How to get resources directory path programmatically

Question

I have the following directory layout:

src

main

java

resources

sql (scripts for database)

spring (configuration)

webapp

Within a ServletContextListener class, I want to access the files under the SQL directory and list them. Basically my problem is with the path, because I know that listing files under a directory in a nutshell is:

File folder = new File(path);
File[] listOfFiles = folder.listFiles();

Maybe I could use the ServletContextEvent Object to try and build a path to resources/sql

public void contextInitialized(ServletContextEvent event) {
    event.getServletContext(); //(getRealPath etc.)
}

Does something exist to set that path in a relative, non-hardcoded way? Something like new File("classpath:sql") (preferably spring if possible) or what should I do with the servletContext to point at resources/sql?

There are three variants of solution, depending on the situation: https://stackoverflow.com/a/56327069/715269 — Gangnus, May 27 '19 at 13:33
See also [How to get the path of src/test/resources directory in JUnit?](https://stackoverflow.com/questions/28673651/how-to-get-the-path-of-src-test-resources-directory-in-junit) — Vadzim, Feb 15 '22 at 02:31

score 86 · Accepted Answer · answered Oct 17 '13 at 01:20

86

I'm assuming the contents of src/main/resources/ is copied to WEB-INF/classes/ inside your .war at build time. If that is the case you can just do (substituting real values for the classname and the path being loaded).

URL sqlScriptUrl = MyServletContextListener.class
                       .getClassLoader().getResource("sql/script.sql");

answered Oct 17 '13 at 01:20

Dev

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Thanks!! this worked for me, with that URL then I build the Path for the new File, and finally get the files in that directory. – yamilmedina Oct 17 '13 at 15:25
2

You don't need the File. You already have a URL. You can't even assume there is a file or a directory at all. The WAR may not have been unpacked. – user207421 Oct 18 '13 at 21:24
What does MyServletContextListener mean here ? – Tejesh Raut Dec 04 '15 at 10:02
@TejeshRaut It is a place holder for a class that implements [`javax.servlet.ServletContextListener`](http://docs.oracle.com/javaee/7/api/javax/servlet/ServletContextListener.html). I used that as an example because the question had a snippet from `contextInitialized` which is a method defined by the `ServletContextListener` interface. – Dev Dec 05 '15 at 01:24
For me it was crucial to prepend the path with `classpath:`, so: `"classpath:/sql/script.sql"`. – isgoed Jan 21 '21 at 15:19

score 76 · Answer 2 · edited May 04 '19 at 09:37

76

Finally, this is what I did:

private File getFileFromURL() {
    URL url = this.getClass().getClassLoader().getResource("/sql");
    File file = null;
    try {
        file = new File(url.toURI());
    } catch (URISyntaxException e) {
        file = new File(url.getPath());
    } finally {
        return file;
    }
}

...

File folder = getFileFromURL();
File[] listOfFiles = folder.listFiles();

edited May 04 '19 at 09:37

Matthias Braun

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answered Oct 18 '13 at 21:16

yamilmedina

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12

Are you sure of using `"/sql"` as the parameter? I have problems with a starting slash, Instead `"sql"` works. – Abdull May 13 '16 at 13:57
2

Please be aware that you can get a different URL with this.getClass().getResource("/sql")); – ShadowGames Oct 15 '18 at 13:59
This won't work on all containers. Tomcat, for example. – user207421 Feb 11 '19 at 23:10

score 19 · Answer 3 · answered Oct 16 '13 at 22:20

import org.springframework.core.io.ClassPathResource;

...

File folder = new ClassPathResource("sql").getFile();
File[] listOfFiles = folder.listFiles();

It is worth noting that this will limit your deployment options, ClassPathResource.getFile() only works if the container has exploded (unzipped) your war file.

score 11 · Answer 4 · answered Feb 11 '19 at 22:43

11

Just use com.google.common.io.Resources class. Example:

 URL url = Resources.getResource("file name")

After that you have methods like: .getContent(), .getFile(), .getPath() etc

answered Feb 11 '19 at 22:43

grep

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Do they also support getting resources while the code is being executed from a jar file? – Displee Dec 22 '19 at 15:56

How to get resources directory path programmatically

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