如何在 symfony2 上添加自定义异常 [英] How to add a custom exception on symfony2
问题描述
我是 Symfony 的新手(使用 2.2 版)并尝试添加自定义异常侦听器.我已阅读以下链接,但无法使其正常工作.
I am new to Symfony(using version 2.2) and trying to add a custom exception listener. I have read the following links but cannot get it to work.
我想要做的是创建一个自定义的错误异常侦听器,并在我的控制器和服务中使用它,
What I'm trying to do is to create a custom Error Exception Listener and use it from my controllers and services like this,
throw new jsonErrorException('invalid_params');
像这样显示 json twig 模板.(我正在为我的原生智能手机应用程序开发一个后台程序)
to display json twig template like this.(I'm developing a background program for my native smartphone applications)
{"status": "error", "message": "invalid_params"}
我已将 CustomEventListener 注册到我的 src/My/Bundle/Resources/config/services.yml 并创建了一个自定义异常类,如以下链接所示(覆盖 Symfony 2 异常?) 但我收到错误
I have registered my CustomEventListener to my src/My/Bundle/Resources/config/services.yml and created a custom exception class as shown on following link(Overriding Symfony 2 exceptions?) but I get the error
symfony exceptions must be valid objects derived from the exception base class
我在这里做错了吗?谢谢.
Am I doing something wrong here? Thanks.
推荐答案
您可以通过symfony 方式"创建自定义异常让我们看看异常或在 symfony 中是如何创建的:
You can create custom exception the "symfony way" let's look at how exception or created in symfony:
首先创建您的自定义异常接口
first create your customExceptionInterface
namespace MySymfonyBundleException;
/**
* Interface for my bundle exceptions.
*/
interface MySymfonyBundleExceptionInterface
{
}
并创建您的 jsonErrorException
namespace MySymfonyBundleException;
class HttpException extends Exception implements MySymfonyBundleExceptionInterface
{
}
不要犹豫,浏览symfony 的异常代码示例github
Don't hesitate to browse symfony's exceptions code examples on github
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