【Java REST API超入門】簡単サーバ構築ガイド｜初心者でもすぐ作れる実用例付き

"I'm studying Java, but I don't understand what goes on behind the scenes of web applications."
"APIs seem difficult... but I want to learn how to use them."
For those who have such concerns,Experience creating a REST API server in Javais recommended.

Starting with the basics of APIs,Receiving requests, JSON responses, and POST/GET processingto
By experiencing the entire process, you will be able to quickly deepen your practical skills and understanding of Java.

In this article, we will introduce how to build a system that even beginners can use.Simple REST API serverHow to make
Easy to understand, detailed, and with codeLet me explain.

table of contents

What is a REST API Server?
1. It is the gateway between the server and the client.
Understand the features and configuration you need
1. You can create a REST API with just two functions
Standard Java classes to use
1. You can create an API without additional libraries
Implementation: Creating a REST API server in Java
Common errors and solutions
1. Resolves common stumbling points
Application: File saving and JSON format support
1. Towards a more practical API
Complete code summary and configuration
Summary: Understanding the basics of APIs in Java

What is a REST API Server?

It is the gateway between the server and the client.

Bottom line: REST APIs are a gateway for connecting apps and web services together.

For example, when getting today's weather forecast using a weather app,
Behind the scenes, the following interaction takes place: "Access the API → The server returns data in JSON format."

You can also implement this mechanism in Java code.

The Ministry of Education, Culture, Sports, Science and Technology also promotes "development of information utilization skills"Understanding the structure and mechanisms of information and communicationsWe place great importance on the following.
(Source: https://www.mext.go.jp/a_menu/shotou/zyouhou/detail/1375607.htm)

Understand the features and configuration you need

You can create a REST API with just two functions

Conclusion: The basic API is complete just by implementing GET and POST operations.

The API we will create will look something like this:

GET /message → Return a list of messages
POST /message Receive and save messages

The setup is very simple:

JavaHttpServer.java：API invocation and routing process
/message → GET/POST compatible
Data is stored in a simple List (in-memory processing)

Standard Java classes to use

You can create an API without additional libraries

Java hascom.sun.net.httpserver.HttpServerThere is a standard class.
If you use this,No server software such as TomcatYou can create an API server.

Required imports:

import com.sun.net.httpserver.HttpServer; import com.sun.net.httpserver.HttpHandler; import com.sun.net.httpserver.HttpExchange;

Implementation: Creating a REST API server in Java

import com.sun.net.httpserver.*; import java.io.*; import java.net.InetSocketAddress; import java.nio.charset.StandardCharsets; import java.util.*; public class JavaHttpServer { private static final List messages = new ArrayList<>(); public static void main(String[] args) throws IOException { HttpServer server = HttpServer.create(new InetSocketAddress(8000), 0); server.createContext("/message", new MessageHandler()); server.setExecutor(null); // Default thread System.out.println("API server running: http://localhost:8000/message"); server.start(); } static class MessageHandler implements HttpHandler { @Override public void handle(HttpExchange exchange) throws IOException { String method = exchange.getRequestMethod(); if ("GET".equalsIgnoreCase(method)) { handleGet(exchange); } else if ("POST".equalsIgnoreCase(method)) { handlePost(exchange); } else { exchange.sendResponseHeaders(405, -1); // Method not allowed } } private void handleGet(HttpExchange exchange) throws IOException { String response = String.join("\n", messages); byte[] bytes = response.getBytes(StandardCharsets.UTF_8); exchange.sendResponseHeaders(200, bytes.length); try (OutputStream os = exchange.getResponseBody()) { os.write(bytes); } } private void handlePost(HttpExchange exchange) throws IOException { InputStream is = exchange.getRequestBody(); String body = new String(is.readAllBytes(), StandardCharsets.UTF_8); messages.add(body); String response = "Message accepted."; byte[] bytes = response.getBytes(StandardCharsets.UTF_8); exchange.sendResponseHeaders(200, bytes.length); try (OutputStream os = exchange.getResponseBody()) { os.write(bytes); } } } }