小编Fra*_*ank的帖子

我想要实现的是一个简单的登录页面,如果用户登录成功,则重定向到主页面,否则保持登录页面.

我有1个名为的控制器login,和1个名为的模型main.当用户单击登录按钮时,将调用login/login_send.

<?php if ( ! defined('BASEPATH')) exit('No direct script access allowed');

class Login extends CI_Controller{

function __construct() {
    parent::__construct();
    $this->load->model('model_login');
}

function index()
{
    if ($this->model_login->is_logged_in())
    {
        redirect('main');
    }
    else
    {
        // load login page
        $data['main'] = 'view_login';
        $data['style'] = 'style_login';
        $this->load->view('template', $data);           
    }
}

function login_send()
{
    $this->form_validation->set_rules('username', 'Username', 'trim|required');
    $this->form_validation->set_rules('password', 'Password', 'trim|required'); 

    if ($this->form_validation->run() == FALSE) 
    {
        $this->index();
    } 
    else 
    {
        if ( $this->model_login->validate_user() )
        {
            $user_session_data = array(
                'username' => $this->input->post('username'),
                'is_logged_in' …

我在 iOS 14 中型小部件中显示 3 行，如下所示：

 row 1
-------
 row 2
-------
 row 3
-------

我的视图结构在这里：

VStack {
    ForEach(records, id: \.id) { item in 
        ZStack {
            // some views
        }
        .widgetURL(URL(string: "wig://\(item.id)"))
    }
    Divider()
}

似乎第一项和第二项的小部件 URL被第三项覆盖，所有深层链接都将打开第三项的内容。

为 ForEach 生成的视图添加深层链接的正确方法是什么？

我试图在drupal中实现hook_menu.

function menufun_menu() {
    $items['menufun'] = array(
        'title' => 'Menu Fun',
        'title callback' => 'menufun_title',
        'page callback' => 'menufun_greeting',
        'file' => 'menufun_greeting.inc',
        'page arguments' => array('aaa', 'bbb', 'ccc', 'ddd'),
        'access callback' => 'user_access',
        'access arguments' => array('receive greeting'),
        'type' => MENU_NORMAL_ITEM,
        'weight' => -1,
    );

    $items['menufun/farewell'] = array(
        'title' => 'Farewell',
        'page callback' => 'menufun_farewell',
        'file' => 'menufun_greeting.inc',
        'access callback' => 'user_access',
        'access agruments' => array('receive greeting'),
        'type' => MENU_NORMAL_ITEM,
    );

    return $items;
}

但是,上面的代码会出现这两个错误:

Notice: Undefined offset: 0 in …

我下载CodeIgniter 2.1.0,然后按照CodeIgniter 2.1.0上的教程进行操作.

问题是当我尝试在此URL中提交表单时:

http://localhost/CodeIgniter/index.php/news/create

该页面将重定向到此URL:

http://localhost/CodeIgniter/index.php/news/localhost/CodeIgniter/index.php/news/create

我厌倦了很多改变route.php的方法,但它不起作用.我怎样才能解决这个问题？？

route.php

$route['news/create'] = 'news/create';
$route['news/(:any)'] = 'news/view/$1';
$route['news'] = 'news';
$route['(:any)'] = 'pages/view/$1';
$route['default_controller'] = 'pages/view';

这是表单页面的代码:

public function create()
{
    $this->load->helper('form');
    $this->load->library('form_validation');

    $data['title'] = 'Create a news item';

    $this->form_validation->set_rules('title', 'Title', 'required');
    $this->form_validation->set_rules('text', 'text', 'required');

    if ($this->form_validation->run() === FALSE)
    {
        $this->load->view('templates/header', $data);   
        $this->load->view('news/create');
        $this->load->view('templates/footer');

    }
    else
    {
        $this->news_model->set_news();
        $this->load->view('news/success');
    }
}

我注意到当页面加载时的表单页面,下面的代码段将被执行,但是,else段从未得到执行的更改.

if ($this->form_validation->run() === FALSE)
    {
        $this->load->view('templates/header', $data);   
        $this->load->view('news/create');
        $this->load->view('templates/footer');

    }

这是表单页面,没什么特别的,只需调用上面显示的create函数即可. create.php

<h2>Create a news item</h2>

<?php echo …