I want to create multiple application contexts in my Tomcat application.
Some of these application contexts have the same package and class names, but they all refer to different jars.
For example:
application0 use service.jar, model.jar
application1 use service-a.jar, model-a.jar
application2 use service-b.jar, model-b.jar
application0 context is OK because is in orign project.
I reference some web page to custom application1, I use my custom classloader to start applicationContext.
File file0 = new File("D://git/project1/service-a.jar");
File file1 = new File("D://git/project1/modele-a.jar");
// convert the file to URL format
URL url0 = file0.toURI().toURL();
URL url1 = file1.toURI().toURL();
List<URL> urls = new LinkedList<>();
List<File> libs = listFilesForFolder(new File("D://protal//apache-tomcat-8.0.39//lib"));
for(File lib : libs) {
urls.add(lib.toURI().toURL());
}
urls.add(url1);
urls.add(url0);
final URLClassLoader customClassLoader = new URLClassLoader(urls.toArray(new URL[urls.size()]));
ClassPathXmlApplicationContext context1 = new ClassPathXmlApplicationContext("applicationContext.xml") {
protected void initBeanDefinitionReader(XmlBeanDefinitionReader reader)
{
super.initBeanDefinitionReader(reader);
reader.setValidationMode(XmlBeanDefinitionReader.VALIDATION_NONE);
reader.setBeanClassLoader(customClassLoader);
}
};
allApplicationContexts.add(context1);
The Spring contexts start OK, but they fail to create the component-scan bean, and PropertyPlaceholderConfigurer isn't working. Everything else seems correct.
I sure my config is correct because it works without the custom classloader. Libs contains all spring lib.
Is it possible to get this working with multiple Spring contexts?
I am upgrading Spring 2.5 to 4.2.
The issue is with one bean which has property type org.springframework.core.io.ClassPathResource. The resource value is defined in xml as p:location="classpath:/<the resource path>"
This worked perfect and bean property was populated with the resource. But in 4.2 the value is not getting set.
So I debugged the code and found that the class org.springframework.beans.BeanWrapperImpl was manipulating the value and removing classpath: string from actual value in Spring 2.5.
However the same is not true in 4.2 and class org.springframework.beans.BeanWrapperImpl isn't modifying the value which results in spring not finding the resource.
Anyone faced similar situation? What solution did you apply?
Thanks,
Hanumant
EDIT 1: code sample
spring config file
<bean class="com.test.sample.TestBean" id="testBean"
p:schemaLocation="classpath:/com/test/sample/Excalibur_combined.xsd" />
TestBean.java
public class TestBean {
private ClassPathResource schemaLocation;
public ClassPathResource getSchemaLocation() {
return schemaLocation;
}
public void setSchemaLocation(ClassPathResource schemaLocation) {
this.schemaLocation = schemaLocation;
}
}
App.java
public class App {
public static void main(String[] args) {
ApplicationContext ap = new ClassPathXmlApplicationContext("classpath:/com/test/sample/spring-config.xml");
TestBean tb = (TestBean) ap.getBean("testBean");
try {
URL url = tb.getSchemaLocation().getURL();
System.out.println(url);
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
Error Message
INFO: Loading XML bean definitions from class path resource
[com/test/sample/spring-config.xml] java.io.FileNotFoundException:
class path resource
[classpath:/com/test/sample/Excalibur_combined.xsd] cannot be resolved
to URL because it does not exist at
org.springframework.core.io.ClassPathResource.getURL(ClassPathResource.java:187)> at com.test.sample.App.main(App.java:20)
However if I remove the classpath: from bean definition it works.
So is classpth: necessary in bean definition xml file?? And why it was working fine in Spring 2.5??
The main issue is that you aren't programming to interfaces. Instead of the concrete org.springframework.core.io.ClassPathResource use org.springframework.core.io.Resource. When doing to the org.springframework.core.io.ResourceEditor will kick in and convert the String into a Resource instance. The location you are providing classpath:/<the resource path> will be passed to the ResourceLoader which will get the resource or throw an error if it doesn't exist.
If, however, you are using the concrete type ClassPathResouce directly this mechanism doesn't kick in and the location is set to what you provide classpath:/<the resource path>. However this is actually not a valid location for the URL class and that will eventually fail with the message you see.
It worked in earlier versions due to a hack/workaround/patch in the BeanWrapperImpl to strip the prefix.
Basically it now fails because you where doing things you shouldn't have been doing in the first place.
Using Apache Commons Configurations 1.9, how to avoid ConfigurationException upon loading a configuration file if the provided file cannot be found?
The Spring app context resembles:
<bean name="foo.config" class="org.apache.commons.configuration.PropertiesConfiguration" init-method="load">
<property name="fileName" value="foo.properties" />
</bean>
However my config file is optional, so I want to make sure the application starts correctly even the file doesn't exist.
How can I achieve this with Commons Configurations? A FactoryBean works, but is there another way?
if (!file.exists()) return new PropertiesConfiguration();
Or using try/catch syntax using an XML configuration:
import org.apache.commons.configuration2.XMLConfiguration;
import org.apache.commons.configuration2.builder.fluent.Configurations;
public class Workspace {
private final XMLConfiguration mConfig;
public Workspace() {
final var configs = new Configurations();
XMLConfiguration config;
try {
config = configs.xml( "filename.xml" );
} catch( final Exception e ) {
config = new XMLConfiguration();
}
mConfig = config;
}
Using a regular properties configuration will work the same way.
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.
I'm writing a jar intended to be used with Spring and Ehcache. Spring requires that there be a cache defined for each element, so I was planning to have an Ehcache defined for the jar, preferably as a resource in the jar that could be imported into the primary Ehcache configuration for the app. However, my reading of the example Ehcache config file and my Google searches have not turned up any way to import a sub Ehcache config file.
Is there a way to import a sub Ehcache config file, or is there some other way to solve this problem?
What I did to do something similar (replace some placeholders in my Ehcache xml file - a import statement is more or less a placeholder if you will) is to extend (more or less copy to be honest) from Springs EhCacheManagerFactoryBean and create the final Ehcache xml config file on the fly.
For creating the CacheManager instance in afterPropertiesSet() you just hand over a InputStream which points to your config.
#Override
public void afterPropertiesSet() throws IOException, CacheException {
if (this.configLocation != null) {
InputStreamSource finalConfig = new YourResourceWrapper(this.configLocation); // put your custom logic here
InputStream is = finalConfig.getInputStream();
try {
this.cacheManager = (this.shared ? CacheManager.create(is) : new CacheManager(is));
} finally {
IOUtils.closeQuietly(is);
}
} else {
// ...
}
// ...
}
For my filtering stuff I internally used a ByteArrayResource to keep the final config.
data = IOUtils.toString(source.getInputStream()); // get the original config (configLocation) as string
// do your string manipulation here
Resource finalConfigResource = new ByteArrayResource(data.getBytes());
For "real" templating one could also think of using a real template engine like FreeMarker (which Spring has support for) to do more fancy stuff.