3
function create()
{
$data['headline'] =" ";
$this->load->module('site_security');
$this->site_security->_make_sure_is_admin();
$update_id = $this->uri->segment(3);
if(!is_numeric($update_id))
{
$data['headline'] = "Add New Product";
//$data['headline'] = print_r($update_id)."if";
}
else
{
$data['headine'] = "Update Product Details";
//$data['headline'] = print_r($update_id)."else";
}
$data['view_module'] = "store_products"; //passing module name
$data['view_file'] = "create"; //passing method name
$this->load->module('templates'); // loads the template module
$this->templates->admin($data); //calls the template controller method admin which loads the admin view
}
我正在传递标题文本,以根据url中第三个第三段的内容在视图中更改标题。第一个if语句完美地设置标题,但它不会进入else语句,并给我错误,因为$ headline在视图中未定义。奇怪的是,如果我使用评论print_r语句一切正常,它会显示在视图中。未将PHP codeigniter变量传递给视图
我认为它必须是行$ data ['headine'] =“更新产品详细信息”;在别的,但我不知道为什么。