Java 的文件路径或文件位置 - new file()
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File path or file location for Java - new file()
提问by Joey
I have the following structure for my project.
我的项目具有以下结构。
In Eclipse:
在日食中：
myPorjectName
src
com.example.myproject
a.java
com.example.myproject.data
b.xml
In a.java, I want to read b.xmlfile. How can I do that? Specifically, in a.java, I used the following code:
在a.java，我想读取b.xml文件。我怎样才能做到这一点？具体来说，在 中a.java，我使用了以下代码：
DocumentBuilderFactory docBuilderFactory = DocumentBuilderFactory.newInstance();
DocumentBuilder docBuilder = docBuilderFactory.newDocumentBuilder();
Document doc = docBuilder.parse (new File("data/b.xml"));
This code cannot find b.xml. However, if I change the path to src/com/example/myproject/data/b.xmlthen it works. The current location seems to be in the root of my project file.
此代码找不到b.xml. 但是，如果我将路径更改为 ，src/com/example/myproject/data/b.xml则它可以工作。当前位置似乎在我的项目文件的根目录中。
But I see other people's examples, if b.xmland a.javaare in the same folder, then we can directly use new File("b.xml").But I try putting b.xmlin the same folder of a.javarather than putting in the sub folder, but it still does not work. If this works, then in my case, I should be able to use new File("data/b.xml"), right? I really do not understand why this is not working.
但是我看别人的例子，如果b.xml和a.java在同一个文件夹下，那么我们可以直接使用new File("b.xml"). 但是我尝试放在b.xml的同一个文件夹中a.java而不是放在子文件夹中，但它仍然不起作用。如果这有效，那么在我的情况下，我应该可以使用new File("data/b.xml")，对吗？我真的不明白为什么这不起作用。
采纳答案by NINCOMPOOP
If it is already in the classpath and in the same package, use
如果它已经在类路径和同一个包中，请使用
URL url = getClass().getResource("b.xml");
File file = new File(url.getPath());
OR , read it as an InputStream:
或者，将其读作InputStream：
InputStream input = getClass().getResourceAsStream("b.xml");
Inside a staticmethod, you can use
在static方法内部，您可以使用
InputStream in = YourClass.class.getResourceAsStream("b.xml");
If your file is not in the same package as the class you are trying to access the file from, then you have to give it relative path starting with '/'.
如果您的文件与您尝试从中访问该文件的类不在同一个包中，那么您必须为其提供以'/'.
ex : InputStream input = getClass().getResourceAsStream
("/resources/somex.cfg.xml");which is in another jar resource folder
