将编码器$data['foo']从控制器传递到视图 - Pass codeigniter $data['foo'] from controller to view

Pass codeigniter $data['foo'] from controller to view

本文关键字：控制器视图编码器 foo data | 更新日期: 2023-09-27

我有这样的代码

 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');
    }

，我用它来删除一些数据从我的表，但我现在需要从控制器传递数据到视图，但我的控制器有一个参数。

不知何故，变量$data['success_or_failure']没有被传递，因为我一直得到这个错误

遇到PHP错误
严重性:注意
消息:未定义变量:success_or_failure
文件名:视图/success_or_failure.php
行号:54

为什么不把$data['success_or_failure'] = 'That dataset already exists.Kindly go back and try again.';传递给视图?

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); }