what's the correct way to send a file from REST web service to client?

Java Problem Overview

I've just started to develop REST services, but I've come across a difficult situation: sending files from my REST service to my client. So far I've gotten the hang of how to send simple data types (strings, integers, etc) but sending a file is a different matter since there are so many file formats that I don't know where I should even begin. My REST service is made on Java and I'm using Jersey, I'm sending all the data using the JSON format.

I've read about base64 encoding, some people say it's a good technique, others say it isn't because of file size issues. What is the correct way? This is how a simple resource class in my project is looking:

import java.sql.SQLException;
import java.util.List;

import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.Context;
import javax.ws.rs.core.MediaType;
import javax.ws.rs.core.Request;
import javax.ws.rs.core.UriInfo;

import com.mx.ipn.escom.testerRest.dao.TemaDao;
import com.mx.ipn.escom.testerRest.modelo.Tema;

@Path("/temas")
public class TemaResource {
		
	@GET
	@Produces({MediaType.APPLICATION_JSON})
	public List<Tema> getTemas() throws SQLException{

		TemaDao temaDao = new TemaDao();		
		List<Tema> temas=temaDao.getTemas();
		temaDao.terminarSesion();
		
		return temas;
	}
}

I'm guessing the code for sending a file would be something like:

import java.sql.SQLException;

import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;

@Path("/resourceFiles")
public class FileResource {
		
	@GET
	@Produces({application/x-octet-stream})
	public File getFiles() throws SQLException{ //I'm not really sure what kind of data type I should return

		// Code for encoding the file or just send it in a data stream, I really don't know what should be done here
		
		return file;
	}
}

What kind of annotations should I use? I've seen some people recommend for a @GET using @Produces({application/x-octet-stream}), is that the correct way? The files I'm sending are specific ones so the client doesn't need to browse through the files. Can anyone guide me into how am I supposed to send the file? Should I encode it using base64 to send it as a JSON object? or the encoding isn't necessary to send it as a JSON object? Thanks for any help you may give.

Java Solutions

Solution 1 - Java

I don't recommend encoding binary data in base64 and wrapping it in JSON. It will just needlessly increase the size of the response and slow things down.

Simply serve your file data using GET and application/octect-streamusing one of the factory methods of javax.ws.rs.core.Response (part of the JAX-RS API, so you're not locked into Jersey):

@GET
@Produces(MediaType.APPLICATION_OCTET_STREAM)
public Response getFile() {
  File file = ... // Initialize this to the File path you want to serve.
  return Response.ok(file, MediaType.APPLICATION_OCTET_STREAM)
      .header("Content-Disposition", "attachment; filename=\"" + file.getName() + "\"" ) //optional
      .build();
}

If you don't have an actual File object, but an InputStream, Response.ok(entity, mediaType) should be able to handle that as well.

Solution 2 - Java

If you want to return a File to be downloaded, specially if you want to integrate with some javascript libs of file upload/download, then the code bellow should do the job:

@GET
@Path("/{key}")
public Response download(@PathParam("key") String key,
                         @Context HttpServletResponse response) throws IOException {
	try {
		//Get your File or Object from wherever you want...
            //you can use the key parameter to indentify your file
            //otherwise it can be removed
		//let's say your file is called "object"
		response.setContentLength((int) object.getContentLength());
		response.setHeader("Content-Disposition", "attachment; filename="
				+ object.getName());
		ServletOutputStream outStream = response.getOutputStream();
		byte[] bbuf = new byte[(int) object.getContentLength() + 1024];
		DataInputStream in = new DataInputStream(
				object.getDataInputStream());
		int length = 0;
		while ((in != null) && ((length = in.read(bbuf)) != -1)) {
			outStream.write(bbuf, 0, length);
		}
		in.close();
		outStream.flush();
	} catch (S3ServiceException e) {
		e.printStackTrace();
	} catch (ServiceException e) {
		e.printStackTrace();
	}
	return Response.ok().build();
}

Solution 3 - Java

Change the machine address from localhost to IP address you want your client to connect with to call below mentioned service.

Client to call REST webservice:

package in.india.client.downloadfiledemo;

import java.io.BufferedInputStream;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.FileOutputStream;
import java.io.IOException;

import javax.ws.rs.core.MediaType;
import javax.ws.rs.core.Response.Status;

import com.sun.jersey.api.client.Client;
import com.sun.jersey.api.client.ClientHandlerException;
import com.sun.jersey.api.client.ClientResponse;
import com.sun.jersey.api.client.UniformInterfaceException;
import com.sun.jersey.api.client.WebResource;
import com.sun.jersey.multipart.BodyPart;
import com.sun.jersey.multipart.MultiPart;

public class DownloadFileClient {

    private static final String BASE_URI = "http://localhost:8080/DownloadFileDemo/services/downloadfile";

    public DownloadFileClient() {

        try {
            Client client = Client.create();
            WebResource objWebResource = client.resource(BASE_URI);
            ClientResponse response = objWebResource.path("/")
                    .type(MediaType.TEXT_HTML).get(ClientResponse.class);

            System.out.println("response : " + response);
            if (response.getStatus() == Status.OK.getStatusCode()
                    && response.hasEntity()) {
                MultiPart objMultiPart = response.getEntity(MultiPart.class);
                java.util.List<BodyPart> listBodyPart = objMultiPart
                        .getBodyParts();
                BodyPart filenameBodyPart = listBodyPart.get(0);
                BodyPart fileLengthBodyPart = listBodyPart.get(1);
                BodyPart fileBodyPart = listBodyPart.get(2);

                String filename = filenameBodyPart.getEntityAs(String.class);
                String fileLength = fileLengthBodyPart
                        .getEntityAs(String.class);
                File streamedFile = fileBodyPart.getEntityAs(File.class);

                BufferedInputStream objBufferedInputStream = new BufferedInputStream(
                        new FileInputStream(streamedFile));

                byte[] bytes = new byte[objBufferedInputStream.available()];

                objBufferedInputStream.read(bytes);

                String outFileName = "D:/"
                        + filename;
                System.out.println("File name is : " + filename
                        + " and length is : " + fileLength);
                FileOutputStream objFileOutputStream = new FileOutputStream(
                        outFileName);
                objFileOutputStream.write(bytes);
                objFileOutputStream.close();
                objBufferedInputStream.close();
                File receivedFile = new File(outFileName);
                System.out.print("Is the file size is same? :\t");
                System.out.println(Long.parseLong(fileLength) == receivedFile
                        .length());
            }
        } catch (UniformInterfaceException e) {
            e.printStackTrace();
        } catch (ClientHandlerException e) {
            e.printStackTrace();
        } catch (FileNotFoundException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }

    }

    public static void main(String... args) {
        new DownloadFileClient();
    }
}

Service to response client:

package in.india.service.downloadfiledemo;

import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;
import javax.ws.rs.core.Response;

import com.sun.jersey.multipart.MultiPart;

@Path("downloadfile")
@Produces("multipart/mixed")
public class DownloadFileResource {

    @GET
    public Response getFile() {

        java.io.File objFile = new java.io.File(
                "D:/DanGilbert_2004-480p-en.mp4");
        MultiPart objMultiPart = new MultiPart();
        objMultiPart.type(new MediaType("multipart", "mixed"));
        objMultiPart
                .bodyPart(objFile.getName(), new MediaType("text", "plain"));
        objMultiPart.bodyPart("" + objFile.length(), new MediaType("text",
                "plain"));
        objMultiPart.bodyPart(objFile, new MediaType("multipart", "mixed"));

        return Response.ok(objMultiPart).build();

    }
}

JAR needed:

jersey-bundle-1.14.jar
jersey-multipart-1.14.jar
mimepull.jar

WEB.XML:

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xmlns="http://java.sun.com/xml/ns/javaee" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
    xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
    id="WebApp_ID" version="2.5">
    <display-name>DownloadFileDemo</display-name>
    <servlet>
        <display-name>JAX-RS REST Servlet</display-name>
        <servlet-name>JAX-RS REST Servlet</servlet-name>
        <servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
        <init-param>
             <param-name>com.sun.jersey.config.property.packages</param-name> 
             <param-value>in.india.service.downloadfiledemo</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>JAX-RS REST Servlet</servlet-name>
        <url-pattern>/services/*</url-pattern>
    </servlet-mapping>
    <welcome-file-list>
        <welcome-file>index.jsp</welcome-file>
    </welcome-file-list>
</web-app>

Solution 4 - Java

Since youre using JSON, I would Base64 Encode it before sending it across the wire.

If the files are large, try to look at BSON, or some other format that is better with binary transfers.

You could also zip the files, if they compress well, before base64 encoding them.

Content Type	Original Author	Original Content on Stackoverflow
Question	Uriel	View Question on Stackoverflow
Solution 1 - Java	Philipp Reichart	View Answer on Stackoverflow
Solution 2 - Java	john 4d5	View Answer on Stackoverflow
Solution 3 - Java	Shankar Saran Singh	View Answer on Stackoverflow
Solution 4 - Java	LarsK	View Answer on Stackoverflow