blocks|key|4692335|text|如果你想以文件的形式阅读，我相信还有一个类似的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4692336|++++ClassLoader+classLoader+=+getClass().getClassLoader();
++++File+file+=+new+File(classLoader.getResource("file/test.xml").getFile());|code-block|syntax|javascript|4692337|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|$]]

If you wanna read as a file, I believe there still is a similar solution:

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

blocks|key|4692391|text|确保使用正确的分隔符。我用File.separator替换了相对路径中的所有/。这在IDE中工作得很好，但是在构建JAR中就不行了。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4692392|entityMap^0|D|E|12|1|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

Make sure that you work with the correct separator. I replaced all <code>/</code> in a relative path with a <code>File.separator</code>. This worked fine in the IDE, however did not work in the build JAR.

blocks|key|4692427|text|如果您使用的是spring，那么您可以使用以下方法从src/main/resources读取文件：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4692428|import+java.io.BufferedReader;
import+java.io.IOException;
import+java.io.InputStream;
import+java.io.InputStreamReader;
import+org.springframework.core.io.ClassPathResource;

++public+String+readFileToString(String+path)+throws+IOException+{

++++StringBuilder+resultBuilder+=+new+StringBuilder("");
++++ClassPathResource+resource+=+new+ClassPathResource(path);

++++try+(
++++++++InputStream+inputStream+=+resource.getInputStream();
++++++++BufferedReader+bufferedReader+=+new+BufferedReader(new+InputStreamReader(inputStream)))+{

++++++String+line;

++++++while+((line+=+bufferedReader.readLine())+!=+null)+{
++++++++resultBuilder.append(line);
++++++}

++++}

++++return+resultBuilder.toString();
++}|code-block|syntax|javascript|4692429|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|$]]

If you are using spring, then you can use the the following method to read file from src/main/resources:

<pre><code>import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import org.springframework.core.io.ClassPathResource;

 public String readFileToString(String path) throws IOException {

 StringBuilder resultBuilder = new StringBuilder("");
 ClassPathResource resource = new ClassPathResource(path);

 try (
 InputStream inputStream = resource.getInputStream();
 BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream))) {

 String line;

 while ((line = bufferedReader.readLine()) != null) {
 resultBuilder.append(line);
 }

 }

 return resultBuilder.toString();
 }
</code></pre>

blocks|key|4692456|text|我以前遇到过这个问题，所以我采取了后备的加载方式。基本上，第一种方法是在.jar文件中工作，第二种方法是在eclipse或其他集成开发环境中工作。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4692457|public+class+MyClass+{

++++public+static+InputStream+accessFile()+{
++++++++String+resource+=+"my-file-located-in-resources.txt";

++++++++//+this+is+the+path+within+the+jar+file
++++++++InputStream+input+=+MyClass.class.getResourceAsStream("/resources/"+%2B+resource);
++++++++if+(input+==+null)+{
++++++++++++//+this+is+how+we+load+file+within+editor+(eg+eclipse)
++++++++++++input+=+MyClass.class.getClassLoader().getResourceAsStream(resource);
++++++++}

++++++++return+input;
++++}
}|code-block|syntax|javascript|4692458|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|$]]

I had this problem before and I made fallback way for loading. Basically first way work within .jar file and second way works within eclipse or other IDE.

<pre><code>public class MyClass {

 public static InputStream accessFile() {
 String resource = "my-file-located-in-resources.txt";

 // this is the path within the jar file
 InputStream input = MyClass.class.getResourceAsStream("/resources/" + resource);
 if (input == null) {
 // this is how we load file within editor (eg eclipse)
 input = MyClass.class.getClassLoader().getResourceAsStream(resource);
 }

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

blocks|key|186718|text|您可以使用类加载器，它将从类路径读取作为根路径(在开头没有"/“)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|186719|InputStream+in+=+getClass().getClassLoader().getResourceAsStream("file.txt");+
BufferedReader+reader+=+new+BufferedReader(new+InputStreamReader(in));|code-block|syntax|javascript|186720|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 class loader which will read from classpath as ROOT path (without "/" in the beginning)

<pre><code>InputStream in = getClass().getClassLoader().getResourceAsStream("file.txt"); 
BufferedReader reader = new BufferedReader(new InputStreamReader(in));
</code></pre>

blocks|key|186770|text|由于某些原因，当我将web应用程序部署到WildFly+14时，classLoader.getResource()总是返回null。从getClass().getClassLoader()或Thread.currentThread().getContextClassLoader()获取classLoader返回null。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|186771|getClass().getClassLoader()应用程序接口文档说，|186772|“返回类的类加载器。某些实现可能使用null来表示引导类加载器。如果该类是由引导类加载器加载的，则此方法在此类实现中将返回null。”|186773|如果您使用的是WildFly和您的web应用程序，请尝试以下内容|186774|request.getServletContext().getResource()返回了资源url。这里的request是ServletRequest的一个对象。|186775|entityMap^0|W|P|1V|R|2N|1A|0|0|R|0|0|0|0|15|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]|$9|S|A|T|B|C]|$9|U|A|V|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|Z|8|@]|D|@]|E|$]]|$1|J|3|K|5|6|7|10|8|@]|D|@]|E|$]]|$1|L|3|M|5|6|7|11|8|@$9|12|A|13|B|C]]|D|@]|E|$]]|$1|N|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|O|$]]

For some reason <code>classLoader.getResource()</code> always returned null when I deployed the web application to WildFly 14. getting classLoader from <code>getClass().getClassLoader()</code> or <code>Thread.currentThread().getContextClassLoader()</code> returns null.

<code>getClass().getClassLoader()</code> API doc says,

"Returns the class loader for the class. Some implementations may use null to represent the bootstrap class loader. This method will return null in such implementations if this class was loaded by the bootstrap class loader."

may be if you are using WildFly and yours web application try this

<code>request.getServletContext().getResource()</code> returned the resource url. Here request is an object of ServletRequest.

blocks|key|4692631|text|问题是某些第三方库需要文件路径名，而不是输入流。大多数答案都没有解决这个问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4692632|在这种情况下，一种解决方法是将资源内容复制到临时文件中。下面的示例使用jUnit的TemporaryFolder。|offset|length|style|CODE|4692633|++++private+List<String>+decomposePath(String+path){
++++++++List<String>+reversed+=+Lists.newArrayList();
++++++++File+currFile+=+new+File(path);
++++++++while(currFile+!=+null){
++++++++++++reversed.add(currFile.getName());
++++++++++++currFile+=+currFile.getParentFile();
++++++++}
++++++++return+Lists.reverse(reversed);
++++}

++++private+String+writeResourceToFile(String+resourceName)+throws+IOException+{
++++++++ClassLoader+loader+=+getClass().getClassLoader();
++++++++InputStream+configStream+=+loader.getResourceAsStream(resourceName);
++++++++List<String>+pathComponents+=+decomposePath(resourceName);
++++++++folder.newFolder(pathComponents.subList(0,+pathComponents.size()+-+1).toArray(new+String[0]));
++++++++File+tmpFile+=+folder.newFile(resourceName);
++++++++Files.copy(configStream,+tmpFile.toPath(),+REPLACE_EXISTING);
++++++++return+tmpFile.getAbsolutePath();
++++}|code-block|syntax|javascript|4692634|entityMap^0|0|15|F|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|$]]

The problem is that certain third party libraries require file pathnames rather than input streams. Most of the answers don't address this issue.

In this case, one workaround is to copy the resource contents into a temporary file. The following example uses jUnit's <code>TemporaryFolder</code>.

<pre><code> private List&lt;String&gt; decomposePath(String path){
 List&lt;String&gt; reversed = Lists.newArrayList();
 File currFile = new File(path);
 while(currFile != null){
 reversed.add(currFile.getName());
 currFile = currFile.getParentFile();
 }
 return Lists.reverse(reversed);
 }

 private String writeResourceToFile(String resourceName) throws IOException {
 ClassLoader loader = getClass().getClassLoader();
 InputStream configStream = loader.getResourceAsStream(resourceName);
 List&lt;String&gt; pathComponents = decomposePath(resourceName);
 folder.newFolder(pathComponents.subList(0, pathComponents.size() - 1).toArray(new String[0]));
 File tmpFile = folder.newFile(resourceName);
 Files.copy(configStream, tmpFile.toPath(), REPLACE_EXISTING);
 return tmpFile.getAbsolutePath();
 }
</code></pre>

blocks|key|4692669|text|下面的代码适用于Spring+boot(kotlin)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4692670|val+authReader+=+InputStreamReader(javaClass.getResourceAsStream("/file1.json"))|code-block|syntax|javascript|4692671|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|$]]

Below code works with Spring boot(kotlin):

<pre><code>val authReader = InputStreamReader(javaClass.getResourceAsStream("/file1.json"))
</code></pre>

blocks|key|186870|text|我找到了一个解决办法|type|unstyled|depth|inlineStyleRanges|entityRanges|data|186871|BufferedReader+br+=+new+BufferedReader(new+InputStreamReader(Main.class.getResourceAsStream(path)));|code-block|syntax|javascript|186872|将"Main“替换为您在其中编写它的java类。将"+path“替换为jar文件中的路径。|186873|例如，如果您将State1.txt放在包com.issac.state中，那么如果您运行Linux或Mac，则输入路径为“/com/issac/state/stat1”。如果运行Windows，则键入路径"\com\issac\state\State1“。除非发生找不到文件异常，否则不要向文件添加.txt扩展名。|186874|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]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I have found a fix

<pre><code>BufferedReader br = new BufferedReader(new InputStreamReader(Main.class.getResourceAsStream(path)));
</code></pre>

Replace "Main" with the java class you coded it in. replace "path" with the path within the jar file.

for example, if you put State1.txt in the package com.issac.state, then type the path as "/com/issac/state/State1" if you run Linux or Mac. If you run Windows then type the path as "\com\issac\state\State1". Don't add the .txt extension to the file unless the File not found exception occurs.

blocks|key|186899|text|这段代码可以在Eclipse和导出的Runnable+JAR中运行|type|unstyled|depth|inlineStyleRanges|entityRanges|data|186900|private+String+writeResourceToFile(String+resourceName)+throws+IOException+{
++++File+outFile+=+new+File(certPath+%2B+File.separator+%2B+resourceName);

++++if+(outFile.isFile())
++++++++return+outFile.getAbsolutePath();
++++
++++InputStream+resourceStream+=+null;
++++
++++//+Java:+In+caso+di+JAR+dentro+il+JAR+applicativo+
++++URLClassLoader+urlClassLoader+=+(URLClassLoader)Cypher.class.getClassLoader();
++++URL+url+=+urlClassLoader.findResource(resourceName);
++++if+(url+!=+null)+{
++++++++URLConnection+conn+=+url.openConnection();
++++++++if+(conn+!=+null)+{
++++++++++++resourceStream+=+conn.getInputStream();
++++++++}
++++}
++++
++++if+(resourceStream+!=+null)+{
++++++++Files.copy(resourceStream,+outFile.toPath(),+StandardCopyOption.REPLACE_EXISTING);
++++++++return+outFile.getAbsolutePath();
++++}+else+{
++++++++System.out.println("Embedded+Resource+"+%2B+resourceName+%2B+"+not+found.");
++++}
++++
++++return+"";
}+++|code-block|syntax|javascript|186901|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|$]]

This code works both in Eclipse and in Exported Runnable JAR
<pre><code>private String writeResourceToFile(String resourceName) throws IOException {
 File outFile = new File(certPath + File.separator + resourceName);

 if (outFile.isFile())
 return outFile.getAbsolutePath();
 
 InputStream resourceStream = null;
 
 // Java: In caso di JAR dentro il JAR applicativo 
 URLClassLoader urlClassLoader = (URLClassLoader)Cypher.class.getClassLoader();
 URL url = urlClassLoader.findResource(resourceName);
 if (url != null) {
 URLConnection conn = url.openConnection();
 if (conn != null) {
 resourceStream = conn.getInputStream();
 }
 }
 
 if (resourceStream != null) {
 Files.copy(resourceStream, outFile.toPath(), StandardCopyOption.REPLACE_EXISTING);
 return outFile.getAbsolutePath();
 } else {
 System.out.println(&quot;Embedded Resource &quot; + resourceName + &quot; not found.&quot;);
 }
 
 return &quot;&quot;;
} 
</code></pre>

blocks|key|186915|text|在我的案例中，我最终做到了|type|unstyled|depth|inlineStyleRanges|entityRanges|data|186916|import+java.lang.Thread;
import+java.io.BufferedReader;
import+java.io.InputStreamReader;

final+BufferedReader+in+=+new+BufferedReader(new+InputStreamReader(
++++++Thread.currentThread().getContextClassLoader().getResourceAsStream("file.txt"))
);+//+no+initial+slash+in+file.txt|code-block|syntax|javascript|186917|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|$]]

In my case I finally made it with
<pre><code>import java.lang.Thread;
import java.io.BufferedReader;
import java.io.InputStreamReader;

final BufferedReader in = new BufferedReader(new InputStreamReader(
 Thread.currentThread().getContextClassLoader().getResourceAsStream(&quot;file.txt&quot;))
); // no initial slash in file.txt
</code></pre>

blocks|key|4692874|text|最后我解决了错误：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4692875|String+input_path+=+"resources\\file.txt";
++++++++
++++++++input_path+=+input_path.replace("\\",+"/");++//+doesn't+work+with+back+slash
++++++++
++++++++URL+file_url+=+getClass().getClassLoader().getResource(input_path);
++++++++String+file_path+=+new+URI(file_url.toString().replace("+","%2520")).getSchemeSpecificPart();
++++++++InputStream+file_inputStream+=+file_url.openStream();|code-block|syntax|javascript|4692876|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|$]]

finally i solved errors:
<pre><code>String input_path = &quot;resources\\file.txt&quot;;
 
 input_path = input_path.replace(&quot;\\&quot;, &quot;/&quot;); // doesn't work with back slash
 
 URL file_url = getClass().getClassLoader().getResource(input_path);
 String file_path = new URI(file_url.toString().replace(&quot; &quot;,&quot;%20&quot;)).getSchemeSpecificPart();
 InputStream file_inputStream = file_url.openStream();
</code></pre>

I would like to read a resource from within my jar like so:
<pre><code>File file;
file = new File(getClass().getResource(&quot;/file.txt&quot;).toURI());
BufferedReader reader = new BufferedReader(new FileReader(file));

//Read the file
</code></pre>
and it works fine when running it in Eclipse, but if I export it to a jar, and then run it, there is an IllegalArgumentException:
<pre><code>Exception in thread &quot;Thread-2&quot;
java.lang.IllegalArgumentException: URI is not hierarchical
</code></pre>
and I really don't know why but with some testing I found if I change
<pre><code>file = new File(getClass().getResource(&quot;/file.txt&quot;).toURI());
</code></pre>
to
<pre><code>file = new File(getClass().getResource(&quot;/folder/file.txt&quot;).toURI());
</code></pre>
then it works the opposite (it works in jar but not eclipse).
I'm using Eclipse and the folder with my file is in a class folder.

Reading a resource file from within jar

集成开发环境

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Eclipse

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Java

我想从我的jar中读取资源，如下所示：File file;file = new File(getClass().getResource("/file.txt").toURI());BufferedReader reader = new BufferedReader(new FileReader(file));//Rea...

问从jar中读取资源文件
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