blocks|key|1122846|text|首先，确保将abc.xml复制到您的输出目录。那么你应该使用getResourceAsStream()|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1122847|InputStream+inputStream+=+
++++Thread.currentThread().getContextClassLoader().getResourceAsStream("test/resources/abc.xml");|code-block|syntax|javascript|1122848|一旦你有了InputStream，你只需要把它转换成一个字符串。这个参考资料详细说明了这一点：http://www.kodejava.org/examples/266.html。但是，我将摘录相关代码：|1122849|public+String+convertStreamToString(InputStream+is)+throws+IOException+{
++++if+(is+!=+null)+{
++++++++Writer+writer+=+new+StringWriter();

++++++++char[]+buffer+=+new+char[1024];
++++++++try+{
++++++++++++Reader+reader+=+new+BufferedReader(
++++++++++++++++++++new+InputStreamReader(is,+"UTF-8"));
++++++++++++int+n;
++++++++++++while+((n+=+reader.read(buffer))+!=+-1)+{
++++++++++++++++writer.write(buffer,+0,+n);
++++++++++++}
++++++++}+finally+{
++++++++++++is.close();
++++++++}
++++++++return+writer.toString();
++++}+else+{++++++++
++++++++return+"";
++++}
}|1122850|entityMap|0|LINK|mutability|MUTABLE|url|http://www.kodejava.org/examples/266.html^0|6|7|U|L|0|0|1B|15|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$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|@]|D|@$9|13|A|14|1|15]]|E|$]]|$1|M|3|N|5|H|7|16|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]]]

First make sure that <code>abc.xml</code> is being copied to your output directory. Then you should use <code>getResourceAsStream()</code>:

<pre><code>InputStream inputStream = 
 Thread.currentThread().getContextClassLoader().getResourceAsStream("test/resources/abc.xml");
</code></pre>

Once you have the InputStream, you just need to convert it into a string. This resource spells it out: <a href="http://www.kodejava.org/examples/266.html" rel="noreferrer">http://www.kodejava.org/examples/266.html</a>. However, I'll excerpt the relevent code:

<pre><code>public String convertStreamToString(InputStream is) throws IOException {
 if (is != null) {
 Writer writer = new StringWriter();

 char[] buffer = new char[1024];
 try {
 Reader reader = new BufferedReader(
 new InputStreamReader(is, "UTF-8"));
 int n;
 while ((n = reader.read(buffer)) != -1) {
 writer.write(buffer, 0, n);
 }
 } finally {
 is.close();
 }
 return writer.toString();
 } else { 
 return "";
 }
}
</code></pre>

blocks|key|1122918|text|最后，我找到了一个简洁的解决方案，多亏了Apache+Commons|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1122919|package+com.example;
import+org.apache.commons.io.IOUtils;
public+class+FooTest+{
++@Test+
++public+void+shouldWork()+throws+Exception+{
++++String+xml+=+IOUtils.toString(
++++++this.getClass().getResourceAsStream("abc.xml"),
++++++"UTF-8"
++++);
++}
}|code-block|syntax|javascript|1122920|效果很好。文件src/test/resources/com/example/abc.xml已加载(我使用的是Maven)。|style|CODE|1122921|比方说，如果用"/foo/test.xml"替换"abc.xml"，则将加载此资源：src/test/resources/foo/test.xml|1122922|您也可以使用Cactoos|1122923|package+com.example;
import+org.cactoos.io.ResourceOf;
import+org.cactoos.io.TextOf;
public+class+FooTest+{
++@Test+
++public+void+shouldWork()+throws+Exception+{
++++String+xml+=+new+TextOf(
++++++new+ResourceOf("/com/example/abc.xml")+//+absolute+path+always!
++++).asString();
++}
}|1122924|entityMap|0|LINK|mutability|MUTABLE|url|https://commons.apache.org/proper/commons-io/|1|http://www.cactoos.org/^0|K|E|0|0|0|7|12|0|7|F|O|9|16|V|0|6|7|1|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@$A|13|B|14|1|15]]|C|$]]|$1|D|3|E|5|F|7|16|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|17|8|@$A|18|B|19|K|L]]|9|@]|C|$]]|$1|M|3|N|5|6|7|1A|8|@$A|1B|B|1C|K|L]|$A|1D|B|1E|K|L]|$A|1F|B|1G|K|L]]|9|@]|C|$]]|$1|O|3|P|5|6|7|1H|8|@]|9|@$A|1I|B|1J|1|1K]]|C|$]]|$1|Q|3|R|5|F|7|1L|8|@]|9|@]|C|$G|H]]|$1|S|3|-4|5|6|7|1M|8|@]|9|@]|C|$]]]|T|$U|$5|V|W|X|C|$Y|Z]]|10|$5|V|W|X|C|$Y|11]]]]

Finally I found a neat solution, thanks to <a href="https://commons.apache.org/proper/commons-io/" rel="noreferrer">Apache Commons</a>:

<pre><code>package com.example;
import org.apache.commons.io.IOUtils;
public class FooTest {
 @Test 
 public void shouldWork() throws Exception {
 String xml = IOUtils.toString(
 this.getClass().getResourceAsStream("abc.xml"),
 "UTF-8"
 );
 }
}
</code></pre>

Works perfectly. File <code>src/test/resources/com/example/abc.xml</code> is loaded (I'm using Maven).

If you replace <code>"abc.xml"</code> with, say, <code>"/foo/test.xml"</code>, this resource will be loaded: <code>src/test/resources/foo/test.xml</code>

You can also use <a href="http://www.cactoos.org" rel="noreferrer">Cactoos</a>:

<pre><code>package com.example;
import org.cactoos.io.ResourceOf;
import org.cactoos.io.TextOf;
public class FooTest {
 @Test 
 public void shouldWork() throws Exception {
 String xml = new TextOf(
 new ResourceOf("/com/example/abc.xml") // absolute path always!
 ).asString();
 }
}
</code></pre>

blocks|key|908010|text|您可以尝试执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|908011|String+myResource+=+IOUtils.toString(this.getClass().getResourceAsStream("yourfile.xml")).replace("\n","");|code-block|syntax|javascript|908012|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|$]]

You can try doing:

<pre><code>String myResource = IOUtils.toString(this.getClass().getResourceAsStream("yourfile.xml")).replace("\n","");
</code></pre>

blocks|key|1123008|text|开门见山：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1123009|ClassLoader+classLoader+=+getClass().getClassLoader();
File+file+=+new+File(classLoader.getResource("file/test.xml").getFile());|code-block|syntax|javascript|1123010|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|$]]

Right to the point :

<pre><code>ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource("file/test.xml").getFile());
</code></pre>

blocks|key|1123032|text|使用Google+Guava：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1123033|import+com.google.common.base.Charsets;
import+com.google.common.io.Resources;

public+String+readResource(final+String+fileName,+Charset+charset)+throws+Exception+{
++++++++try+{
++++++++++++return+Resources.toString(Resources.getResource(fileName),+charset);
++++++++}+catch+(IOException+e)+{
++++++++++++throw+new+IllegalArgumentException(e);
++++++++}
}|code-block|syntax|javascript|1123034|示例：|1123035|String+fixture+=+this.readResource("filename.txt",+Charsets.UTF_8)|1123036|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

With the use of Google Guava:

<pre><code>import com.google.common.base.Charsets;
import com.google.common.io.Resources;

public String readResource(final String fileName, Charset charset) throws Exception {
 try {
 return Resources.toString(Resources.getResource(fileName), charset);
 } catch (IOException e) {
 throw new IllegalArgumentException(e);
 }
}
</code></pre>

Example:

<pre><code>String fixture = this.readResource("filename.txt", Charsets.UTF_8)
</code></pre>

blocks|key|1123085|text|下面是我用来获取包含文本的文本文件的方法。我使用了commons的IOUtils和guava的资源。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1123086|public+static+String+getString(String+path)+throws+IOException+{
++++try+(InputStream+stream+=+Resources.getResource(path).openStream())+{
++++++++return+IOUtils.toString(stream);
++++}
}|code-block|syntax|javascript|1123087|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|$]]

Here's what i used to get the text files with text. I used commons' IOUtils and guava's Resources.

<pre><code>public static String getString(String path) throws IOException {
 try (InputStream stream = Resources.getResource(path).openStream()) {
 return IOUtils.toString(stream);
 }
}
</code></pre>

blocks|key|1123120|text|您可以使用Junit+Rule为您的测试创建此临时文件夹：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1123121|@Rule+public+TemporaryFolder+temporaryFolder+=+new+TemporaryFolder();
File+file+=+temporaryFolder.newFile(".src/test/resources/abc.xml");|code-block|syntax|javascript|1123122|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|$]]

You can use a Junit Rule to create this temporary folder for your test:
<pre class="lang-java prettyprint-override"><code>@Rule public TemporaryFolder temporaryFolder = new TemporaryFolder();
File file = temporaryFolder.newFile(&quot;.src/test/resources/abc.xml&quot;);
</code></pre>

blocks|key|908181|text|好的，对于JAVA+8，经过大量的调试后，我发现|type|unstyled|depth|inlineStyleRanges|entityRanges|data|908182|URL+tenantPathURI+=+getClass().getResource("/test_directory/test_file.zip");|code-block|syntax|javascript|908183|和|908184|URL+tenantPathURI+=+getClass().getResource("test_directory/test_file.zip");|908185|是的，路径开头的/没有它，我得到的是null！|offset|length|style|CODE|908186|并且test_directory位于test目录下。|908187|entityMap^0|0|0|0|0|8|1|I|4|0|2|E|I|4|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|Y|8|@$M|Z|N|10|O|P]|$M|11|N|12|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|13|8|@$M|14|N|15|O|P]|$M|16|N|17|O|P]]|9|@]|A|$]]|$1|S|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|T|$]]

OK, for JAVA 8, after a lot of debugging
I found that there's a difference between
<pre><code>URL tenantPathURI = getClass().getResource(&quot;/test_directory/test_file.zip&quot;);
</code></pre>
and
<pre><code>URL tenantPathURI = getClass().getResource(&quot;test_directory/test_file.zip&quot;);
</code></pre>
Yes, the <code>/</code> at the beginning of the path without it I was getting <code>null</code>!
and the <code>test_directory</code> is under the <code>test</code> directory.

I have a unit test that needs to work with XML file located in <code>src/test/resources/abc.xml</code>. What is the easiest way just to get the content of the file into <code>String</code>?

How to read a text-file resource into Java unit test?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有一个单元测试，需要处理位于src/test/resources/abc.xml中的XML文件。将文件内容导入String的最简单方法是什么

问如何将文本文件资源读入Java单元测试？
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