I am working on a projecj with codeigniter, and it's my first time using GroceryCRUD, and I need to make that a patient can have any number of reports, I have this on my database.
Where idpatients is the foreign key.
I am using this on my controller
public function reports()
{
$crud=$this->grocery_crud;
$crud->set_table('reports');
$crud->set_subject('Reports');
$crud->set_language('english');
$crud->set_relation('idpatients','patients','Patient');
$output=$crud->render();
$this->load->view('admin_reports', $output);
}
And I get this error
Any help would be appreciated to fix it, How can I declare the relationship so when I add a new report I can choose the patient from a dropdownbox or something?
UPDATE
When I change db_debug for false I get this error
Based on documentation:
void set_relation( string $field_name , string $related_table, string $related_title_field [, mixed $where [, string $order_by ] ] )
Set a relation 1-n database relation. This will automatically create a dropdown list to the fields and show the actual name of the field and not just a primary key to the list.
Which means, it will display the field from the table other then primary key (and you are trying to display field Patient
which is not present in patients
).
Solution for your issue - replace Patient
with Name
(or some other field which is present in patinets
table):
$crud->set_relation('idpatients','patients','Name');