我刚刚开始开发REST服务,但我遇到了一个困难的情况:从我的REST服务向我的客户端发送文件。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,我不知道我应该在哪里开始。我的REST服务是基于Java的,我使用的是Jersey,我使用JSON格式发送所有数据。
我读过有关base64编码的内容,有人说这是一种很好的技术,有人说这不是因为文件大小问题。什么是正确的方法?这就是我项目中一个简单的资源类的样子:
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;
}
}
我猜测发送文件的代码类似于:
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?我见过一些人推荐@GET
使用@Produces({application/x-octet-stream})
,这是正确的方法吗?我发送的文件是特定文件,因此客户端不需要浏览文件。任何人都可以指导我如何发送文件?我应该使用base64对其进行编码以将其作为JSON对象发送吗?或者不需要编码将其作为JSON对象发送?感谢您提供的任何帮助。
答案 0 :(得分:118)
我不建议在base64中编码二进制数据并将其包装在JSON中。它会不必要地增加响应的大小并减慢速度。
使用GET和application/octect-stream
使用javax.ws.rs.core.Response
的工厂方法(JAX-RS API的一部分,因此您不会被锁定到Jersey),只需使用GET和@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();
}
提供文件数据:
File
如果您没有实际的InputStream
对象,但{{1}},Response.ok(entity, mediaType)
也应该能够处理该对象。
答案 1 :(得分:5)
如果要返回要下载的文件,特别是如果要与文件上传/下载的某些javascript库集成,那么下面的代码应该可以完成这项工作:
@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();
}
答案 2 :(得分:5)
将机器地址从localhost更改为您希望客户端连接的IP地址,以便调用下面提到的服务。
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();
}
}
响应客户服务:
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需要:
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>
答案 3 :(得分:-2)
由于您使用的是JSON,我会在通过网络发送之前对Base64进行编码。
如果文件很大,请尝试查看BSON,或其他一些二进制传输更好的格式。
如果文件压缩得很好,你也可以压缩文件,然后再对base64进行编码。