Java Jar文件：使用资源错误：URI不是分层的 [英] Java Jar file: use resource errors: URI is not hierarchical

问题描述

我已将我的应用程序部署到jar文件。当我需要将数据从一个资源文件复制到jar文件外部时，我做这个代码：

  URL resourceUrl = getClass （）.getResource（ /资源/ data.sav）; 
文件src = new File（resourceUrl.toURI（））; // ERROR HERE 
文件dst = new File（CurrentPath（）+data.sav）; // CurrentPath：jar文件的路径不包括jar文件名
 FileInputStream in = new FileInputStream（src）; 
 FileOutputStream out = new FileOutputStream（dst）; 
 //这里的一些excute代码
  

我遇到的错误是：$ code> URI不是层次结构的。这个错误在IDE中运行时不符合要求。

如果我在StackOverFlow上更改了上述代码作为其他帖子的一些帮助：

  InputStream in = Model.class.getClassLoader（）。getResourceAsStream（/ resource / data.sav）; 
文件dst = new File（CurrentPath（）+data.sav）; 
 FileOutputStream out = new FileOutputStream（dst）; 
 // .... 
 byte [] buf = new byte [1024]; 
 int len; 
 while（（len = in.read（buf））> 0）{// NULL POINTER EXCEPTION 
 // .... 
} 
  

解决方案

你不能这样做

 文件src = new File（resourceUrl.toURI（））; // ERROR HERE 
  

它不是一个文件！
当您从ide运行时，您没有任何错误，因为您不运行jar文件。在IDE中，文件系统中提取了类和资源。

但是您可以以这种方式打开一个 InputStream ：

  InputStream in = Model.class.getClassLoader（）。getResourceAsStream（/ data.sav）; 
  

删除/ resource。通常，IDE在文件系统类和资源上分离。但是当创建该jar时，它们将被放在一起。因此，文件夹级别/ resource仅用于类和资源分离。

获取资源时从classloader，您必须指定资源在jar内的路径，即真正的包层次结构。

I have deployed my app to jar file. When I need to copy data from one file of resource to outside of jar file, I do this code:

URL resourceUrl = getClass().getResource("/resource/data.sav");
File src = new File(resourceUrl.toURI()); //ERROR HERE
File dst = new File(CurrentPath()+"data.sav");  //CurrentPath: path of jar file don't include jar file name
FileInputStream in = new FileInputStream(src);
FileOutputStream out = new FileOutputStream(dst);
 // some excute code here

The error I have met is: URI is not hierarchical. this error I don't meet when run in IDE.

If I change above code as some help on other post on StackOverFlow:

InputStream in = Model.class.getClassLoader().getResourceAsStream("/resource/data.sav");
File dst = new File(CurrentPath() + "data.sav");
FileOutputStream out = new FileOutputStream(dst);
//....
byte[] buf = new byte[1024];
int len;
while ((len = in.read(buf)) > 0) { //NULL POINTER EXCEPTION
  //....
}

解决方案

You cannot do this

File src = new File(resourceUrl.toURI()); //ERROR HERE

it is not a file! When you run from the ide you don't have any error, because you don't run a jar file. In the IDE classes and resources are extracted on the file system.

But you can open an InputStream in this way:

    InputStream in = Model.class.getClassLoader().getResourceAsStream("/data.sav");

Remove "/resource". Generally the IDEs separates on file system classes and resources. But when the jar is created they are put all together. So the folder level "/resource" is used only for classes and resources separation.

When you get a resource from classloader you have to specify the path that the resource has inside the jar, that is the real package hierarchy.

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