blocks|key|138455|text|尝试使用ClassLoader类：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|138456|ClassLoader+classLoader+=+getClass().getClassLoader();
File+file+=+new+File(classLoader.getResource("somefile").getFile());
System.out.println(file.getAbsolutePath());|code-block|syntax|javascript|138457|ClassLoader负责在类中加载。每个类都有一个对ClassLoader的引用。这段代码从资源目录返回一个File。对它调用getAbsolutePath()将返回它的绝对Path。|138458|ClassLoader的Javadoc：http://docs.oracle.com/javase/7/docs/api/java/lang/ClassLoader.html|138459|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/7/docs/api/java/lang/ClassLoader.html^0|4|B|0|0|0|B|R|B|1J|4|1S|H|2G|4|0|0|B|K|1V|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@$9|11|A|12|B|C]|$9|13|A|14|B|C]|$9|15|A|16|B|C]|$9|17|A|18|B|C]|$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|1B|8|@$9|1C|A|1D|B|C]]|D|@$9|1E|A|1F|1|1G]]|E|$]]|$1|O|3|-4|5|6|7|1H|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]]]

Try working with the <code>ClassLoader</code> class: 

<pre><code>ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource("somefile").getFile());
System.out.println(file.getAbsolutePath());
</code></pre>

A <code>ClassLoader</code> is responsible for loading in classes. Every class has a reference to a <code>ClassLoader</code>. This code returns a <code>File</code> from the resource directory. Calling <code>getAbsolutePath()</code> on it returns its absolute <code>Path</code>.

Javadoc for <code>ClassLoader</code>: <a href="http://docs.oracle.com/javase/7/docs/api/java/lang/ClassLoader.html" rel="noreferrer">http://docs.oracle.com/javase/7/docs/api/java/lang/ClassLoader.html</a>

blocks|key|140885|text|你不需要和类加载器打交道。实际上，这是一个不好的习惯，因为当类加载器资源在jar归档中时，它们是而不是+java.io.File对象。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|140886|140887|++++File+resourcesDirectory+=+new+File("src/test/resources");|code-block|syntax|javascript|140888|如果您确实需要的话，resourcesDirectory.getAbsolutePath()将返回正确的值。|CODE|140889|这让你清楚地知道你在做什么。|140890|entityMap^0|1C|3|0|0|0|A|10|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|V|8|@]|D|@]|E|$]]|$1|G|3|H|5|I|7|W|8|@]|D|@]|E|$J|K]]|$1|L|3|M|5|6|7|X|8|@$9|Y|A|Z|B|N]]|D|@]|E|$]]|$1|O|3|P|5|6|7|10|8|@]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|R|$]]

You don't need to mess with class loaders. In fact it's a bad habit to get into because class loader resources are not java.io.File objects when they are in a jar archive.

Maven automatically sets the current working directory before running tests, so you can just use:

<pre><code> File resourcesDirectory = new File("src/test/resources");
</code></pre>

<code>resourcesDirectory.getAbsolutePath()</code> will return the correct value if that is what you really need.

I recommend creating a <code>src/test/data</code> directory if you want your tests to access data via the file system. This makes it clear what you're doing.

blocks|key|139767|text|我有一个使用JUnit+4.12和Java8的Maven3项目。为了获得src/test/resources下一个名为myxml.xml的文件的路径，我在测试用例中执行以下操作：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|139768|@Test
public+void+testApp()
{
++++File+inputXmlFile+=+new+File(this.getClass().getResource("/myxml.xml").getFile());
++++System.out.println(inputXmlFile.getAbsolutePath());
++++...
}|code-block|syntax|javascript|139769|在带有IntelliJ集成开发环境的Ubuntu14.04上测试。参考here。|139770|entityMap|0|LINK|mutability|MUTABLE|url|http://jlorenzen.blogspot.de/2007/06/proper-way-to-access-file-resources-in.html^0|10|I|1N|9|0|0|Z|4|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|D|@$9|11|A|12|1|13]]|E|$]]|$1|M|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

I have a Maven3 project using JUnit 4.12 and Java8.
In order to get the path of a file called <code>myxml.xml</code> under <code>src/test/resources</code>, I do this from within the test case:

<pre><code>@Test
public void testApp()
{
 File inputXmlFile = new File(this.getClass().getResource("/myxml.xml").getFile());
 System.out.println(inputXmlFile.getAbsolutePath());
 ...
}
</code></pre>

Tested on Ubuntu 14.04 with IntelliJ IDE.
Reference <a href="http://jlorenzen.blogspot.de/2007/06/proper-way-to-access-file-resources-in.html" rel="noreferrer">here</a>.

blocks|key|140892|text|@Steve+C和@ashosborne1提供的选项存在差异和限制。我相信，它们必须被明确说明。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|140893|什么时候可以使用：File+resourcesDirectory+=+new+File("src/test/resources");|offset|length|style|CODE|140894|当测试将只通过maven运行而不是通过集成开发环境运行时，|140895|1。当测试将通过maven或|unordered-list-item|140896|2.2通过集成开发环境运行，并且只有一个项目导入到集成开发环境中时，使用|140897|2.1。(我使用“导入”一词，因为它在IntelliJ+IDEA中使用。我认为eclipse的用户也会导入他们的maven项目)。|140898|3.1当测试通过集成开发环境运行，并且多个项目被导入到集成开发环境中(当您不是学生时，通常导入几个项目)，和之前，您需要手动配置测试的工作目录。该工作目录应该引用包含测试的导入项目。默认情况下，导入到IDE中的所有项目的工作目录只有一个。这可能只是IntelliJ+IDEA的一个限制，但我认为所有IDE都是这样工作的。而这种必须手动完成的配置，一点也不好。使用存在于不同maven项目中的几个测试，但是导入到一个大的“集成开发环境”项目中，迫使我们记住这一点，并且不允许放松并从您的工作中获得乐趣。|BOLD|140899|140900|@ashosborne1提供的解决方案(我个人更喜欢这个)需要两个额外的要求，这两个要求必须在运行测试之前完成。以下是使用此解决方案的步骤列表：|140901|140902|在“src/+test+/|140903|/”中创建一个测试文件夹(“teva”)和文件(“readme”)：|140904|140905|src/test/resources/teva/readme|140906|必须在文件夹中创建测试文件，否则将无法工作。Maven忽略空folders.|140907|140908|至少一次通过mvn+clean+install构建项目。它还将运行测试。通过maven只运行您的测试类/方法而不构建整个项目可能就足够了。因此，您的测试资源将被复制到测试类中，路径如下：target/test-classes/teva/readme|140909|140910|140911|之后，您可以使用@ashosborne1提供的代码访问该文件夹(很抱歉，我无法在此项目列表中正确编辑此代码)：|140912|140913|140914|公共静态最终字符串TEVA_FOLDER+=+"teva"；...URL+tevaUrl+=+YourTest.class.getClassLoader().getResource(TEVA_FOLDER)；String+tevaTestFolder+=新建文件(tevaUrl.toURI()).getAbsolutePath()；|blockquote|140915|140916|140917|现在，您可以通过IDE多次运行测试。直到你干净地运行mvn。它将删除目标文件夹。|140918|在通过IDE运行测试之前，在test文件夹中创建文件和第一次运行maven是必要的步骤。如果没有这些步骤，如果您只是在IDE中创建测试资源，然后编写测试并仅通过IDE运行它，您将得到一个错误。通过mvn运行测试将测试资源复制到target/+test+-classes/teva/readme中，这样类加载器就可以访问它们了。|140919|你可能会问，为什么我需要在IDE中导入多个maven项目，为什么会有这么多复杂的事情？对我来说，主要的动机之一是:使与IDA相关的文件远离代码。我首先在IDE中创建一个新项目。这是一个假项目，只是IDE相关文件的持有者。然后，我导入已经存在的maven项目。我强制这些导入的项目只在我的原始假项目中保留IDEA文件。因此，我在代码中看不到与IDE相关的文件。SVN应该看不到它们(请不要提供配置svn/git忽略这些文件)。这也是非常方便的。|140920|entityMap^0|0|9|1L|0|0|0|0|0|1H|1|3G|D|0|0|0|0|0|0|0|0|3|Z|0|0|6|H|2L|V|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1T|8|@$D|1U|E|1V|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|1W|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|1X|8|@]|9|@]|A|$]]|$1|M|3|N|5|L|7|1Y|8|@]|9|@]|A|$]]|$1|O|3|P|5|L|7|1Z|8|@]|9|@]|A|$]]|$1|Q|3|R|5|L|7|20|8|@$D|21|E|22|F|S]|$D|23|E|24|F|G]]|9|@]|A|$]]|$1|T|3|-4|5|6|7|25|8|@]|9|@]|A|$]]|$1|U|3|V|5|6|7|26|8|@]|9|@]|A|$]]|$1|W|3|-4|5|6|7|27|8|@]|9|@]|A|$]]|$1|X|3|Y|5|L|7|28|8|@]|9|@]|A|$]]|$1|Z|3|10|5|L|7|29|8|@]|9|@]|A|$]]|$1|11|3|-4|5|6|7|2A|8|@]|9|@]|A|$]]|$1|12|3|13|5|6|7|2B|8|@]|9|@]|A|$]]|$1|14|3|15|5|6|7|2C|8|@$D|2D|E|2E|F|S]]|9|@]|A|$]]|$1|16|3|-4|5|6|7|2F|8|@]|9|@]|A|$]]|$1|17|3|18|5|L|7|2G|8|@$D|2H|E|2I|F|G]|$D|2J|E|2K|F|G]]|9|@]|A|$]]|$1|19|3|-4|5|6|7|2L|8|@]|9|@]|A|$]]|$1|1A|3|-4|5|6|7|2M|8|@]|9|@]|A|$]]|$1|1B|3|1C|5|L|7|2N|8|@]|9|@]|A|$]]|$1|1D|3|-4|5|6|7|2O|8|@]|9|@]|A|$]]|$1|1E|3|-4|5|6|7|2P|8|@]|9|@]|A|$]]|$1|1F|3|1G|5|1H|7|2Q|8|@]|9|@]|A|$]]|$1|1I|3|-4|5|6|7|2R|8|@]|9|@]|A|$]]|$1|1J|3|-4|5|6|7|2S|8|@]|9|@]|A|$]]|$1|1K|3|1L|5|6|7|2T|8|@]|9|@]|A|$]]|$1|1M|3|1N|5|6|7|2U|8|@]|9|@]|A|$]]|$1|1O|3|1P|5|6|7|2V|8|@]|9|@]|A|$]]|$1|1Q|3|-4|5|6|7|2W|8|@]|9|@]|A|$]]]|1R|$]]

There are differences and constraints in options offered by @Steve C and @ashosborne1. They must be specified, I believe.
When can we can use: <code>File resourcesDirectory = new File(&quot;src/test/resources&quot;);</code>?
<ul>
<li>1 When tests are going to be run via maven only but not via IDE.</li>
<li>2.1 When tests are going to be run via maven or</li>
<li>2.2 via IDE and only one project is imported into IDE. (I use “imported” term, cause it is used in IntelliJ IDEA. I think users of eclipse also import their maven project). This will work, cause working directory when you run tests via IDE is the same as your project.</li>
<li>3.1 When tests are going to be run via maven or</li>
<li>3.2 via IDE, and more than one projects are imported into IDE (when you are not a student, you usually import several projects), AND before you run tests via IDE, you manually configure working directory for your tests. That working directory should refer to your imported project that contains the tests. By default, working directory of all projects imported into IDE is only one. Probably it is a restriction of <code>IntelliJ IDEA</code> only, but I think all IDEs work like this. And this configuration that must be done manually, is not good at all. Working with several tests existing in different maven projects, but imported into one big “IDE” project, force us to remember this and don’t allow to relax and get pleasure from your work.</li>
</ul>
Solution offered by @ashosborne1 (personally I prefer this one) requires 2 additional requirements that must be done before you run tests. Here is a list of steps to use this solution:
<ul>
<li>Create a test folder (“teva”) and file (“readme”) inside of “src/test/resources/”:
src/test/resources/teva/readme
File must be created in the test folder, otherwise, it will not work. Maven ignores empty folders.
</li>
<li>At least once build project via <code>mvn clean install</code>. It will run tests also. It may be enough to run only your test class/method via maven without building a whole project. As a result your test resources will be copied into test-classes, here is a path: <code>target/test-classes/teva/readme</code>
</li>
<li>After that, you can access the folder using code, already offered by @ashosborne1 (I'm sorry, that I could not edit this code inside of this list of items correctly):
</li>
</ul>
<blockquote>
<pre><code>public static final String TEVA_FOLDER = &quot;teva&quot;; ... 
URL tevaUrl = YourTest.class.getClassLoader().getResource(TEVA_FOLDER); 
String tevaTestFolder = new File(tevaUrl.toURI()).getAbsolutePath();
</code></pre>
</blockquote>
Now you can run your test via IDE as many times as you want. Until you run mvn clean. It will drop the target folder.
Creating file inside a test folder and running maven first time, before you run tests via IDE are needed steps. Without these steps, if you just in your IDE create test resources, then write test and run it via IDE only, you'll get an error. Running tests via mvn copies test resources into target/test-classes/teva/readme and they become accessible for a classloader.
You may ask, why do I need import more than one maven project in IDE and why so many complicated things? For me, one of the main motivation: keeping IDA-related files far from code. I first create a new project in my IDE. It is a fake project, that is just a holder of IDE-related files. Then, I import already existing maven projects. I force these imported projects to keep IDEA files in my original fake project only. As a result I don't see IDE-related files among the code. SVN should not see them (don't offer to configure svn/git to ignore such files, please). Also it is just very convenient.

blocks|key|142014|text|如果是spring项目，我们可以使用下面的代码从src/test/resource文件夹中获取文件。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|142015|File+file+=+ResourceUtils.getFile(this.getClass().getResource("/some_file.txt"));|code-block|syntax|javascript|142016|entityMap^0|O|H|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

If it's a spring project, we can use the below code to get files from src/test/resource folder.

<pre><code>File file = ResourceUtils.getFile(this.getClass().getResource("/some_file.txt"));
</code></pre>

blocks|key|142018|text|我将简单地使用Java7中的Path|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|142019|Path+resourceDirectory+=+Paths.get("src","test","resources");|code-block|syntax|javascript|142020|整洁干净！|142021|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.oracle.com/javase/7/docs/api/java/nio/file/Path.html^0|E|4|E|4|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@$9|X|A|Y|1|Z]]|E|$]]|$1|F|3|G|5|H|7|10|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|11|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

I would simply use <a href="https://docs.oracle.com/javase/7/docs/api/java/nio/file/Path.html" rel="noreferrer"><code>Path</code></a> from Java 7

<pre><code>Path resourceDirectory = Paths.get("src","test","resources");
</code></pre>

Neat and clean!

blocks|key|138593|text|我使用的最简单干净的解决方案是，假设测试类的名称是TestQuery1，并且在test文件夹中有一个resources目录，如下所示：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|138594|├──+java
│  +└──+TestQuery1.java
└──+resources
++++└──+TestQuery1
++++++++├──+query.json
++++++++└──+query.rq|code-block|syntax|javascript|138595|要获取TestQuery1的URI，请执行以下操作：|138596|URL+currentTestResourceFolder+=+getClass().getResource("/"%2BgetClass().getSimpleName());|138597|要获取其中一个文件TestQuery1的URI，请执行以下操作：|138598|File+exampleDir+=+new+File(currentTestResourceFolder.toURI());
URI+queryJSONFileURI+=+exampleDir.toURI().resolve("query.json");|138599|entityMap^0|P|A|13|4|1E|9|0|0|3|A|0|0|9|A|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|16|8|@$9|17|A|18|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|H|7|19|8|@]|D|@]|E|$I|J]]|$1|S|3|-4|5|6|7|1A|8|@]|D|@]|E|$]]]|T|$]]

The simplest and clean solution I uses, suppose the name of the test class is <code>TestQuery1</code> and there is a <code>resources</code> directory in your <code>test</code> folder as follows:

<pre><code>├── java
│   └── TestQuery1.java
└── resources
 └── TestQuery1
 ├── query.json
 └── query.rq
</code></pre>

To get the URI of <code>TestQuery1</code> do:

<pre><code>URL currentTestResourceFolder = getClass().getResource("/"+getClass().getSimpleName());
</code></pre>

To get the URI of one of the file <code>TestQuery1</code>, do:

<pre><code>File exampleDir = new File(currentTestResourceFolder.toURI());
URI queryJSONFileURI = exampleDir.toURI().resolve("query.json");
</code></pre>

blocks|key|138601|text|在文件对象上使用.getAbsolutePath()。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|138602|getClass().getResource("somefile").getFile().getAbsolutePath()|code-block|syntax|javascript|138603|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Use .getAbsolutePath() on your File object.

<pre><code>getClass().getResource("somefile").getFile().getAbsolutePath()
</code></pre>

blocks|key|2712119|text|使用以下内容将Hibernate与Spring一起注入到您的单元测试中：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2712120|@Bean
public+LocalSessionFactoryBean+getLocalSessionFactoryBean()+{
++++LocalSessionFactoryBean+localSessionFactoryBean+=+new+LocalSessionFactoryBean();
++++localSessionFactoryBean.setConfigLocation(new+ClassPathResource("hibernate.cfg.xml"));
++++localSessionFactoryBean.setPackagesToScan("com.example.yourpackage.model");
++++return+localSessionFactoryBean;
}|code-block|syntax|javascript|2712121|如果您的src/test/resources文件夹中没有该hibernate.cfg.xml，它将自动回退到您的src/main/resources文件夹中。|offset|length|style|CODE|2712122|entityMap^0|0|0|4|I|T|H|1K|I|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]|$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|X|8|@]|9|@]|A|$]]]|N|$]]

Use the following to inject Hibernate with Spring in your unit tests:

<pre><code>@Bean
public LocalSessionFactoryBean getLocalSessionFactoryBean() {
 LocalSessionFactoryBean localSessionFactoryBean = new LocalSessionFactoryBean();
 localSessionFactoryBean.setConfigLocation(new ClassPathResource("hibernate.cfg.xml"));
 localSessionFactoryBean.setPackagesToScan("com.example.yourpackage.model");
 return localSessionFactoryBean;
}
</code></pre>

If you don't have the <code>hibernate.cfg.xml</code> present in your <code>src/test/resources</code> folder it will automatically fall back to the one in your <code>src/main/resources</code> folder.

blocks|key|142044|text|src/test/resources中的所有内容都将复制到target/test-classes文件夹中。因此，要在maven构建期间从测试资源中获取文件，您必须从test-classes文件夹加载它，如下所示：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|142045|Paths.get(
++++getClass().getProtectionDomain().getCodeSource().getLocation().toURI()
).resolve(
++++Paths.get("somefile")
).toFile()|code-block|syntax|javascript|142046|细分：|142047|142048|getClass().getProtectionDomain().getCodeSource().getLocation().toURI()+-为您提供target/test-classes.|ordered-list-item|142049|resolve(Paths.get("somefile"))的URI+-将someFile解析为target/test-classes文件夹。|142050|142051|原始答案取自this|142052|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/12178523/1251549^0|0|I|T|J|2A|C|0|0|0|0|0|1Y|24|J|0|0|U|11|8|1C|J|0|0|6|4|0|0^^$0|@$1|2|3|4|5|6|7|13|8|@$9|14|A|15|B|C]|$9|16|A|17|B|C]|$9|18|A|19|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|1A|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1B|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]|$1|N|3|O|5|P|7|1D|8|@$9|1E|A|1F|B|C]|$9|1G|A|1H|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|P|7|1I|8|@$9|1J|A|1K|B|C]|$9|1L|A|1M|B|C]|$9|1N|A|1O|B|C]]|D|@]|E|$]]|$1|S|3|-4|5|6|7|1P|8|@]|D|@]|E|$]]|$1|T|3|U|5|6|7|1Q|8|@]|D|@$9|1R|A|1S|1|1T]]|E|$]]|$1|V|3|-4|5|6|7|1U|8|@]|D|@]|E|$]]]|W|$X|$5|Y|Z|10|E|$11|12]]]]

All content in <code>src/test/resources</code> is copied into <code>target/test-classes</code> folder. So to get file from test resources during maven build you have to load it from <code>test-classes</code> folder, like that:

<pre><code>Paths.get(
 getClass().getProtectionDomain().getCodeSource().getLocation().toURI()
).resolve(
 Paths.get("somefile")
).toFile()
</code></pre>

Break down:

<ol>
<li><code>getClass().getProtectionDomain().getCodeSource().getLocation().toURI()</code> - give you URI to <code>target/test-classes</code>.</li>
<li><code>resolve(Paths.get("somefile"))</code> - resolves <code>someFile</code> to <code>target/test-classes</code> folder.</li>
</ol>

Original anwser is taken from <a href="https://stackoverflow.com/a/12178523/1251549">this</a>

blocks|key|141031|text|在常见情况下，您不能使用资源文件夹中的文件进行测试。原因是资源文件夹中的资源文件存储在jar中。所以它们在文件系统中没有真正的路径。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|141032|最简单的解决方案可以是：|141033|141034|将文件从资源复制到临时文件夹，并获取该临时文件的路径。|ordered-list-item|141035|使用临时路径进行测试。|141036|删除临时文件。|141037|141038|来自JUnit的TemporaryFolder可用于创建临时文件，并在测试完成后将其删除。guava库中的类用于从资源文件夹中复制文件。|offset|length|style|CODE|141039|请注意，如果我们使用资源文件夹中的子文件夹，比如good+one，我们不必在资源路径中添加前导/。|141040|public+class+SomeTest+{

++++@Rule
++++public+TemporaryFolder+tmpFolder+=+new+TemporaryFolder();


++++@Test
++++public+void+doSomethinge()+throws+IOException+{
++++++++File+file+=+createTmpFileFromResource(tmpFolder,+"file.txt");
++++++++File+goodFile+=+createTmpFileFromResource(tmpFolder,+"good/file.txt");

++++++++//+do+testing+here
++++}

++++private+static+File+createTmpFileFromResource(TemporaryFolder+folder,
++++++++++++++++++++++++++++++++++++++++++++++++++String+classLoaderResource)+throws+IOException+{
++++++++URL+resource+=+Resources.getResource(classLoaderResource);

++++++++File+tmpFile+=+folder.newFile();
++++++++Resources.asByteSource(resource).copyTo(Files.asByteSink(tmpFile));
++++++++return+tmpFile;
++++}

}|code-block|syntax|javascript|141041|entityMap^0|0|0|0|0|0|0|0|2|5|8|F|19|5|0|O|4|1B|1|0|0^^$0|@$1|2|3|4|5|6|7|11|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|12|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|13|8|@]|9|@]|A|$]]|$1|E|3|F|5|G|7|14|8|@]|9|@]|A|$]]|$1|H|3|I|5|G|7|15|8|@]|9|@]|A|$]]|$1|J|3|K|5|G|7|16|8|@]|9|@]|A|$]]|$1|L|3|-4|5|6|7|17|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|18|8|@$O|19|P|1A|Q|R]|$O|1B|P|1C|Q|R]|$O|1D|P|1E|Q|R]]|9|@]|A|$]]|$1|S|3|T|5|6|7|1F|8|@$O|1G|P|1H|Q|R]|$O|1I|P|1J|Q|R]]|9|@]|A|$]]|$1|U|3|V|5|W|7|1K|8|@]|9|@]|A|$X|Y]]|$1|Z|3|-4|5|6|7|1L|8|@]|9|@]|A|$]]]|10|$]]

You can't use a file from a resource folder for tests in a common case. The reason is that resource files in the resource folder are stored inside a jar. So they don't have a real path in the file system.

The most simple solution can be:

<ol>
<li>Copy a file from resources to the temporary folder and get a path to that temporary file.</li>
<li>Do tests using a temporary path.</li>
<li>Delete the temporary file.</li>
</ol>

<code>TemporaryFolder</code> from <code>JUnit</code> can be used to create temporary files and delete it after test is complited. Classes from <code>guava</code> library are used to copy a file form resource folder. 

Please, notice that if we use a subfolder in the resources folder, like <code>good</code> one, we don't have to add leading <code>/</code> to the resource path.

<pre><code>public class SomeTest {

 @Rule
 public TemporaryFolder tmpFolder = new TemporaryFolder();


 @Test
 public void doSomethinge() throws IOException {
 File file = createTmpFileFromResource(tmpFolder, "file.txt");
 File goodFile = createTmpFileFromResource(tmpFolder, "good/file.txt");

 // do testing here
 }

 private static File createTmpFileFromResource(TemporaryFolder folder,
 String classLoaderResource) throws IOException {
 URL resource = Resources.getResource(classLoaderResource);

 File tmpFile = folder.newFile();
 Resources.asByteSource(resource).copyTo(Files.asByteSink(tmpFile));
 return tmpFile;
 }

}
</code></pre>

blocks|key|142069|text|使用Spring，您可以很容易地从资源文件夹(+main/resources或test/+resources+)中读取它：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|142070|例如，创建一个文件：test/resources/subfolder/sample.json|offset|length|style|CODE|142071|@Test
public+void+testReadFile()+{
++++String+json+=+this.readFile("classpath:subfolder/sample.json");
++++System.out.println(json);
}

public+String+readFile(String+path)+{
++++try+{
++++++++File+file+=+ResourceUtils.getFile(path);
++++++++return+new+String(Files.readAllBytes(file.toPath()));
++++}+catch+(IOException+e)+{
++++++++e.printStackTrace();
++++}

++++return+null;
}|code-block|syntax|javascript|142072|entityMap^0|0|A|10|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@$D|Q|E|R|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|S|8|@]|9|@]|A|$K|L]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

With Spring you could easily read it from the resources folder (either main/resources or test/resources):

For example create a file: <code>test/resources/subfolder/sample.json</code>

<pre><code>@Test
public void testReadFile() {
 String json = this.readFile("classpath:subfolder/sample.json");
 System.out.println(json);
}

public String readFile(String path) {
 try {
 File file = ResourceUtils.getFile(path);
 return new String(Files.readAllBytes(file.toPath()));
 } catch (IOException e) {
 e.printStackTrace();
 }

 return null;
}
</code></pre>

blocks|key|2924778|text|List<String>+lines+=+Files.readAllLines(Paths.get("src/test/resources/foo.txt"));
lines.forEach(System.out::println);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2924779|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>List&lt;String&gt; lines = Files.readAllLines(Paths.get("src/test/resources/foo.txt"));
lines.forEach(System.out::println);
</code></pre>

blocks|key|2712263|text|在Spring中，您可以使用以下代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2712264|import+org.springframework.core.io.ClassPathResource;

//+Don't+worry+when+use+a+not+existed+directory+or+a+empty+directory
//+It+can+be+used+in+@before
String+dir+=+new+ClassPathResource(".").getFile().getAbsolutePath()%2B"/"%2B"Your+Path";|code-block|syntax|javascript|2712265|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

With Spring, you can use this:
<pre class="lang-java prettyprint-override"><code>import org.springframework.core.io.ClassPathResource;

// Don't worry when use a not existed directory or a empty directory
// It can be used in @before
String dir = new ClassPathResource(&quot;.&quot;).getFile().getAbsolutePath()+&quot;/&quot;+&quot;Your Path&quot;;
</code></pre>

blocks|key|2924897|text|你可以到达你所在的地方|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2924898|new+File(".").getAbsolutePath()|code-block|syntax|javascript|2924899|然后，您可以派生到src/test/resources的路径|2924900|通常情况下它只是|2924901|new+File("src/test/resources")|2924902|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

You can get where you are through
<pre><code>new File(&quot;.&quot;).getAbsolutePath()
</code></pre>
then you can derive path to src/test/resources 
usually it is just
<pre><code>new File(&quot;src/test/resources&quot;)
</code></pre>

blocks|key|2924917|text|哇，正确答案还没出来呢！|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2924918|MyClass.class.getResource("/somefile");
MyClass.class.getResourceAsStream("/somefile");|code-block|syntax|javascript|2924919|https://javachannel.org/posts/how-to-access-static-resources/|offset|length|2924920|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|0|1P|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|T|8|@]|9|@$I|U|J|V|1|W]]|A|$]]|$1|K|3|-4|5|6|7|X|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|H]]]]

Wow the correct answer is not here yet!
<pre><code>MyClass.class.getResource(&quot;/somefile&quot;);
MyClass.class.getResourceAsStream(&quot;/somefile&quot;);
</code></pre>
<a href="https://javachannel.org/posts/how-to-access-static-resources/" rel="nofollow noreferrer">https://javachannel.org/posts/how-to-access-static-resources/</a>

I know I can load a file from src/test/resources with:

<pre><code>getClass().getResource("somefile").getFile()
</code></pre>

But how can I get the full path to the src/test/resources directory, i.e. I don't want to load a file, I just want to know the path of the directory?

How to get the path of src/test/resources directory in JUnit?

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我知道我可以使用以下命令从src/test/resources加载文件：getClass().getResource("somefile").getFile()但是，我如何才能获得src//目录的完整路径，也就是说，我不想加载文件，我只想知道该目录的路径？

问如何获取JUnit中src/test/resources目录的路径？
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