What's the difference between if(!Variable) and if(isset($variable))?

if(!Variable)和if(isset($ variable))之间有什么区别? Well, the answer is pretty simple. isset( $var ) returns whether or not a variable exists and is not null, where !$var tells you if that variable is true, or anything that evaluates to true (such as a string). Also, using !$var will output a notice that you're using an undefined variable, whereas isset won't do that. Mind yo

if(!Variable)和if(isset($ variable))之间有什么区别?

if(!Variable)和if(isset($ variable))之间有什么区别? 那么,答案很简单。 isset($ var)返回一个变量是否存在并且不是null,其中!$ var告诉你该变量是否为真或者任何计算结果为true的变量(如字符串)。 另外,使用!$ var会输出一个通知,说明您使用的是未定义的变量,而isset不会这样做。 请注意:他们是两码事: <?php var_dump( isset( $foo ) ); // false. var_dump( !$foo ); // true, but with a war

Doing HTTP requests FROM Laravel to an external API

What I want is get an object from an API with a HTTP (eg, jQuery's AJAX) request to an external api. How do I start? I did research on Mr Google but I can't find anything helping. Im starting to wonder is this is even possible? In this post Laravel 4 make post request from controller to external url with data it looks like it can be done. But there's no example nor any source wh

从Laravel执行HTTP请求到外部API

我想要的是从具有HTTP(例如,jQuery的AJAX)请求的API获取对象到外部API。 我如何开始? 我对Google先生进行了研究,但找不到任何帮助。 我开始怀疑这是甚至可能的吗? 在这篇文章中,Laravel 4用控制器向外部网址发送请求,看起来好像可以完成。 但是没有例子,也没有任何来源可以找到一些文档。 请帮助我? 根据类似问题的答案在这里:https://stackoverflow.com/a/22695523/1412268 看看Guzzle $client = new Gu

What does sprintf stand for?

I tried looking it up on google and wikipedia but couldn't find an answer... Does anyone know what 'sprintf' or 'printf' stands for? Is it an abbreviation for something??? Thanks String PRINT Format(ed). Ie print to string using a given format. The various members of the printf family, derived from C where they first appeared (though they hark back to the olden days o

sprintf代表什么?

我试图在谷歌和维基百科上查找它,但无法找到答案......有谁知道'sprintf'或'printf'代表什么? 它是什么的缩写? 谢谢 字符串打印格式(编辑)。 即打印到字符串使用给定的格式。 printf家族的各个成员都是从他们第一次出现的C开始的(虽然他们回顾了BCPL在这个特定谱系中的writef调用),包括: printf - print formatted (to standard output). fprintf - file printf (to a file handle). sp

PHP Elseif always returning false, multiple conditions

I'm trying to make an helper php-function that's going to return true/false if the user got right access level. The access level is set when the user logs in. The problem is that the function always return false. The function is located in a "class" php-file that's included(with include_once) on the page I want to use it. I'm kinda new to php, but the if conditions s

PHP Elseif总是返回错误,多个条件

我试图做一个帮助器php函数,如果用户访问权限正确,它将返回true / false。 访问级别在用户登录时设置。问题是该函数始终返回false。 该函数位于我要使用它的页面中的“class”php文件中(包含include_once)。 我有点新的PHP,但如果条件似乎是正确的。 我已经测试过以管理员和“经济”身份登录,但它并没有返回true。 我也尝试回显作为参数发送的值,甚至在elseif语句之前检查访问级别是否正确(通过回显值)。 const

admin role only via Auth

I had implemented entrust for roles and permissions. I have 3 Roles, super-admin, admin and customer. Super Admin has access to Web-app (eg. www.myurl.com) Admin has access through api only ie mobile app (eg. www.myurl.com/api/login) via api.php route customer had access through api ie mobile app Now, I found a bug that when admin tries to login via www.myurl.com.login with his credentia

管理员角色只能通过身份验证

我已经实施了角色和权限entrust 。 我有3个角色,超级管理员,管理员和客户。 超级管理员可以访问Web-app(例如www.myurl.com) 管理员只能通过api访问,即通过api.php路由访问移动应用程序(例如www.myurl.com/api/login) 客户通过api即移动应用访问 现在,我发现一个错误,当管理员试图通过www.myurl.com.login用他的凭据登录时,他被允许登录! 在进一步调查中,我发现我需要更改login方法并在login提供角色检查

When should I return?

I am struggling to create an access object to sections stored in the Database. This is a skellington of the process, this contains static data until I can get the principle working. class User { const IS_ADMIN = 1; const IS_MODERATOR = 2; const IS_MEMBER = 4; } This class will auto-load data from the database eventually but for the time being, this class has default values. class

我应该什么时候回来?

我正在努力为存储在数据库中的部分创建一个访问对象。 这是一个过程的skellington,它包含静态数据,直到我能够得到原理的工作。 class User { const IS_ADMIN = 1; const IS_MODERATOR = 2; const IS_MEMBER = 4; } 这个类最终会自动从数据库中加载数据,但是目前这个类有默认值。 class Scope { private $priv = []; public function __construct() { $this->priv = [1]; } publi

Do all php types return a bool?

I'm currently building a website and I come across the case where I can do: if (myString) if (myArray) , and it seems to "return" true whenever there is data inside the variable. At least, that's what I think. Eg $testVar = "test"; if ($testVar) echo $testVar; else echo "Empty"; When i assert $testVar = "" , then it echos "Empty". I'm

所有的PHP类型都返回一个布尔值吗?

我目前正在建立一个网站,我遇到了我可以做的事情: if (myString) if (myArray) , 并且当变量内部有数据时它似乎“返回”真。 至少,这是我的想法。 例如 $testVar = "test"; if ($testVar) echo $testVar; else echo "Empty"; 当我断言$testVar = "" ,那么它回声“空”。 我想知道这是否是PHP的一个定义的特性,如果它不为null或空,任何类型都会返回true,就像在其他语言中需要做的一样if($test

PHP MySQL query returning NULL

I have a MySQL table that looks like this: index | tag | posts ------------------------- 1 | cats | 9,10 2 | a cat | 9,10 3 | kitty | 9,10 4 | meow | 9,10 I am trying to just return the row that matches a search query. I passed the search parameter using a simple ?search=cats . This is the PHP that I'm using: $search = $_GET['search']; $query = mysql_qu

PHP MySQL查询返回NULL

我有一个如下所示的MySQL表: index | tag | posts ------------------------- 1 | cats | 9,10 2 | a cat | 9,10 3 | kitty | 9,10 4 | meow | 9,10 我试图只返回匹配搜索查询的行。 我使用简单的?search=cats传递了搜索参数。 这是我正在使用的PHP: $search = $_GET['search']; $query = mysql_query("SELECT * FROM tags WHERE tag = '$search'"); echo(mysql_num_rows($query)); $r

Set variable in if statement expression

I ran across some interesting code today. I tried to find out if this is a feature of PHP or if I am missing something, but was unable to find anything on Google. Probably because I don't know the name of it. Code if($logo = RepositoriesLogo::getLogoData($id)){ $logo_href = $logo->link; } The variable $logo is not being set anywhere else. It seems like the expression in this if

在if语句表达式中设置变量

今天我遇到了一些有趣的代码。 我试图找出这是PHP的一项功能,或者我错过了一些东西,但无法在Google上找到任何东西。 可能是因为我不知道它的名字。 码 if($logo = RepositoriesLogo::getLogoData($id)){ $logo_href = $logo->link; } 变量$logo并未在其他地方设置。 看起来这个if语句中的表达式正在检查该类方法是否正在返回任何内容,并同时设置要在语句中使用的变量$logo 。 这是真的? 如果是这样,那么

php variables won't save properly

I'm posting to a php script, and I want to update some already existing variables in the script. My javascript is this : $('.submit_html').click( function(e) { e.preventDefault(); var requested = $('.page_select').val(); var text = new Array($('.edit_html').val(), $('.header_val').val()); $.post('data_handlers/get_content.php', {page: requ

PHP变量将无法正常保存

我张贴到一个PHP脚本,我想更新脚本中的一些已经存在的变量。 我的JavaScript是这样的: $('.submit_html').click( function(e) { e.preventDefault(); var requested = $('.page_select').val(); var text = new Array($('.edit_html').val(), $('.header_val').val()); $.post('data_handlers/get_content.php', {page: requested, action: 'update', text: text}, func