Is there a way we can lookup file resources using relative path in application.properties file in Spring boot application as specified below
spring.datasource.url=jdbc:hsqldb:file:${project.basedir}/db/init
I'm using spring boot to build a upload sample, and meet the same problem, I only want to get the project root path. (e.g. /sring-boot-upload)
I find out that below code works:
upload.dir.location=${user.dir}\\uploadFolder
#membersound answer is just breaking up the hardcoded path in 2 parts, not dynamically resolving the property. I can tell you how to achieve what you're looking for, but you need to understand is that there is NO project.basedir when you're running the application as a jar or war. Outside the local workspace, the source code structure doesn't exist.
If you still want to do this for testing, that's feasible and what you need is to manipulate the PropertySources. Your simplest option is as follows:
Define an ApplicationContextInitializer, and set the property there. Something like the following:
public class MyApplicationContextInitializer implements ApplicationContextInitializer<ConfigurableApplicationContext> {
#Override
public void initialize(ConfigurableApplicationContext appCtx) {
try {
// should be /<path-to-projectBasedir>/build/classes/main/
File pwd = new File(getClass().getResource("/").toURI());
String projectDir = pwd.getParentFile().getParentFile().getParent();
String conf = new File(projectDir, "db/init").getAbsolutePath();
Map<String, Object> props = new HashMap<>();
props.put("spring.datasource.url", conf);
MapPropertySource mapPropertySource = new MapPropertySource("db-props", props);
appCtx.getEnvironment().getPropertySources().addFirst(mapPropertySource);
} catch (URISyntaxException e) {
throw new RuntimeException(e);
}
}}
Looks like you're using Boot, so you can just declare context.initializer.classes=com.example.MyApplicationContextInitializer in your application.properties and Boot will run this class at startup.
Words of caution again:
This will not work outside the local workspace as it depends on the source code structure.
I've assumed a Gradle project structure here /build/classes/main. If necessary, adjust according to your build tool.
If MyApplicationContextInitializer is in the src/test/java, pwd will be <projectBasedir>/build/classes/test/, not <projectBasedir>/build/classes/main/.
your.basedir=${project.basedir}/db/init
spring.datasource.url=jdbc:hsqldb:file:${your.basedir}
#Value("${your.basedir}")
private String file;
new ClassPathResource(file).getURI().toString()
I am trying to apply the checkstyle plugin to my gradle project. The configuration of which is in a seperate, shared jar dependency:
project.apply plugin: StaticAnalysisPlugin
class StaticAnalysisPlugin implements Plugin<Project> {
#Override
void apply(Project project) {
project.apply plugin: 'checkstyle'
project.configurations {
codingStandardsConfig
}
project.dependencies {
codingStandardsConfig 'com.sample.tools:coding-standards:1.+:#jar'
}
def checkstyleConfigFileLocation = "classpath:sample-checkstyle-config.xml"
project.checkstyle {
toolVersion = '6.3'
project.logger.debug "$project Using checkstyle version $toolVersion."
project.logger.debug "$project Using checkstyle config from: ${checkstyleConfigFileLocation}"
config = project.resources.text.fromFile(checkstyleConfigFileLocation)
}
project.checkstyleMain.source = "src/main/java"
project.checkstyleTest.exclude "**/*"
}
}
the config file is located in the coding-standards jar, but I am unsure how to wire this in to the checkstyle config.
I think you should use the configFile option (which expects a File object) rather than the config option (which expects a text resource with your actual configuration).
Then the problem comes down to determining the correct File object. From this answer, it seems you can do
ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource("file/test.xml").getFile());
Alternative solutions can be found on StackOverflow. Make sure that the class loader's class path includes your coding-standards.jar. You may have to put that dependency in a buildscript block to make it available, well, to the build script.
I have a code like follows
public LocalFileStorage(String storageUrl, Resource storageDirectory) {
this.storageUrl = storageUrl;
try {
this.storageDirectory = storageDirectory.getFile();
this.storageDirectory.deleteOnExit();
this.storageDirectory.createNewFile();
} catch (IOException e) {
throw new RuntimeException(e);
}
}
I call the class the follows.
private ResourceLoader resourceLoader; // from spring
LocalFileStorage pictureStorage = new LocalFileStorage(Url+ "/resources/", resourceLoader.getResource("/resources/"));
call to
resourceLoader.getResource("/resources/")
throws exception. I thought ResourceLoader loads directory also because after all directory is also a file.
My structure
Typically, only anything inside /WEB-INF/classes, /WEB-INF/lib, /WEB-INF/... will be added to the classpath and accessible through ClassLoader.getResource(). The folder you are trying to access is not in WEB-INF and will therefore not appear in the classpath.
Assuming you are using something similar to Maven, you should put resource files under /src/main/resources. When your project is built, those files will end up in WEB-INF/classes.
I've developed a maven plugin, which can scan classes within a module, to find some specific classes and do something about them.
The problem is that, when I'm using this plugin in a maven module, It's not able to find classes within that module.
I've checked the plugin classpath and it only contains plugin classes and it's dependencies.
Is there any way to automatically include module classes into plugin classpath?
I took this approach and apparently it's working:
1 - a MavenProject parameter is needed within your Mojo class:
#Parameter(defaultValue = "${project}", required = true, readonly = true)
private MavenProject project;
2 - and then you can get the classpath elements from project instance:
try {
Set<URL> urls = new HashSet<>();
List<String> elements = project.getTestClasspathElements();
//getRuntimeClasspathElements()
//getCompileClasspathElements()
//getSystemClasspathElements()
for (String element : elements) {
urls.add(new File(element).toURI().toURL());
}
ClassLoader contextClassLoader = URLClassLoader.newInstance(
urls.toArray(new URL[0]),
Thread.currentThread().getContextClassLoader());
Thread.currentThread().setContextClassLoader(contextClassLoader);
} catch (DependencyResolutionRequiredException e) {
throw new RuntimeException(e);
} catch (MalformedURLException e) {
throw new RuntimeException(e);
}
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?
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.
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().
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!
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);
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...
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;
}
}
For kotlin users, I solved it like this:
val url = ResourceUtils.getURL("classpath:$fileName")
val response = url.openStream().bufferedReader().readText()
The answer by #sbk is the way we should do it in spring-boot environment (apart from #Value("${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>);
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);
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);
I had the same issue, ended up using the much more convenient Guava Resources:
Resources.getResource("my.file")
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.
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.