scala Scala获取资源文件夹中文件的文件路径
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Scala get file path of file in resources folder
提问by user1893354
I am using the Stanford CRFClassifier and in order to run, it requires a file that is the trained classifier model. I have put this file in the resources directory. From the Javadocs for the CRFClassifier http://nlp.stanford.edu/nlp/javadoc/javanlp/edu/stanford/nlp/ie/crf/CRFClassifier.html#getClassifier(java.lang.String)the path to the file must be an input to CRFClassifier.getClassifier() and it is a java.lang.String object. So my question is how do I tell .getClassifier() that the file is in the resources directory? i.e. how do I get the file path of a file in the resources directory?
我正在使用斯坦福 CRFClassifier，为了运行，它需要一个经过训练的分类器模型的文件。我已将此文件放在资源目录中。从 CRFClassifier 的 Javadocs http://nlp.stanford.edu/nlp/javadoc/javanlp/edu/stanford/nlp/ie/crf/CRFClassifier.html#getClassifier(java.lang.String)文件的路径必须是 CRFClassifier.getClassifier() 的输入，它是一个 java.lang.String 对象。所以我的问题是如何告诉 .getClassifier() 该文件在资源目录中？即如何获取资源目录中文件的文件路径？
I have tried simply
我试过简单
val classifier = CRFClassifier.getClassifier("./src/main/resources/my_model.ser.gz")
But this returns a FileNotFoundException.
但这会返回 FileNotFoundException。
I have also tried
我也试过
Source.fromURL(getClass.getResource("/my_model.ser.gz"))
which returns a BufferedSource object, but I do not know how to get a file path from this.
它返回一个 BufferedSource 对象，但我不知道如何从中获取文件路径。
Any help would be greatly appreciated.
任何帮助将不胜感激。
回答by user1893354
I managed to be able to get the file path by doing the following
我设法通过执行以下操作来获取文件路径
val url=getClass.getResource("/my_model.ser.gz")
val url=getClass.getResource("/my_model.ser.gz")
val classifier = CRFClassifier.getClassifier(url.getPath())
val classifier = CRFClassifier.getClassifier(url.getPath())
