如何在Java，Eclipse中找到当前项目的路径？

本文介绍了如何在Java，Eclipse中找到当前项目的路径？的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 我试图用Java编程方式找到当前运行/调试的项目的路径，我在Google中找到了，我发现是 System.getProperty（user.id），这没有得到我的项目路径。

我知道命令 Environment.currentDirectory in C＃给出了当前运行/调试的项目的路径，所以我确信Java中还有一个类似的方式。

所以我问是否有人可以告诉或者给我一个代码，以编程方式找到当前运行/调试项目的路径？

编辑：i am writing eclipse启动加载的插件。 插件添加一个按钮到eclipse的工作台，在插件的代码当我点击按钮我搜索当前的目录路径。

$ b $我的目的是，例如我调试一个项目，然后点击按钮，会弹出一个窗口，显示调试项目的路径，谢谢

解决方式

两种方式

System.getProperty（user.dir）;

或这个

code> File currentDirFile = new File（。）; String helper = currentDirFile.getAbsolutePath（）; String currentDir = helper.substring（0，helper.length - currentDirFile.getCanonicalPath（）。length）;

想法是使用。获取当前文件夹，然后获取绝对位置并从其中删除文件名，因此从

/ home / shark / eclipse / workspace / project / src / com / package / name / bin / Class.class

当你删除Class.class你会得到

/ home / shark / eclipse / workspace / project / src / com / package / name / bin /

这是你想要的。

I am trying to locate the path of the current running/debugged project programmatically in Java, I looked in Google and what I found was System.getProperty("user.id"), which didn't get me the project's path.

I know the command Environment.currentDirectory in C# gives the path of the current running/debugged project,so I am sure there must be a similar way in Java, too.

So I am asking if anyone could tell me or give me a code to how locate the path of the currently running/debugged project programmatically?

edit: i am writing plugin which loads with eclipse start. the plugin adds a button to the workbench of eclipse, with in the code of the plugin when i click the button i search for the current directory path.

my purpose is that when for example i debug a project and then click the button, a window will pop up presenting the debugged project's path, thanks

解决方案

Two ways

System.getProperty("user.dir");

or this

File currentDirFile = new File("."); String helper = currentDirFile.getAbsolutePath(); String currentDir = helper.substring(0, helper.length - currentDirFile.getCanonicalPath().length);

The idea is to get the current folder with ".", and then fetch the absolute position to it and remove the filename from it, so from something like

/home/shark/eclipse/workspace/project/src/com/package/name/bin/Class.class

when you remove Class.class you'd get

/home/shark/eclipse/workspace/project/src/com/package/name/bin/

which is kinda what you want.