Java的文件路径或文件位置 - new file() [英] File path or file location for Java - new file()
问题描述
<$> $ p
$ b在我的项目中有以下结构。 c $ c> myPorjectName
src
com.example.myproject
a.java
com.example.myproject.data
b.xml
code>
在 a.java
中,我想读 b.xml
文件。我怎样才能做到这一点?具体来说,在 a.java
中,我使用了下面的代码:
DocumentBuilderFactory docBuilderFactory = DocumentBuilderFactory.newInstance();
DocumentBuilder docBuilder = docBuilderFactory.newDocumentBuilder();
Document doc = docBuilder.parse(new File(data / b.xml));
这段代码找不到 b.xml
。但是,如果我将路径更改为 src / com / example / myproject / data / b.xml
,那么它可以工作。目前的位置似乎在我的项目文件的根。
但是我看到其他人的例子, if b.xml
和 a .java
在同一个文件夹中,那么我们可以直接使用 new File(b.xml)
。把 b.xml
放在 a.java
的文件夹中,而不是放在子文件夹中, strong>不起作用。如果这个工作,那么在我的情况下,我应该能够使用新的文件(data / b.xml)
,对不对?我真的不明白为什么这是行不通的。
如果它已经在类路径中,并在相同的包中,使用
URL url = getClass()。getResource(b.xml);
File file = new File(url.getPath());
或者,读取它作为 InputStream
:
InputStream input = getClass()。getResourceAsStream(b.xml);
在静态
方法中,可以使用
InputStream in = YourClass.class.getResourceAsStream(b.xml);
如果您的文件与您尝试访问该文件的类不在同一个包中,那么你必须给它的相对路径从'/'
开始。
ex:InputStream input = getClass()。getResourceAsStream
(/resources/somex.cfg.xml\");这是另一个jar资源文件夹中的
I have the following structure for my project.
In Eclipse:
myPorjectName
src
com.example.myproject
a.java
com.example.myproject.data
b.xml
In a.java
, I want to read b.xml
file. How can I do that? Specifically, in a.java
, I used the following code:
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.xml
then it works. The current location seems to be in the root of my project file.
But I see other people's examples, if b.xml
and a.java
are in the same folder, then we can directly use new File("b.xml")
. But I try putting b.xml
in the same folder of a.java
rather 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.
If it is already in the classpath and in the same package, use
URL url = getClass().getResource("b.xml");
File file = new File(url.getPath());
OR , read it as an InputStream
:
InputStream input = getClass().getResourceAsStream("b.xml");
Inside a static
method, you can use
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 '/'
.
ex : InputStream input = getClass().getResourceAsStream
("/resources/somex.cfg.xml");which is in another jar resource folder
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