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Php 消息:在将数据从cotroller传递到视图时尝试获取非对象的属性_Php_Codeigniter - Fatal编程技术网

Php 消息:在将数据从cotroller传递到视图时尝试获取非对象的属性

Php 消息:在将数据从cotroller传递到视图时尝试获取非对象的属性,php,codeigniter,Php,Codeigniter,我将模型中的函数编写为: public function select() { $this->db->select('*'); $this->db->from('students'); $query = $this->db->get(); return $query->result(); } 和控制器为: public function edit() { $data['row']=$this->st

我将模型中的函数编写为:

public function select()
{
    $this->db->select('*');    
    $this->db->from('students');
    $query = $this->db->get();
    return $query->result();
}
和控制器为:

public function edit()
{
    $data['row']=$this->student_model->select();
    //print_r($data['row']);
    $this->load->view('edit.php',$data);
}
它可以在控制器功能中打印数据..但在进入查看页面时,我得到的错误消息如下:

Message: Trying to get property of non-object

对于每个字段,都需要像这样在编辑页面中回显数据

echo $row[0]->field_name;

你的视图代码在哪里???