從控制器傳遞codeigniter $ data ['foo']以查看

Question

我有這個代碼

 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');
    }

我正在使用它從表中刪除一些數據，但是現在我需要將數據從控制器傳遞到視圖，但是我的控制器有一個參數。

不知何故，變量$data['success_or_failure']沒有被傳遞，因為我不斷收到此錯誤

遇到PHP錯誤

嚴重程度：注意

消息：未定義的變量：success_or_failure

文件名：views / success_or_failure.php

行號：54

'為什么$data['success_or_failure'] = 'That dataset already exists.Kindly go back and try again.'; 被傳遞給視圖？

Answer 1

發現我沒有傳遞$data變量

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); }