Maven exec：java：如何在资源目录中打开和读取文件？ [英] Maven exec:java : how to open and read a file in the resources directory?

问题描述

这是我项目的结构。

proj
  ---src
    ----main
        ----java
            ----Main.java
        ----resources
             ----res.txt

我在Eclipse中使用m2eclipse插件。
在Main.java中，我有

I am using m2eclipse plugin with Eclipse. In Main.java, I have

File f = new File("res.txt");  System.out.println(f.getAbsolutePath());

当我运行mvn exec：java时，打印出来的路径是... \ proj \res.txt。如何在... \ proj\target\classes目录中查找资源文件？

When I run mvn exec:java, the path got printed out is "...\proj\res.txt". How can I make it look for the resource file in "...\proj\target\classes" directory?

编辑：

感谢我原来问题的答案。我有一个后续问题：

Thanks for the answers to my original question. I have a follow-up questions:

所以基本上我想做的是让Main类读取res.txt然后生成一个新的newres .txt到资源目录，以便我可以在以后的包阶段将这个新的newres.txt打包到jar文件中。目前我将这个exec：java挂载到prepare-package阶段。如何在资源目录中创建这个newres.txt而没有硬编码的绝对路径或者取决于Maven的目录结构？

So basically what I want to do is to have the Main class read the "res.txt" and then generate a new "newres.txt" to the resources directory so that I can package this new "newres.txt" to the jar file in the package phase later. Currently I mounted this exec:java to the prepare-package phase. How should I create this "newres.txt" in the resources directory without a hard-coded absolute path or depending on the directory structure of Maven?

推荐答案

我想我会回答我自己的问题， Thread.currentThread（）。getContextClassLoader（）。getResourceAsStream（）
对我来说效果最好，特别是当项目为另一个Web项目生成jar依赖项时。

I guess I will answer my own question, Thread.currentThread().getContextClassLoader().getResourceAsStream() works the best for me, especially when the project produces a jar dependency for another web project.

这篇关于Maven exec：java：如何在资源目录中打开和读取文件？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！