Play框架(2.3.5)-访问内部类的模板失败 [英] Play framework (2.3.5) - Template accessing an inner class fails

问题描述

我有一个这样定义的类(伪代码):

I have a class defined like so (pseudo code):

package com.some.namespace

public class SomeClass {

    protected SomeClass() {}

    // stuff...

    public class SomeInnerClass {

        public SomeInnerClass() {}

        // more stuff (accesses instance variables from SomeClass)...

    }

}

然后在模板中，我使用可重用的块定义，如下所示:

Then in my template I use the reusable block defines like so:

@doSomething(val: com.some.namespace.SomeClass.SomeInnerClass) = {

    // even more stuff...

}

但是我得到了错误:

类型SomeInnerClass不是对象的成员 com.some.namespace.SomeClass

type SomeInnerClass is not a member of object com.some.namespace.SomeClass

我无法访问模板中的内部类吗?还是应该这样做? (如果应该可以，我可能需要发布实际的代码)

Am I not able to access inner classes within the templates or is this supposed to work? (if it is supposed to work I might need to post my actual code)

推荐答案

它应该可以工作，但是由于SomeInnerClass是依赖的类型，即依赖于您的单例外部类实例，因此您需要像这样写:

It should work, but because SomeInnerClass is a dependent type, i.e. dependent on your singleton outer class instance, you need to write it like:

@doSomething(value: com.some.namespace.SomeClass#SomeInnerClass) = {
    // even more stuff...
}

Scala中的SomeClass#SomeInnerClass语法含糊地表示 SomeClass的任何实例中的SomeInnerClass.有关更多详细信息，请参见此答案.

The SomeClass#SomeInnerClass syntax in Scala means, vaguely, a SomeInnerClass from any instance of SomeClass. See this answer for more detail.

但是，如果内部类是 static ，那么您当前的SomeClass.SomeInnerClass语法将是解决之道.

If the inner class was static, however, your current SomeClass.SomeInnerClass syntax would be the way to go.

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