Tom*_*Tom 2 action symfony1 view configuration-files
我想知道是否有可能在一个动作中从view.yml获取样式表的名称,理想情况下使用如下简单的东西:
sfConfig::get('......');
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我想访问view.yml中的现有声明,而不是硬编码或复制它,如app.yml.
谢谢.
小智 5
如果要访问当前模块的配置,可以使用:
sfViewConfigHandler::getConfiguration(array(dirname(__DIR__) . '/config/view.yml'));
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它应该返回这样的东西:
Array
(
[indexSuccess] => Array
(
[javascripts] => Array
(
[0] => mission-control.js
)
[stylesheets] => Array
(
[0] => control-box.css
[1] => question.css
)
)
[newSuccess] => Array
(
[javascripts] => Array
(
[0] => box-checker.js
[1] => topic.js
)
[stylesheets] => Array
(
[0] => question.css
[1] => topic.css
)
)
[searchSuccess] => Array
(
[javascripts] => Array
(
[0] => topic.js
)
[stylesheets] => Array
(
[0] => topic.css
)
)
[showSuccess] => Array
(
[javascripts] => Array
(
[0] => mission-control.js
)
[stylesheets] => Array
(
[0] => control-box.css
[1] => question.css
)
)
[editSuccess] => Array
(
[javascripts] => Array
(
[0] => box-checker.js
[1] => topic.js
)
[stylesheets] => Array
(
[0] => question.css
[1] => topic.css
)
)
[all] => Array
(
[stylesheets] => Array
(
)
[javascripts] => Array
(
)
)
)
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