I'm defining an Usuario entity that extends from FOSUserBundle BaseUser and I'm do it as follow: namespace UsuarioBundleEntity; use FOSUserBundleModelUser as BaseUser; use DoctrineORMMapping as ORM; use GedmoMappingAnnotation as Gedmo; use GedmoTimestampableTraitsTimestampableEntity; use SymfonyBridgeDoctrineValidatorConstraintsUniqueEntity; use DoctrineCommonCollectionsArrayCollection;
我正在定义一个从FOSUserBundle BaseUser扩展的Usuario实体，我按照如下方式执行： namespace UsuarioBundleEntity; use FOSUserBundleModelUser as BaseUser; use DoctrineORMMapping as ORM; use GedmoMappingAnnotation as Gedmo; use GedmoTimestampableTraitsTimestampableEntity; use SymfonyBridgeDoctrineValidatorConstraintsUniqueEntity; use DoctrineCommonCollectionsArrayCollection; use MisdPhoneNumberBundleV
I have an Entity, where I want to use trait "TimestampableEntity" for mapping some properties: namespace WbudowiePortalBundleEntity; use DoctrineORMMapping as ORM; use GedmoMappingAnnotation as Gedmo; use WbudowiePortalBundleTraitsTimestampableEntity; /** * Category * * @GedmoTree(type="materializedPath") * @GedmoSoftDeleteable(fieldName="deletedAt", timeAware=false) * @ORMTabl
我有一个实体，我想在其中使用特征“TimestampableEntity”来映射一些属性： namespace WbudowiePortalBundleEntity; use DoctrineORMMapping as ORM; use GedmoMappingAnnotation as Gedmo; use WbudowiePortalBundleTraitsTimestampableEntity; /** * Category * * @GedmoTree(type="materializedPath") * @GedmoSoftDeleteable(fieldName="deletedAt", timeAware=false) * @ORMTable(name="categories") * @ORMEntity(
I'm using symfony 2.3.4 and in security configuration I want to to load Security Users from the Database accordingly to this Cookbook instruction. But I only get this error: Fatal error: Class MyAppAdminBundleEntityUsers contains 3 abstract methods and must therefore be declared abstract or implement the remaining methods (SymfonyComponentSecurityCoreUserUserInterface::getRoles, SymfonyCom
我正在使用symfony 2.3.4，并且在安全配置中，我希望从数据库中加载安全用户，因此符合本Cookbook指令。 但我只会得到这个错误： 致命错误：类MyApp AdminBundle Entity Users包含3个抽象方法，因此必须声明为抽象或实现其余方法（Symfony Component Security Core User UserInterface :: getRoles，Symfony Component Security Core User UserInterface :: getUsername，Symfony Component Security Core U
I'm working on a MVC application in which the Model is implemented using an abstract base class that all actual models have to extend. In every model there is some info about that model, currently implemented as an array, let call that protected static $info . So, every model has a different $info array. Now, the base class has lots of functions that use data from that array, and at the mo
我正在开发一个MVC应用程序，在该应用程序中，使用抽象基类来实现模型，所有实际模型都必须扩展。 在每个模型中都有关于该模型的一些信息，目前以数组的形式实现，让我们调用protected static $info 。 所以，每个模型都有一个不同的$info数组。 现在，基类有很多使用该数组的数据的函数，并且此刻每个函数都以类似下面的示例save（）函数开始。 abstract class BaseModel { function save(){ $className = get_c
I have my UserBundle who extends the FOSUserBundle and works fine. But now, I want create multiple users entities with différents properties. But the problem is when I create my user enttity who extends my main User entity like this : class User extends BaseUser { protected $id; // The main user class who extends FOSUser entity } class UserB extends User { // } When I do this, I&
我有我的UserBundle扩展了FOSUserBundle并且工作正常。 但是现在，我想创建具有différents属性的多个用户实体。 但问题是，当我创建我的用户Entity扩展我的主用户实体像这样： class User extends BaseUser { protected $id; // The main user class who extends FOSUser entity } class UserB extends User { // } 当我这样做时，我有一个错误：` MyApp UserBundle Entity UserB :: $ id的访问级别必须受
I want to know if there is a solution on how to unit-test a PHP trait. I know we can test a class which is using the trait. but I was wondering if there are better approaches. Thanks for any advice in advance :) EDIT One alternative is to use the Trait in the test class itself as I'm going to demonstrate bellow. But I`m not that keen on this approach since there is no guaranty ther
我想知道是否有关于如何单元测试PHP特征的解决方案。 我知道我们可以测试一个正在使用这个特性的类。 但我想知道是否有更好的方法。 预先感谢您的任何建议:) 编辑 另一种方法是在测试课程中使用Trait，因为我要演示下面的内容。 但我并不那么热衷于这种方法，因为没有保证，在特征，类和PHPUnit_Framework_TestCase（在本例中）之间没有类似的方法名称： 这是一个示例特征： trait IndexableTrait { /** @var i
I've search all over this site and google and I ended up creating this account... I need some help with php, traits and classes. I have this 2 different traits, that have some methods with the same name. The problem relies on that I need both of them! I can't use insteadof... Here goes an example code: http://sandbox.onlinephpfunctions.com/code/b69f8e73cccd2dfd16f9d24c5d8502b21083
我搜索了这个网站和谷歌，我最终创建了这个帐户... 我需要一些关于php，特性和类的帮助。 我有这2种不同的特征，有一些同名的方法。 问题依赖于我需要他们两个！ 我不能使用... 这里有一个示例代码：http://sandbox.onlinephpfunctions.com/code/b69f8e73cccd2dfd16f9d24c5d8502b21083c1a3 trait traitA { protected $myvar; public function myfunc($a) { $this->myvar = $a; } } trait traitB { publi
I'm getting a division by zero error on this line of code: $ratio['p{W}'] = ($ratio['{W}'] === 0) ? 0 : ($colorTotal === 0) ? 0 : ($ratio['{W}'] / $colorTotal) * 100; I've tested the above with: echo '$ratio[{W}]:'.$ratio['{W}'].', $colorTotal:'.$colorTotal; if($ratio['{W}'] === 0) { echo('$ratio[{W}]: zero'); } else { echo('$ratio[{W}]: not zero'); } if($colorTotal === 0) {
在这行代码中，我得到了一个零错误的除法： $ratio['p{W}'] = ($ratio['{W}'] === 0) ? 0 : ($colorTotal === 0) ? 0 : ($ratio['{W}'] / $colorTotal) * 100; 我已经用上面的测试过了： echo '$ratio[{W}]:'.$ratio['{W}'].', $colorTotal:'.$colorTotal; if($ratio['{W}'] === 0) { echo('$ratio[{W}]: zero'); } else { echo('$ratio[{W}]: not zero'); } if($colorTotal === 0) { echo('$colorTotal: zero')
I have add below code for percentage discount badge(It automatically calculates the percentage between the old and new price.) for product. So, It is working fine. But When, add 0.00 price in product. so. get error. Warning: Division by zero in /home/...... So, any solution for this error. How can i fix it? below code add in .php file 'saving' => round((($product_info['price'] -
我为产品添加了以下折扣百分比代码（它会自动计算新旧价格之间的百分比）。 所以，它工作正常。 但是，在产品中加上0.00的价格。 所以。 得到错误。 警告：在/ home /中划分为零 所以，这个错误的任何解决方案。 我该如何解决它？ 下面的代码添加.php文件 'saving' => round((($product_info['price'] - $product_info['special'])/$product_info['price'])*100, 0) 下面的代码添加在.tpl文件中 <span cla
I am using Swiftmailer for sending mails in my application. While everything works good, I want to be able to suppress all exceptions thrown by Swiftmailer and simply return false (for instance, running the application on localhost with my GMail SMTP settings throws a timeout fatal exception). I have the error_reporting set to 0, and I also tried adding a try-catch block, but I have come to un
我正在使用Swiftmailer在我的应用程序中发送邮件。 虽然一切正常，但我希望能够抑制Swiftmailer引发的所有异常，并简单地返回false（例如，使用GMail SMTP设置在本地主机上运行应用程序会引发超时致命异常）。 我有error_reporting设置为0，我也尝试添加一个try-catch块，但我已经明白，catch块不会真的捕获异常。 我怎样才能完全抑制所有的错误，如果出现问题就返回false？