代码之家 › 专栏 › 技术社区 › HemalHerath

如何从两个函数向同一视图传递数据?

codeigniter php

0

HemalHerath · 技术社区 · 8 年前

我的函数如下

public function view($slug = NULL){

        $data['book'] = $this->Book_Model->get_books($slug);
        $data['categories'] = $this->Category_Model->get_categories();

        //current book view
        $book_id = $data['book']['book_id'];
        echo "current book-".$data['book']['book_id']." ";

        if(empty($data['book'])) {
            show_404();
        }

        $this->add_views($book_id);
        $this->get_viewed_books($book_id);

        $this->load->view('templates/header', $data);
        $this->load->view('books/view', $data);
        $this->load->view('templates/footer');


    }

查看\u书籍 功能如下

public function get_viewed_books($book_id) {

$data['results'] = $this->Book_Model->get_viewed_books($book_id);

}

$data['results'] 到 $this->load->view('books/view', $data);

1 回复 | 直到 8 年前

1

u_mulder 8 年前

分配结果 get_viewed_books 一些钥匙 $data

$this->add_views($book_id);
$data['viewed_books'] = $this->get_viewed_books($book_id); // here

$this->load->view('templates/header', $data);
$this->load->view('books/view', $data);
$this->load->view('templates/footer');

现在,在模板中你有一个 viewed_books

当然还有你的 get_viewed_books($book_id) 应该 返回结果 ,而不是分配给某个未知的 $数据

public function get_viewed_books($book_id) {
    return $this->Book_Model->get_viewed_books($book_id);
}