You can use the Path api:
Path p = Paths.get(yourFileNameUri);
Path folder = p.getParent();
Answer from assylias on Stack OverflowVideos
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Oracle
docs.oracle.com โบ javase โบ 8 โบ docs โบ api โบ java โบ nio โบ file โบ Path.html
Path (Java Platform SE 8 )
April 21, 2026 - In addition, the toFile method is useful to construct a File from the String representation of a Path. Implementations of this interface are immutable and safe for use by multiple concurrent threads. ... Returns the file system that created this object. ... Tells whether or not this path is ...
Dev.java
dev.java โบ learn โบ java-io โบ file-system โบ file-path
Accessing Resources using Paths - Dev.java
January 25, 2023 - At the appropriate time, you can use the methods in the Files class to check the existence of the file corresponding to the Path, create the file, open it, delete it, change its permissions, and so on. Perhaps you have legacy code that uses java.io.File and would like to take advantage of the java.nio.file.Path functionality with minimal impact to your code.
DigitalOcean
digitalocean.com โบ community โบ tutorials โบ java-file-path-absolute-canonical
Java File Path, Absolute Path and Canonical Path | DigitalOcean
August 3, 2022 - I need to pass the full path and file name for the selected files to apache pdf box class to get the excel extracts. /* * To change this license header, choose License Headers in Project Properties. * To change this template file, choose Tools | Templates * and open the template in the editor. */ package com.pdfet.pdfext; import javax.swing.*; import java.awt.*; import java.awt.event.*; import java.awt.event.ActionListener; //import com.pdfet.pdfext.ExtensionFileFilter; import java.io.File; import javax.swing.filechooser.FileFilter; /** * * @author vak_salem */ class ExtensionFileFilter extend
Baeldung
baeldung.com โบ home โบ java โบ java io โบ list files in a directory in java
List Files in a Directory in Java | Baeldung
January 8, 2024 - Java 8 introduced a new list() method in java.nio.file.Files. The list method returns a lazily populated Stream of entries in the directory. ... public Set<String> listFilesUsingFilesList(String dir) throws IOException { try (Stream<Path> stream = Files.list(Paths.get(dir))) { return stream .filter(file -> !Files.isDirectory(file)) .map(Path::getFileName) .map(Path::toString) .collect(Collectors.toSet()); } }
Docker Hub
hub.docker.com โบ _ โบ eclipse-temurin
eclipse-temurin - Official Image | Docker Hub
To run a pre-built jar file with the latest OpenJDK 25, use the following Dockerfile: FROM eclipse-temurin:25 RUN mkdir /opt/app COPY japp.jar /opt/app CMD ["java", "-jar", "/opt/app/japp.jar"] Copy
Render
render.com
Render | The cloud for builders
Store and retrieve files from your services with lightweight SDKs.
W3Schools
w3schools.com โบ java โบ java_files.asp
Java Files
Java Examples Java Videos Java Compiler Java Exercises Java Quiz Java Code Challenges Java Practice Problems Java Server Java Syllabus Java Study Plan Java Interview Q&A ... File handling is an important part of any application.
W3Schools
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Java Getting Started
In Java, every application begins with a class name, and that class must match the filename.
GeeksforGeeks
geeksforgeeks.org โบ java โบ how-to-extract-file-extension-from-a-file-path-string-in-java
How to Extract File Extension From a File Path String in Java? - GeeksforGeeks
July 23, 2025 - Otherwise, it extracts the substring from the index after the last dot to the end of the file path, effectively giving the file extension. Example file paths are provided to showcase the method in action. The results are printed, demonstrating how the method accurately extracts file extensions. Handles cases where no extension is present, returning an appropriate message. ... import java.io.File; // Driver Class public class Code2 { // Main Function public static void main(String[] args) { String filepath = "example.txt"; // It will print file name and extension printFileNameAndExtension(filep
Apache Kafka
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Documentation Redirect | Apache Kafka
May 22, 2026 - Redirecting ยท Security | Donate | Thanks | Events | License | Privacy
Mkyong
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Java - Read a file from resources folder - Mkyong.com
September 4, 2020 - In Java, we can use getResourceAsStream or getResource to read a file or multiple files from a resources folder or root of the classpath.