dimanche 26 juillet 2020

CodeIgniter - displaying view based on variable from database

I've been learning codeigniter recently and trying to push an application out. im running into problems with the if else statements displaying views. i am trying to pull in_party from the databaseci_admin_info and display views depending if its set to 1 or 0. its bypassing any condition and just displaying the first view set.

Model

function get_in_par(){
        
        $this->db->select('in_party');
        $this->db->from('ci_admin_info');
        $query = $this->db->get();
        $ret = $query->row();
        return $ret->in_party;
        
            
    }

Controller

public function index(){

        $data['title'] = 'Party';
        $data['party'] = $this->party_model->get_party();
        $data['prov'] = $this->party_model->get_prov_name();
        $data['info'] = $this->party_model->get_all();
        $data['res'] = $this->party_model->get_party_res();
        $data['rank'] = $this->party_model->get_pa_rank();


        $this->load->view('admin/includes/_header', $data);
        
        
        
        $inpar = $this->party_model->get_in_par();
        
        if(isset($inpar['in_party'])==0){
            
        $this->load->view('admin/party/join');
            
        } else {
            
        $this->load->view('admin/party/index'); 
        
        }

        $this->load->view('admin/includes/_footer');

    }

I've tried rearranging it into the parent construct as well and it just bypassed it completely. just trying to make it load the join view if in_party = 0 . i've tried without isset bypasses it as well. i've tried switch but don't think it registered that as well. i think i'm missing something in my model

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