将文件从 REST Web 服务发送到客户端的正确方法是什么？

我刚刚开始开发 REST 服务，但是我遇到了一个困难的情况: 将文件从 REST 服务发送到我的客户端。到目前为止，我已经掌握了如何发送简单数据类型(字符串、整数等) ，但发送文件是另一回事，因为有太多的文件格式，我甚至不知道从哪里开始。我的 REST 服务是用 Java 制作的，我使用 Jersey，我使用 JSON 格式发送所有数据。

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;
}
}

我猜发送文件的代码是这样的:

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;
}
}

我应该使用哪种注释？我看到一些人推荐使用 @Produces({application/x-octet-stream})的 @GET，这是正确的方法吗？我发送的文件是特定的，所以客户端不需要浏览这些文件。有人能告诉我该怎么发送文件吗？我是否应该使用 base64对其进行编码，以将其作为 JSON 对象发送？或者不需要编码就可以将其作为 JSON 对象发送？谢谢你的帮助。

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小开

因为您使用的是 JSON，所以在通过网络发送之前，我会先对它进行 Base64编码。

如果文件很大，请尝试查看 BSON 或其他更适合二进制传输的格式。

您还可以在 base64对文件进行编码之前压缩这些文件(如果压缩得很好)。

小开

如果你想返回一个要下载的文件，特别是如果你想集成一些文件上传/下载的 javascript 库，那么下面的代码就可以完成这项工作:

@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();
}
小开
最佳答案

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.

只需使用 GET 和 application/octect-stream使用 javax.ws.rs.core.Response的一种工厂方法(JAX-RS API 的一部分，这样您就不会被锁定到 Jersey)提供文件数据:

@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();
}

如果你没有一个实际的 File对象，但一个 InputStream，Response.ok(entity, mediaType)应该能够处理这一点。

小开

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

客户端调用 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>

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