Java的绝对路径项目的文件夹

问题描述

这个话题中有许多混乱。已经提出了几个问题。事情似乎还不清楚。
ClassLoader，绝对文件路径等等

假设我有一个项目目录结构，

 
 MyProject-- 
 --dist 
 --lib 
 --src 
 --test 
 

我有一个资源在lib / txt目录中显示txtfile.txt。我想以系统独立的方式访问它。我需要项目的绝对路径。
所以我可以将路径编码为abspath +/ lib / Dictionary / txtfile.txt

假设我这样做

  java.io.File file = new java.io.File（）; //虚拟文件
 String abspath = file.getAbsolutePath（）; 
  

我得到当前工作目录，这不一定是项目根目录。假设我从'dist'文件夹中执行最终的'prj.jar'，它还包含lib / txt / txtfile.txt目录结构和资源，应该也在这里工作。我应该是dist文件夹的绝对路径。

希望问题清楚。

解决方案>

您应该真正使用或使用您的类加载器为这种事情。特别地，这些方法使用您的ClassLoader来确定项目中资源的搜索上下文。

指定类似于 getClass（）的getResource（ lib / txtfile.txt）以获取文本文件。

要澄清：而不是考虑如何获取路径您应该考虑获取资源的资源 - 在这种情况下，某个目录（可能位于您的JAR内）的文件。在这种情况下不需要知道绝对路径，只有一些URL可以获取文件，ClassLoader将为您返回此URL。如果要打开文件流，您可以直接使用 getResourceAsStream 直接执行此操作。

您尝试通过ClassLoader访问的资源需要位于Class-Path（在JAR文件的清单中配置）。这是至关重要的！ ClassLoader使用Class-Path来查找资源，因此如果您在Class-Path中没有提供足够的上下文，则无法找到任何内容。如果你添加，那么ClassLoader应该根据你引用该资源的方式来解决JAR内部或外部的任何内容，尽管你当然可以更具体。

引用前缀为的资源。将导致ClassLoader还查找JAR之外的文件，而不是资源前缀具有一段时间的路径将指示ClassLoader仅在JAR文件中查找。

这意味着如果您有一个文件目录 lib ，名称为 foo.txt ，而您要获取资源，然后运行 getResource（lib / foo.txt）;

如果相同的资源在JAR之外，您将运行 getResource（./ lib / foo.txt）;

Lots of confusion in this topic. Several Questions have been asked. Things still seem unclear.ClassLoader, Absolute File Paths etc etc

Suppose I have a project directory structure as,

MyProject--
            --dist
            --lib
            --src
            --test

I have a resource say "txtfile.txt" in "lib/txt" directory. I want to access it in a system independent way. I need the absolute path of the project.So I can code the path as abspath+"/lib/Dictionary/txtfile.txt"

Suppose I do this

 java.io.File file = new java.io.File("");   //Dummy file
    String  abspath=file.getAbsolutePath();

I get the current working directory which is not necessarily project root.

Suppose I execute the final 'prj.jar' from the 'dist' folder which also contains "lib/txt/txtfile.txt" directory structure and resource,It should work here too. I should absolute path of dist folder.

Hope the problem is clear.

解决方案

You should really be using getResource() or getResourceAsStream() using your class loader for this sort of thing. In particular, these methods use your ClassLoader to determine the search context for resources within your project.

Specify something like getClass().getResource("lib/txtfile.txt") in order to pick up the text file.

To clarify: instead of thinking about how to get the path of the resource you ought to be thinking about getting the resource -- in this case a file in a directory somewhere (possibly inside your JAR). It's not necessary to know some absolute path in this case, only some URL to get at the file, and the ClassLoader will return this URL for you. If you want to open a stream to the file you can do this directly without messing around with a URL using getResourceAsStream.

The resources you're trying to access through the ClassLoader need to be on the Class-Path (configured in the Manifest of your JAR file). This is critical! The ClassLoader uses the Class-Path to find the resources, so if you don't provide enough context in the Class-Path it won't be able to find anything. If you add . the ClassLoader should resolve anything inside or outside of the JAR depending on how you refer to the resource, though you can certainly be more specific.

Referring to the resource prefixed with a . will cause the ClassLoader to also look for files outside of the JAR, while not prefixing the resource path with a period will direct the ClassLoader to look only inside the JAR file.

That means if you have some file inside the JAR in a directory lib with name foo.txt and you want to get the resource then you'd run getResource("lib/foo.txt");

If the same resource were outside the JAR you'd run getResource("./lib/foo.txt");

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