File inside jar is not visible for 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    lt bean id       class       gt       lt property name  resource  value  classpath my config    gt   lt  bean gt    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

User · Answer

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 lt URL gt  systemResources   resourceLoader getClassLoader   getResources fileNameWithoutExt   FILE EXT

User · Answer

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  quot classpath file txt quot   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

User · Answer

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 lt Resource String gt  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 lt Resource  String gt  getResourceContentsInResourceFolder        String folder  String extension        Resource   resources   getResourcesInResourceFolder folder  extension        HashMap lt Resource  String gt  result   new HashMap lt  gt         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

User · Answer

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  lt relative path of the resource from resource directory gt

User · Answer

I had the same issue  ended up using the much more convenient Guava Resources    Resources getResource  my file

User · Answer

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

User · Answer

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  quot classpath some resource in a dependency quot    new BufferedReader    new InputStreamReader resource openStream      lines   forEach System out  println

User · Answer

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

User · Answer

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

[spring] File inside jar is not visible for spring

Examples related to spring

Examples related to jar

Examples related to classpath