File inside jar is not visible for spring - spring

Question

All
I created a jar file with the following MANIFEST.MF inside:
Manifest-Version: 1.0
Ant-Version: Apache Ant 1.8.3
Created-By: 1.6.0_25-b06 (Sun Microsystems Inc.)
Main-Class: my.Main
Class-Path: . lib/spring-core-3.2.0.M2.jar lib/spring-beans-3.2.0.M2.jar
In its root there is a file called my.config which is referenced in my spring-context.xml like this:
<bean id="..." class="...">
<property name="resource" value="classpath:my.config" />
</bean>
If I run the jar, everything looks fine escept the loading of that specific file:
Caused by: java.io.FileNotFoundException: class path resource [my.config] cannot be resolved to absolute file path because it does not reside in the file system: jar:file:/D:/work/my.jar!/my.config
at org.springframework.util.ResourceUtils.getFile(ResourceUtils.java:205)
at org.springframework.core.io.AbstractFileResolvingResource.getFile(AbstractFileResolvingResource.java:52)
at eu.stepman.server.configuration.BeanConfigurationFactoryBean.getObject(BeanConfigurationFactoryBean.java:32)
at eu.stepman.server.configuration.BeanConfigurationFactoryBean.getObject(BeanConfigurationFactoryBean.java:1)
at org.springframework.beans.factory.support.FactoryBeanRegistrySupport.doGetObjectFromFactoryBean(FactoryBeanRegistrySupport.java:142)
... 22 more
classes are loaded the from inside the jar
spring and other dependencies are loaded from separated jars
spring context is loaded (new ClassPathXmlApplicationContext("spring-context/applicationContext.xml"))
my.properties is loaded into PropertyPlaceholderConfigurer ("classpath:my.properties")
if I put my .config file outside the file system, and change the resource url to 'file:', everything seems to be fine...
Any tips?

Answer 1

If your spring-context.xml and my.config files are in different jars then you will need to use classpath*:my.config?
More info here
Also, make sure you are using resource.getInputStream() not resource.getFile() when loading from inside a jar file.

Answer 2

In the spring jar package, I use new ClassPathResource(filename).getFile(), which throws the exception:
cannot be resolved to absolute file path because it does not reside in the file system: jar
But using new ClassPathResource(filename).getInputStream() will solve this problem. The reason is that the configuration file in the jar does not exist in the operating system's file tree,so must use getInputStream().

Answer 3

I know this question has already been answered. However, for those using spring boot, this link helped me - https://smarterco.de/java-load-file-classpath-spring-boot/
However, the resourceLoader.getResource("classpath:file.txt").getFile(); was causing this problem and sbk's comment:
That's it. A java.io.File represents a file on the file system, in a
directory structure. The Jar is a java.io.File. But anything within
that file is beyond the reach of java.io.File. As far as java is
concerned, until it is uncompressed, a class in jar file is no
different than a word in a word document.
helped me understand why to use getInputStream() instead. It works for me now!
Thanks!

Answer 4

The error message is correct (if not very helpful): the file we're trying to load is not a file on the filesystem, but a chunk of bytes in a ZIP inside a ZIP.
Through experimentation (Java 11, Spring Boot 2.3.x), I found this to work without changing any config or even a wildcard:
var resource = ResourceUtils.getURL("classpath:some/resource/in/a/dependency");
new BufferedReader(
new InputStreamReader(resource.openStream())
).lines().forEach(System.out::println);

Answer 5

I had similar problem when using Tomcat6.x and none of the advices I found was helping.
At the end I deleted work folder (of Tomcat) and the problem gone.
I know it is illogical but for documentation purpose...

Answer 6

I was having an issue recursively loading resources in my Spring app, and found that the issue was I should be using resource.getInputStream. Here's an example showing how to recursively read in all files in config/myfiles that are json files.
Example.java
private String myFilesResourceUrl = "config/myfiles/**/";
private String myFilesResourceExtension = "json";
ResourceLoader rl = new ResourceLoader();
// Recursively get resources that match.
// Big note: If you decide to iterate over these,
// use resource.GetResourceAsStream to load the contents
// or use the `readFileResource` of the ResourceLoader class.
Resource[] resources = rl.getResourcesInResourceFolder(myFilesResourceUrl, myFilesResourceExtension);
// Recursively get resource and their contents that match.
// This loads all the files into memory, so maybe use the same approach
// as this method, if need be.
Map<Resource,String> contents = rl.getResourceContentsInResourceFolder(myFilesResourceUrl, myFilesResourceExtension);
ResourceLoader.java
import java.io.IOException;
import java.io.InputStream;
import java.nio.charset.Charset;
import java.util.HashMap;
import java.util.Map;
import org.springframework.core.io.Resource;
import org.springframework.core.io.support.PathMatchingResourcePatternResolver;
import org.springframework.core.io.support.ResourcePatternResolver;
import org.springframework.util.StreamUtils;
public class ResourceLoader {
public Resource[] getResourcesInResourceFolder(String folder, String extension) {
ResourcePatternResolver resolver = new PathMatchingResourcePatternResolver();
try {
String resourceUrl = folder + "/*." + extension;
Resource[] resources = resolver.getResources(resourceUrl);
return resources;
} catch (IOException e) {
throw new RuntimeException(e);
}
}
public String readResource(Resource resource) throws IOException {
try (InputStream stream = resource.getInputStream()) {
return StreamUtils.copyToString(stream, Charset.defaultCharset());
}
}
public Map<Resource, String> getResourceContentsInResourceFolder(
String folder, String extension) {
Resource[] resources = getResourcesInResourceFolder(folder, extension);
HashMap<Resource, String> result = new HashMap<>();
for (var resource : resources) {
try {
String contents = readResource(resource);
result.put(resource, contents);
} catch (IOException e) {
throw new RuntimeException("Could not load resource=" + resource + ", e=" + e);
}
}
return result;
}
}

Answer 7

For kotlin users, I solved it like this:
val url = ResourceUtils.getURL("classpath:$fileName")
val response = url.openStream().bufferedReader().readText()

Answer 8

The answer by #sbk is the way we should do it in spring-boot environment (apart from #Value("${classpath*:})), in my opinion. But in my scenario it was not working if the execute from standalone jar..may be I did something wrong.
But this can be another way of doing this,
InputStream is = this.getClass().getClassLoader().getResourceAsStream(<relative path of the resource from resource directory>);

Answer 9

I was having an issue more complex because I have more than one file with same name, one is in the main Spring Boot jar and others are in jars inside main fat jar.
My solution was getting all the resources with same name and after that get the one I needed filtering by package name.
To get all the files:
ResourceLoader resourceLoader = new FileSystemResourceLoader();
final Enumeration<URL> systemResources = resourceLoader.getClassLoader().getResources(fileNameWithoutExt + FILE_EXT);

Answer 10

In Spring boot 1.5.22.RELEASE Jar packaging this worked for me:
InputStream resource = new ClassPathResource("example.pdf").getInputStream();
"example.pdf" is in src/main/resources.
And then to read it as byte[]
FileCopyUtils.copyToByteArray(resource);

Answer 11

I had the same issue, ended up using the much more convenient Guava Resources:
Resources.getResource("my.file")

Answer 12

While this is a very old thread, but I also faced the same issue while adding FCM in a Spring Boot Application.
In development, the file was getting opened and no errors but when I deployed the application to AWS Elastic beanstalk , the error of FileNotFoundException was getting thrown and FCM was not working.
So here's my solution to get it working on both development env and jar deployment production.
I have a Component class FCMService which has a method as follows:
#PostConstruct
public void initialize() {
log.info("Starting FCM Service");
InputStream inputStream;
try {
ClassPathResource resource = new ClassPathResource("classpath:fcm/my_project_firebase_config.json");
URL url = null;
try {
url = resource.getURL();
} catch (IOException e) {
}
if (url != null) {
inputStream = url.openStream();
} else {
File file = ResourceUtils.getFile("classpath:fcm/my_project_firebase_config.json");
inputStream = new FileInputStream(file);
}
FirebaseOptions options = FirebaseOptions.builder().setCredentials(GoogleCredentials.fromStream(inputStream))
.build();
FirebaseApp.initializeApp(options);
log.info("FCM Service started");
} catch (IOException e) {
log.error("Error starting FCM Service");
e.printStackTrace();
}
}
Hope this helps someone looking for a quick fix with implementing FCM.

Answer 13

Can be handled like:
var serviceAccount = ClassLoader.getSystemResourceAsStream(FB_CONFIG_FILE_NAME);
FirebaseOptions options = new FirebaseOptions.Builder()
.setCredentials(GoogleCredentials.fromStream(serviceAccount))
.build();
Where FB_CONFIG_FILE_NAME is name of file in your 'resources' folder.