从控制器传递codeigniter $ data ['foo']以查看

Question

I have this code

 public function delete($delete_candindate){
    $this->load->database();
    $this->load->dbforge();
    $delete_candindate = $this->uri->segment(3);
    $this->dbforge->drop_table($delete_candindate);
    $this->db->where('dataset_name', $delete_candindate);
    $this->db->delete('all_datasets'); 
    $data['success_or_failure'] = 'That dataset already exists.Kindly go back and try again.';
    $this->load->view('success_or_failure');
    }

which i am using to delete some data from my tables but i now need to pass data from the controller to the view but my controller has a parameter.

Somehow,the variable $data['success_or_failure'] is not being passed since i keep getting this error

A PHP Error was encountered

Severity: Notice

Message: Undefined variable: success_or_failure

Filename: views/success_or_failure.php

Line Number: 54

Why isn't $data['success_or_failure'] = 'That dataset already exists.Kindly go back and try again.'; being passed to the view?.

Answer 1

发现我没有传递$data变量

public function delete($delete_candindate){ $this->load->database(); $this->load->dbforge(); $delete_candindate = $this->uri->segment(3); $this->dbforge->drop_table($delete_candindate); $this->db->where('dataset_name', $delete_candindate); $this->db->delete('all_datasets'); $data['success_or_failure'] = 'That dataset already exists.Kindly go back and try again.'; $this->load->view('success_or_failure',$data); }