YAML可以继承吗？
这个问题涉及很多 symfony，但对于只知道 YAML 而不懂 symfony 的人来说应该很容易理解。
我的 symfony 模型分为三个步骤：首先，我在 MySQL 中创建表。其次，我运行 symfony 命令 (symfonydoctrine:build-schema) 将表结构转换为 YAML 文件。第三，我运行另一个 symfony 命令 (symfonydoctrine:build-model) 将 YAML 文件转换为 PHP 代码。
问题是：数据库中有一些表我不希望出现在我的 symfony 代码中。例如，假设我有两个表：一个名为 my_table，另一个名为 wordpress。我最终得到的 YAML 文件可能如下所示：
MyTable:
connection: doctrine
tableName: my_table
Wordpress:
connection: doctrine
tableName: wordpress
很好，除了 wordpress 表与我的 symfony 模型无关。结果是，每次我对数据库进行更改并生成此 YAML 文件时，我都必须手动删除 wordpress。真烦人！
我希望能够创建一个名为 baseConfig.php 的文件或如下所示的文件：
$config = array(
'MyTable' => array(
'connection' => 'doctrine',
'tableName' => 'my_table',
),
'Wordpress' => array(
'connection' => 'doctrine',
'tableName' => 'wordpress',
),
);
然后我可以有一个名为 config.php 的单独文件或可以对基本配置进行修改的文件：
unset($config['Wordpress']);
所以我的问题是：有什么方法可以将 YAML 转换为可执行的 PHP 代码（而不是像 sfYaml::load() 那样将 YAML 加载到 PHP 代码中）来实现这种事情？或者是否有其他方法来实现 YAML 继承？ 谢谢， 贾森
This question involves a lot of symfony but it should be easy enough for someone to follow who only knows YAML and not symfony.
My symfony models come from a three-step process: First, I create the tables in MySQL. Second, I run a symfony command (symfony doctrine:build-schema) to convert my table structure into a YAML file. Third, I run another symfony command (symfony doctrine:build-model) to convert the YAML file into PHP code.
Here's the problem: there are some tables in the database that I don't want to end up in my symfony code. For example, let's say I have two tables: one called my_table and another called wordpress. The YAML file I end up with might look like this:
MyTable:
connection: doctrine
tableName: my_table
Wordpress:
connection: doctrine
tableName: wordpress
That's great except the wordpress table has nothing to do with my symfony models. The result is that every single time I make a change to my database and generate this YAML file, I have to manually remove wordpress. It's annoying!
I'd like to be able to create a file called baseConfig.php or something that looks like this:
$config = array(
'MyTable' => array(
'connection' => 'doctrine',
'tableName' => 'my_table',
),
'Wordpress' => array(
'connection' => 'doctrine',
'tableName' => 'wordpress',
),
);
And then I could have a separate file called config.php or something where I could make modifications to the base config:
unset($config['Wordpress']);
So my question is: is there any way to convert YAML into executable PHP code (as opposed to load YAML INTO PHP code like what sfYaml::load() does) to achieve this sort of thing? Or is there maybe some other way to achieve YAML inheritance?
Thanks,
Jason
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另一种方法：
我不确定 Symphony 是否会让您这样做，但是 Doctrine 支持手工构建的架构文件。您需要手动保持同步，但是根据您对表的更改程度，这可能是可以管理的。
An alternative approach:
I'm not sure if Symphony will let you do this, but Doctrine supports hand-built schema files. You'll need to keep things in sync manually, but depending on how much you change your tables, this might be managable.
正确答案（据我所知）：不。
顺便说一句，这是我最终编写的插件：
http://jasonswett.net/jsdoctrineschemaoverriderplugin/
Correct answer (as far as I can tell): no.
By the way, here's the plugin I ended up writing:
http://jasonswett.net/jsdoctrineschemaoverriderplugin/