eclipse 在不同的包中类 getResourceAsStream() 资源
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原文地址: http://stackoverflow.com/questions/16320170/
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Class getResourceAsStream() resource in a different package
提问by Sotirios Delimanolis
I have something like the following project structure
我有类似以下项目结构的东西
-project
-src/main/java // source folder
-com.mypackage
-Utility.java
-some.properties
-src/main/resources // source folder
-configuration_files
-configuration_report.txt
I'm generating the jarwith Eclipse's Exportfunction. The above translates to a jarwith the structure:
我正在jar用 Eclipse 的Export函数生成。以上转换为jar具有结构的a：
-myjar.jar
-com
-mypackage
-Utility.class
-some.properties
-configuration_files
-configuration_report.txt
Within the Utilityclass, I'm trying to get the InputStreamfrom configuration_report.txt.
在Utility课堂上，我试图InputStream从configuration_report.txt.
I was doing this (it's within a staticmethod):
我正在这样做（它在一个static方法中）：
Utility.class.getResourceAsStream("/configuration_files/configuration_report.txt");
which seems to be validated by the answer to this question, since the file is not in the same package as the class requesting it.
这似乎得到了这个问题的答案的验证，因为该文件与请求它的类不在同一个包中。
When I run it, however, it returns null. Why?
但是，当我运行它时，它返回null. 为什么？
Note that for the some.propertiesfile, I could do both
请注意，对于some.properties文件，我可以同时执行
Utility.class.getResourceAsStream("/com/package/some.properties");
Utility.class.getResourceAsStream("some.properties");
and I would get the stream.
我会得到流。
I'm running this from the command line if it makes any difference.
如果它有任何不同，我将从命令行运行它。
java [-agentlib:jdwp=transport=dt_socket,server=y,suspend=y,address=1044] -cp myjar.jar MyMain
If I try to run it within Eclipse, it works...
如果我尝试在 Eclipse 中运行它，它会起作用......
Edit:
编辑：
This code is relevant. With or without the preceding /, it return null
此代码是相关的。有或没有前面的/，它返回null
String sourceFilePath = "/configuration_files" + File.separator + "configuration_report.txt";
InputStream in = Utility.class.getResourceAsStream(sourceFilePath);
回答by Sotirios Delimanolis
After many tries, the solution comes down to this little javadoc entry for ClassLoader#getResource(String):
经过多次尝试，解决方案归结为以下 javadoc 小条目ClassLoader#getResource(String)：
The name of a resource is a '/'-separated path name that identifies the resource.
资源的名称是一个以“/”分隔的路径名，用于标识资源。
The snippet I had originally posted would've worked
我最初发布的片段会起作用
Utility.class.getResourceAsStream("/configuration_files/configuration_report.txt");
but I wasn't using /, I was using File.separatorlike in my edit:
但我没有使用/，我File.separator在编辑中使用了like：
String sourceFilePath = "/configuration_files" + File.separator + "configuration_report.txt";
thinking the class loader worked like OS specific File operations. The first /was matched, but the File.separatorevaluated to \\since I'm on Windows.
认为类加载器像操作系统特定的文件操作一样工作。第一个/匹配，但由于我在 Windows 上被File.separator评估为\\。
The javadoc for Class#getResource(String)says more of the same.
javadoc forClass#getResource(String)说明了更多相同的内容。
Where the modified_package_name is the package name of this object with '/' substituted for '.' ('\u002e').
其中 modified_package_name 是这个对象的包名，用“/”代替“.” ('\u002e')。
As to why this worked on Eclipse, but not from the command line, I'm still unsure, but look at @andy's comment below.
至于为什么这在 Eclipse 上起作用，而不是从命令行起作用，我仍然不确定，但请看下面@andy 的评论。
This articlegives is a good explanation of this situation.
这篇文章对这种情况给出了很好的解释。
