Exception when passing a Path argument to WebCassandraDirectoryProvider about embedded-cassandra HOT 6 CLOSED

That seems to have done the trick, thanks :)

from embedded-cassandra.

Hello,

  new CassandraBuilder()
            .workingDirectoryInitializer(new DefaultWorkingDirectoryInitializer(new CassandraDirectoryProvider() {

                @Override
                public Path getDirectory(Version version) throws IOException {
                    //here it means '.' directory must contain all necessary Apache Cassandra files
                    return Paths.get(".");
                }
            }));

WebCassandraDirectoryProvider is being used for downloading Apache Cassandra and extracts an archive into user.home directory and later uses this directory for Working Directory initializing (/tmp/<...>) by default.

from embedded-cassandra.

But actually, I tried your example:

	private final CassandraBuilder builder = new CassandraBuilder().workingDirectoryInitializer(
			new DefaultWorkingDirectoryInitializer(new WebCassandraDirectoryProvider(Paths.get("."))));

and got the same exception.

from embedded-cassandra.

also by default in WebCassandraDirectory we have the following:

	Path downloadDirectory = this.downloadDirectory.resolve(".embedded-cassandra").resolve("cassandra")
				.resolve(version.toString());
				

from embedded-cassandra.

So, just to clarify, do you want to be able to set the download directory is .? right?

from embedded-cassandra.

You can try this option for now:

  	new DefaultWorkingDirectoryInitializer(new WebCassandraDirectoryProvider(Paths.get(System.getProperty("user.dir")))));

from embedded-cassandra.