Upload file using REST webservice

In this tutorial I am going to show you how we can upload file using REST or RESTful web service.

The most important concept in REST is resources, which are identified by global IDs — typically using URIs. Client applications use HTTP methods (GET/ POST/ PUT/ DELETE) to manipulate the resource or collection of resources. A RESTful Web service is implemented using HTTP and the principles of REST. Typically, a RESTful Web service should define the following aspects:

The base/root URI for the Web service such as http://<host>/<appcontext/contextpath>/<url pattern>/<resources>.
The MIME type of the response data supported, which are JSON/XML/TEXT/HTML etc.
The set of operations supported by the service. (for example, POST, GET, PUT or DELETE).

Prerequisites
Eclipse, JDK 1.8, JAX-RS jars 2.23.2
Have maven installed and configured
JAX-RS dependencies in pom.xml

For this tutorial we will create a web maven project in Eclipse. If you already have an idea on how to create a maven project in Eclipse will be great otherwise I will tell you here how to create a maven project in Eclipse.

Step 1. Create a web maven project in Eclipse

Go to File -> New -> Other. On popup window under Maven select Maven Project. Then click on Next. Select the workspace location – either default or browse the location. Click on Next. Now in next window select the row as highlighted from the below list of archtypes and click on Next button.

maven-arctype-webapp
Now enter the required fields (Group Id, Artifact Id) as shown below
Group Id : com.roytuts
Artifact Id : rest
Step 2. Modify the pom.xml file as shown below.

<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
	xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/maven-v4_0_0.xsd">
	<modelVersion>4.0.0</modelVersion>
	<groupId>com.roytuts</groupId>
	<artifactId>rest</artifactId>
	<packaging>war</packaging>
	<version>0.0.1-SNAPSHOT</version>
	<name>rest Maven Webapp</name>
	<url>http://maven.apache.org</url>
	<properties>
		<java.version>1.8</java.version>
		<jersey.version>2.23.2</jersey.version>
	</properties>
	<dependencies>
		<dependency>
			<groupId>org.glassfish.jersey.containers</groupId>
			<artifactId>jersey-container-servlet</artifactId>
			<version>${jersey.version}</version>
		</dependency>
	</dependencies>
	<build>
		<finalName>rest</finalName>
		<plugins>
			<plugin>
				<groupId>org.apache.maven.plugins</groupId>
				<artifactId>maven-compiler-plugin</artifactId>
				<configuration>
					<source>${java.version}</source>
					<target>${java.version}</target>
				</configuration>
			</plugin>
		</plugins>
	</build>
</project>

Step 3. If you see JRE System Library[J2SE-1.5] then change the version by below process

Do right-click on the project and go to Build -> Configure build path, under Libraries tab click on JRE System Library[J2SE-1.5], click on Edit button and select the appropriate jdk 1.8 from the next window. Click on Finish then Ok.

Step 4. Modify web.xml file to use the jersey servlet

<!DOCTYPE web-app PUBLIC
 "-//Sun Microsystems, Inc.//DTD Web Application 2.3//EN"
 "http://java.sun.com/dtd/web-app_2_3.dtd" >
<web-app>
	<display-name>REST Service</display-name>
	<servlet>
		<servlet-name>REST</servlet-name>
		<servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
		<init-param>
			<param-name>jersey.config.server.provider.packages</param-name>
			<param-value>com.roytuts.rest.resources</param-value>
		</init-param>
		<load-on-startup>1</load-on-startup>
	</servlet>
	<!-- Map /rest/* to Jersey -->
	<servlet-mapping>
		<servlet-name>REST</servlet-name>
		<url-pattern>/*</url-pattern>
	</servlet-mapping>
</web-app>

Step 5. Create a REST resource class as shown below

package com.roytuts.rest.resources;
import java.io.File;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import javax.ws.rs.Consumes;
import javax.ws.rs.GET;
import javax.ws.rs.POST;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.Context;
import javax.ws.rs.core.HttpHeaders;
import javax.ws.rs.core.MediaType;
import javax.ws.rs.core.MultivaluedMap;
import javax.ws.rs.core.Response;
import javax.ws.rs.core.Response.ResponseBuilder;
@Path("/file")
public class RestService {
	@POST
	@Path("/upload")
	@Consumes(MediaType.APPLICATION_OCTET_STREAM)
	@Produces(MediaType.TEXT_PLAIN)
	public Response sendFile(@Context HttpHeaders headers, InputStream fileInputStream) {
		MultivaluedMap<String, String> map = headers.getRequestHeaders();
		String fileName = getFileName(map);
		OutputStream out = null;
		String filePath = "D:/" + fileName;
		try {
			out = new FileOutputStream(new File(filePath));
			byte[] buf = new byte[1024];
			int len;
			while ((len = fileInputStream.read(buf)) > 0) {
				out.write(buf, 0, len);
			}
		} catch (Exception e) {
			e.printStackTrace();
		} finally {
			try {
				out.close();
				fileInputStream.close();
			} catch (IOException e) {
				e.printStackTrace();
			}
		}
		return Response.status(Response.Status.OK).entity("File '" + filePath + "' uploaded successfully")
				.type(MediaType.TEXT_PLAIN).build();
	}
	private String getFileName(MultivaluedMap<String, String> headers) {
		String[] contentDisposition = headers.getFirst("Content-Disposition").split(";");
		for (String filename : contentDisposition) {
			if ((filename.trim().startsWith("filename"))) {
				String[] name = filename.split("=");
				String finalFileName = name[1].trim().replaceAll(""", "");
				return finalFileName;
			}
		}
		return "";
	}
}

In the above class, the file will be uploaded to the directory D:

Step 6. Create a main class which will be used for testing the file upload fucntionality

package com.roytuts.rest.test;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.InputStream;
import javax.ws.rs.client.Client;
import javax.ws.rs.client.ClientBuilder;
import javax.ws.rs.client.Entity;
import javax.ws.rs.core.MediaType;
import javax.ws.rs.core.Response;
public class RestTest {
	public static void main(String[] args) {
		try {
			File file = new File("D:/Documents/JavaJ2EE.doc");
			InputStream fileInStream = new FileInputStream(file);
			String contentDisposition = "attachment; filename="" + file.getName() + """;
			Client client = ClientBuilder.newClient();
			Response response = client.target("http://localhost:8080/rest/file").path("upload").request()
					.header("Content-Disposition", contentDisposition)
					.post(Entity.entity(fileInStream, MediaType.APPLICATION_OCTET_STREAM));
			System.out.println(response.readEntity(String.class));
		} catch (FileNotFoundException e) {
			e.printStackTrace();
		}
	}
}

Step 7. Now run the application on Tomcat 8. Run the above main class. Make sure you have a file JavaJ2EE.doc under D:Documents directory because this file will be uploaded using the REST service. Once the file is uploaded to D: drive then you will see the following output

REST file upload
Thanks for reading.

2 thoughts on “Upload file using REST webservice

Leave a Reply

Your email address will not be published. Required fields are marked *