
but i want to get $new_dbemail['name'] in my controller, how i can get the value of $new_dbemail['name'] in controller rather than view ???

Answer 1

To do so... you have to fetch data into a row_array not in result_array in Model.
Controller
function checking()
{
$email=$this->input->post('email');
$this->load->model('login_model');
$data['dbemail']=$this->login_model->email();
$database_field = $data['dbemail']['database_field']; //to get result in controller
$date['database_field'] = $data['dbemail']['database_field']; //to get result in view
$this->load->view('home', $data);
}
Model
public function email()
{
$query = $this->db->query("SELECT email FROM change_password");
return $query->row_array();
}
View
<p> Database Field value : <?=$database_field;?></p>

Answer 2

You are Calling the Model In A Variable :-
$data['dbemail']=$this->login_model->email();
So in this you have the results which you are fetching from Database. Than you passing this variable data to your View.
$this->load->view('home', $data);
So you already have the value in controller which u fetched from model which is stored in $data.
You can View the Values of $data , using
echo "<pre>";
print_r($data);
and it will show what data you are receiving

Answer 3

Try like this...
In controller...
function checking()
{
$email=$this->input->post('email');
$this->load->model('login_model');
$dbemail=$this->login_model->email();//array containing all emails
//For each loop here for displaying emails
foreach($dbemail as $email){
$mail = $email; //assigning to variable
echo $mail."<br/>";//displaying email
}
$data['dbemail'] = $dbemail;//for passing to view
$this->load->view('home', $data); //passing your value to view
}

Answer 4

First you need to fetch the name from the database in the model. Change the query line:
$query = $this->db->query("SELECT name, email FROM change_password");
Then to use the name data in the controller you could try this:
function checking()
{
$email=$this->input->post('email');
$this->load->model('login_model');
$data['dbemail']=$this->login_model->email();
$name = $data['dbemail']['name']; // example usage to retrieve name in controller
$this->load->view('home', $data);
}
Now the name data is available as $name in controller.

Answer 5

Controller:
public function checking()
{
$email=$this->input->post('email');
$this->load->model('login_model');
$data['dbemails']=$this->login_model->email();
$this->load->view('home', $data); //passing your value to view
}
Model
public function email()
{
$this->db->select("email"); // the row of the table you want to select.
$this->db->get('change_password'); //table name. you can still use your query, its just much neat when you use a specific query builder.
$result = $this->result(); //giving you a result of type object
return $result;
}
View
foreach ( $dbemails as $dbemail )
{
echo $dbemail->email; // the '->' will be the identifier of which row you want to get or display, i assume its email beacause you specify it in the model.
}
Hope this what your looking for.