blocks|key|1527472|text|这些解决方案都不考虑具有多个父模块的常量。例如：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1527473|A::B::C|code-block|syntax|javascript|1527474|从Rails+3.2.x开始，您可以简单地：|1527475|"A::B::C".deconstantize+#=>+"A::B"|1527476|从Rails+3.1.x开始，您可以：|1527477|constant_name+=+"A::B::C"
constant_name.gsub(+"::#{constant_name.demodulize}",+''+)|1527478|这是因为#demodulize是#deconstantize的对立面：|1527479|"A::B::C".demodulize+#=>+"C"|1527480|如果您确实需要手动执行此操作，请尝试以下操作：|1527481|constant_name+=+"A::B::C"
constant_name.split(+'::'+)[0,constant_name.split(+'::'+).length-1]|1527482|entityMap^0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|12|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|14|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|16|8|@]|9|@]|A|$]]|$1|U|3|V|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|W|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|X|$]]

None of these solutions consider a constant with multiple parent modules. For instance:

<pre><code>A::B::C
</code></pre>

As of Rails 3.2.x you can simply:

<pre><code>"A::B::C".deconstantize #=&gt; "A::B"
</code></pre>

As of Rails 3.1.x you can:

<pre><code>constant_name = "A::B::C"
constant_name.gsub( "::#{constant_name.demodulize}", '' )
</code></pre>

This is because #demodulize is the opposite of #deconstantize:

<pre><code>"A::B::C".demodulize #=&gt; "C"
</code></pre>

If you really need to do this manually, try this:

<pre><code>constant_name = "A::B::C"
constant_name.split( '::' )[0,constant_name.split( '::' ).length-1]
</code></pre>

blocks|key|1742599|text|对于简单的情况，您可以使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1742600|self.class.parent|code-block|syntax|javascript|1742601|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

For the simple case, You can use :

<pre><code>self.class.parent
</code></pre>

blocks|key|1742318|text|这应该可以做到：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1742319|++def+get_module_name
++++@module_name+=+self.class.to_s.split("::").first
++end|code-block|syntax|javascript|1742320|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This should do it:

<pre><code> def get_module_name
 @module_name = self.class.to_s.split("::").first
 end
</code></pre>

blocks|key|1527325|text|如果控制器有模块名称，这将起作用，但如果没有模块名称，则返回控制器名称。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1527326|class+ApplicationController+<+ActionController::Base
++def+get_module_name
++++@module_name+=+self.class.name.split("::").first
++end
end|code-block|syntax|javascript|1527327|但是，如果我们将其更改为：|1527328|class+ApplicatioNController+<+ActionController::Base
++def+get_module_name
++++my_class_name+=+self.class.name
++++if+my_class_name.index("::").nil?+then
++++++@module_name+=+nil
++++else
++++++@module_name+=+my_class_name.split("::").first
++++end
++end
end|1527329|您可以确定类是否有模块名，并返回可以测试的类名以外的其他内容。|1527330|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

This would work if the controller did have a module name, but would return the controller name if it did not.

<pre><code>class ApplicationController &lt; ActionController::Base
 def get_module_name
 @module_name = self.class.name.split("::").first
 end
end
</code></pre>

However, if we change this up a bit to:

<pre><code>class ApplicatioNController &lt; ActionController::Base
 def get_module_name
 my_class_name = self.class.name
 if my_class_name.index("::").nil? then
 @module_name = nil
 else
 @module_name = my_class_name.split("::").first
 end
 end
end
</code></pre>

You can determine if the class has a module name or not and return something else other than the class name that you can test for.

blocks|key|1527406|text|我知道这是一个很老的线程，但我只是遇到了需要根据控制器的名称空间进行单独导航的问题。我想出的解决方案是在我的应用程序布局中：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1527407|<%25=+render+"#{controller.class.name[/%5E(\w*)::\w*$/,+1].try(:downcase)}/nav"+%25>|code-block|syntax|javascript|1527408|它看起来有点复杂，但基本上做了以下工作-它接受控制器类名，例如，对于非命名空间的控制器，可以是"People“，对于命名空间的控制器，可以是"Admin::Users”。将[]字符串方法与正则表达式一起使用，该正则表达式返回两个冒号之前的任何内容，如果没有任何内容，则返回nil。然后将其更改为小写(如果没有名称空间，则会返回nil，则会出现"try“)。这就给我们留下了命名空间或nil。然后，它简单地呈现带有或不带有名称空间的部分，例如没有名称空间：|1527409|app/views/_nav.html.erb|1527410|或在admin名称空间中：|1527411|app/views/admin/_nav.html.erb|1527412|当然，对于每个名称空间，这些分词必须存在，否则就会出现错误。现在，将为每个控制器显示每个命名空间的导航，而不必更改任何控制器或视图。|1527413|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

I know this is an old thread, but I just came across the need to have separate navigation depending on the namespace of the controller. The solution I came up with was this in my application layout:

<pre><code>&lt;%= render "#{controller.class.name[/^(\w*)::\w*$/, 1].try(:downcase)}/nav" %&gt;
</code></pre>

Which looks a bit complicated but basically does the following - it takes the controller class name, which would be for example "People" for a non-namespaced controller, and "Admin::Users" for a namespaced one. Using the [] string method with a regular expression that returns anything before two colons, or nil if there's nothing. It then changes that to lower case (the "try" is there in case there is no namespace and nil is returned). This then leaves us with either the namespace or nil. Then it simply renders the partial with or without the namespace, for example no namespace:

<pre><code>app/views/_nav.html.erb
</code></pre>

or in the admin namespace:

<pre><code>app/views/admin/_nav.html.erb
</code></pre>

Of course these partials have to exist for each namespace otherwise an error occurs. Now the navigation for each namespace will appear for every controller without having to change any controller or view.

blocks|key|1742576|text|my_class.name.underscore.split('/').slice(0..-2)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1742577|或|1742578|my_class.name.split('::').slice(0..-2)|1742579|entityMap^0|0|1C|0|0|0|12|0^^$0|@$1|2|3|4|5|6|7|L|8|@$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|O|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|J|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|K|$]]

<code>my_class.name.underscore.split('/').slice(0..-2)</code>

or

<code>my_class.name.split('::').slice(0..-2)</code>

blocks|key|1742665|text|没有人提到过使用rpartition|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1742666|const_name+=+'A::B::C'
namespace,+_sep,+module_name+=+const_name.rpartition('::')
#+or+if+you+just+need+the+namespace
namespace+=+const_name.rpartition('::').first|code-block|syntax|javascript|1742667|entityMap^0|8|A|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

No one has mentioned using <code>rpartition</code>?

<pre><code>const_name = 'A::B::C'
namespace, _sep, module_name = const_name.rpartition('::')
# or if you just need the namespace
namespace = const_name.rpartition('::').first
</code></pre>

blocks|key|1527515|text|我推荐gsub而不是split。它比split更有效，因为您不需要任何其他模块名称。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1527516|class+ApplicationController+<+ActionController::Base
++def+get_module_name
++++@module_name+=+self.class.to_s.gsub(/::.*/,+'')
++end
end|code-block|syntax|javascript|1527517|entityMap^0|3|4|A|5|I|5|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|U|8|@]|D|@]|E|$]]]|L|$]]

I recommend <code>gsub</code> instead of <code>split</code>. It's more effective that <code>split</code> given that you don't need any other module name.

<pre><code>class ApplicationController &lt; ActionController::Base
 def get_module_name
 @module_name = self.class.to_s.gsub(/::.*/, '')
 end
end
</code></pre>

blocks|key|1742644|text|包含多个子模块：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1742645|module+ApplicationHelper
++def+namespace
++++controller.class.name.gsub(/(::)?\w%2BController$/,+'')
++end
end|code-block|syntax|javascript|1742646|示例：Foo::Bar::BazController+=>+Foo::Bar|offset|length|style|CODE|1742647|entityMap^0|0|0|3|N|U|8|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|N|$]]

With many sub-modules:

<pre><code>module ApplicationHelper
 def namespace
 controller.class.name.gsub(/(::)?\w+Controller$/, '')
 end
end
</code></pre>

Example: <code>Foo::Bar::BazController</code> => <code>Foo::Bar</code>

How do I find the name of the namespace or module 'Foo' in the filter below?

<pre><code>class ApplicationController &lt; ActionController::Base
 def get_module_name
 @module_name = ???
 end
end


class Foo::BarController &lt; ApplicationController
 before_filter :get_module_name
end
</code></pre>

How do you find the namespace/module name programmatically in Ruby on Rails?

如何在下面的筛选器中找到命名空间或模块'Foo‘的名称？class ApplicationController < ActionController::Base def get_module_name @module_name = ??? endendclass Foo::BarController < ApplicationController before_filter :get_

问如何在Ruby on Rails中以编程方式找到命名空间/模块名称？
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回答 9

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问如何在Ruby on Rails中以编程方式找到命名空间/模块名称？EN

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