blocks|key|2053201|text|ClassLoader.getResourceAsStream()。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2053202|正如下面的评论所述，如果您处于多ClassLoader环境中(例如单元测试、webapps等)。您可能需要使用Thread.currentThread().getContextClassLoader()。参见http://stackoverflow.com/questions/2308188/getresourceasstream-vs-fileinputstream/2308388#comment21307593_2308388。|2053203|entityMap|0|LINK|mutability|MUTABLE|url|http://download.oracle.com/javase/6/docs/api/java/lang/ClassLoader.html#getResourceAsStream(java.lang.String))|1|http://stackoverflow.com/questions/2308188/getresourceasstream-vs-fileinputstream/2308388#comment21307593_2308388^0|0|X|0|X|0|0|G|B|1J|1A|2W|35|1|0^^$0|@$1|2|3|4|5|6|7|R|8|@$9|S|A|T|B|C]]|D|@$9|U|A|V|1|W]]|E|$]]|$1|F|3|G|5|6|7|X|8|@$9|Y|A|Z|B|C]|$9|10|A|11|B|C]]|D|@$9|12|A|13|1|14]]|E|$]]|$1|H|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|O]]|P|$5|K|L|M|E|$N|Q]]]]

<a href="http://download.oracle.com/javase/6/docs/api/java/lang/ClassLoader.html#getResourceAsStream(java.lang.String)" rel="noreferrer"><code>ClassLoader.getResourceAsStream()</code></a>.

As stated in the comment below, if you are in a multi-<code>ClassLoader</code> environment (such as unit testing, webapps, etc.) you may need to use <code>Thread.currentThread().getContextClassLoader()</code>. See <a href="http://stackoverflow.com/questions/2308188/getresourceasstream-vs-fileinputstream/2308388#comment21307593_2308388">http://stackoverflow.com/questions/2308188/getresourceasstream-vs-fileinputstream/2308388#comment21307593_2308388</a>.

blocks|key|1839073|text|someClassWithinYourSourceDir.getClass().getResourceAsStream()；|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1839074|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

someClassWithinYourSourceDir.getClass().getResourceAsStream();

blocks|key|2053335|text|这取决于XML文件的确切位置。它是在sources文件夹(在"default+package“或”root“中)还是在与类相同的文件夹中？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2053336|对于前一种情况，您必须使用"/file.xml“(注意前导斜杠)来查找文件，使用哪个类来查找它并不重要。|offset|length|style|CODE|2053337|如果XML文件紧挨着某个类，那么只需要使用带有文件名的SomeClass.class.getResourceAsStream()。|2053338|entityMap^0|0|E|9|0|R|11|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@$D|N|E|O|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|P|8|@$D|Q|E|R|F|G]]|9|@]|A|$]]|$1|J|3|-4|5|6|7|S|8|@]|9|@]|A|$]]]|K|$]]

That depends on where exactly the XML file is. Is it in the sources folder (in the "default package" or the "root") or in the same folder as the class?

In for former case, you must use "<code>/file.xml</code>" (note the leading slash) to find the file and it doesn't matter which class you use to try to locate it.

If the XML file is next to some class, <code>SomeClass.class.getResourceAsStream()</code> with just the filename is the way to go.

blocks|key|1839188|text|并确保您的编译脚本将ClassLoader.class.getResourceAsStream("/path/to/your/xml")文件复制到您的CLASSPATH中的位置。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1839189|entityMap^0|A|1M|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

<code>ClassLoader.class.getResourceAsStream("/path/to/your/xml")</code> and make sure that your compile script is copying the xml file to where in your CLASSPATH.

blocks|key|1839194|text|ClassLoader.class.getResourceAsStream("/path/file.ext");|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1839195|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>ClassLoader.class.getResourceAsStream("/path/file.ext");
</code></pre>

blocks|key|1839233|text|这个答案中的一些"getResourceAsStream()“选项对我不起作用，但这个选项起作用了：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1839234|SomeClassWithinYourSourceDir.class.getClassLoader().getResourceAsStream("yourResource")；|1839235|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

Some of the "getResourceAsStream()" options in this answer didn't work for me, but this one did:

SomeClassWithinYourSourceDir.class.getClassLoader().getResourceAsStream("yourResource");

blocks|key|1839340|text|我尝试了建议的解决方案，但文件名中的正斜杠对我无效，例如：...().getResourceAsStream("/my.properties")；返回null|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1839341|去掉斜杠：....getResourceAsStream("my.properties")；|1839342|这里来自doc+API:在委托之前，使用以下算法从给定的资源名称构造绝对资源名称：|1839343|If+the+name+begins+with+a+'/'+('\u002f'),+then+the+absolute+name+of+the+resource+is+the+portion+of+the+name+following+the+'/'.
Otherwise,+the+absolute+name+is+of+the+following+form:

++++modified_package_name/name+

Where+the+modified_package_name+is+the+package+name+of+this+object+with+'/'+substituted+for+'.'+('\u002e').+|code-block|syntax|javascript|1839344|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|O|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|P|8|@]|9|@]|A|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I tried proposed solution and forward slash in the file name did not work for me, example: ...().getResourceAsStream("/my.properties"); null was returned

Removing the slash worked: ....getResourceAsStream("my.properties"); 

Here is from doc API:
 Before delegation, an absolute resource name is constructed from the given resource name using this algorithm:

<pre><code>If the name begins with a '/' ('\u002f'), then the absolute name of the resource is the portion of the name following the '/'.
Otherwise, the absolute name is of the following form:

 modified_package_name/name 

Where the modified_package_name is the package name of this object with '/' substituted for '.' ('\u002e'). 
</code></pre>

In Java web application, Suppose if I want to get the InputStream of an XML file, which is placed in the CLASSPATH (i.e. inside the sources folder), how do I do it?

Getting the inputstream from a classpath resource (XML file)

单元测试

Java

在Java web应用程序中，假设我想要获取一个XML文件的InputStream，它被放在CLASSPATH中(即在sources文件夹中)，我该怎么做呢？

问从类路径资源(XML文件)获取输入流
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