小编Bra*_*eed的帖子

谷歌在这种情况下是无用的,似乎我是第一个遇到这个错误:/在我的Mac上工作正常,但在Windows 8装备上使用相同的文件在尝试请求静态资产(如CSS)时会在日志中出现以下错误文件和图像.以下是错误的片段:

INFO     2014-06-08 14:42:28,431 module.py:639] default: "GET /css/rootStyles.css HTTP/1.1" 200 5454
ERROR    2014-06-08 14:42:28,431 module.py:714] Request to '/css/rootStyles.css' failed
Traceback (most recent call last):
  File "C:\Program Files (x86)\Google\google_appengine\google\appengine\tools\devappserver2\module.py", line 710, in _handle_request
    return handler.handle(match, environ, wrapped_start_response)
  File "C:\Program Files (x86)\Google\google_appengine\google\appengine\tools\devappserver2\static_files_handler.py", line 369, in handle
    return self._handle_path(full_path, environ, start_response)
  File "C:\Program Files (x86)\Google\google_appengine\google\appengine\tools\devappserver2\static_files_handler.py", line 182, in _handle_path
    start_response('200 OK', headers)
  File "C:\Program Files (x86)\Google\google_appengine\google\appengine\tools\devappserver2\module.py", line 640, in wrapped_start_response
    return start_response(status, response_headers, exc_info)
  File "C:\Program Files (x86)\Google\google_appengine\lib\cherrypy\cherrypy\wsgiserver\wsgiserver2.py", line 2155, in start_response …

我正在尝试检查模型的构造函数是否允许当前经过身份验证的用户访问给定的模型，但我发现构造函数上下文中的 $this 为空。Laravel 中分配给模型的属性在哪里？加载所有属性后我应该如何调用方法？

public function __construct(array $attributes = [])
{
    parent::__construct($attributes);
    var_dump($this); // empty model
    $this->checkAccessible();

}

提前欢呼

在我的控制器中,我有

    /**
 * @return array action filters
 */
public function filters()
{
    return array(
        'accessControl', // perform access control for CRUD operations
    );
}

/**
 * Specifies the access control rules.
 * This method is used by the 'accessControl' filter.
 * @return array access control rules
 */
public function accessRules()
{
    return array(
        array('allow',  // allow all users to perform 'index' and 'view' actions
            'actions'=>array('index','view'),
            'users'=>array('*'),
        ),
        array('allow', // allow players to comment on games
            'actions'=>array('createComment'),
            'roles'=>array('createComment'),
        ),
  array('allow', // …

我正在尝试将我在项目中使用的Pod转换为Swift 3.我没有写它,但是原作者没有更新它所以我把它分叉,我自己试图做.但...

尝试将扩展转换NSTimer为Swift 3时出现此错误: Cannot convert value of type '(Timer!) -> Void' to expected argument type '((CFRunLoopTimer?) -> Void)!

似乎Swift 3处理程序类型(Timer!) -> Void与旧的学校CFRunLoop样式处理程序不兼容,但我不确定如何在保持与iOS 9的兼容性的同时将其转换.

我正在粘贴下面的代码,由Xcode转换.您可以在https://github.com/entotsu/TKSubmitTransition/blob/master/SubmitTransition/Classes/NSTimerEx.swift找到原始代码.

干杯

import Foundation
extension Timer {
    class func schedule(delay delay: TimeInterval, handler: (Timer!) -> Void) -> NSTimer {
        let fireDate = delay + CFAbsoluteTimeGetCurrent()
        let timer = CFRunLoopTimerCreateWithHandler(kCFAllocatorDefault, fireDate, 0, 0, 0, handler) // Error on this line
        CFRunLoopAddTimer(CFRunLoopGetCurrent(), timer, kCFRunLoopCommonModes)
        return timer
    }

    class func schedule(repeatInterval …