Question

您如何设计基于REST的Web服务，以InputStream的形式接收图像文件？如果将InputStream发布到REST端点，那么该端点如何接收它以便它可以创建图像文件？

Answer 1

在JAX-RS中可以接收InputStream。您只需将InputStream参数放在没有注释的地方：

@POST
public void uploadImage(InputStream stream) {
    // store image
}

请注意它适用于任何内容类型。

虽然它会起作用，但我建议采用更多“JAX-RS方式”：

1创建将从InputStream创建图像类（例如java.awt.Image）的提供程序：

@Provider
@Consumes("image/jpeg")
class ImageProvider implements MessageBodyReader<Image> {

    public Image readFrom(Class<Image> type,
                                Type genericType,
                                Annotation[] annotations,
                                MediaType mediaType,
                                MultivaluedMap<String, String> httpHeaders,
                                InputStream entityStream) throws IOException,
        WebApplicationException {
       // create Image from stream
    }
}

2以与注册资源相同的方式注册提供商 3使您的资源类接收Image而不是InputStream。

为什么这种方法更好？
您将反序列化逻辑与资源类分开。因此，如果将来您希望支持更多图像格式，您只需添加其他提供程序，而资源将保持不变。

Answer 2

示例REST Web服务Java类以接收图像

import java.io.ByteArrayInputStream;
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.FormParam;
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.MediaType;
import javax.ws.rs.core.Response;
import javax.ws.rs.core.UriInfo;

import com.sun.jersey.core.header.FormDataContentDisposition;
import com.sun.jersey.multipart.FormDataParam;
import com.sun.org.apache.xml.internal.security.exceptions.Base64DecodingException;
import com.sun.org.apache.xml.internal.security.utils.Base64;


@Path("/upload")
public class Upload_Documents {


    private static final String UPLOAD_FOLDER = "c:/uploadedFiles/";

    @Context
    private UriInfo context;
    /**
     * Returns text response to caller containing uploaded file location
     * 
     * @return error response in case of missing parameters an internal
     *         exception or success response if file has been stored
     *         successfully
     */
    @POST
    @Path("/pic")
    @Consumes(MediaType.MULTIPART_FORM_DATA)
    public Response uploadFile(
            @FormDataParam("file") InputStream uploadedInputStream,
            @FormDataParam("file") FormDataContentDisposition fileDetail) {

        System.out.println("Called Upload Image");
        // check if all form parameters are provided
        if (uploadedInputStream == null || fileDetail == null)
            return Response.status(400).entity("Invalid form data").build();
        // create our destination folder, if it not exists
        try {
            createFolderIfNotExists(UPLOAD_FOLDER);
        } catch (SecurityException se) {
            return Response.status(500)
                    .entity("Can not create destination folder on server")
                    .build();
        }
        String uploadedFileLocation = UPLOAD_FOLDER + fileDetail.getFileName();
        try {
            saveToFile(uploadedInputStream, uploadedFileLocation);
        } catch (IOException e) {
            return Response.status(500).entity("Can not save file").build();
        }
        return Response.status(200)
                .entity("File saved to " + uploadedFileLocation).build();
    }
    /**
     * Utility method to save InputStream data to target location/file
     * 
     * @param inStream
     *            - InputStream to be saved
     * @param target
     *            - full path to destination file
     */
    private void saveToFile(InputStream inStream, String target)
            throws IOException {
        OutputStream out = null;
        int read = 0;
        byte[] bytes = new byte[1024];
        out = new FileOutputStream(new File(target));
        while ((read = inStream.read(bytes)) != -1) {
            out.write(bytes, 0, read);
        }
        out.flush();
        out.close();
    }
    /**
     * Creates a folder to desired location if it not already exists
     * 
     * @param dirName
     *            - full path to the folder
     * @throws SecurityException
     *             - in case you don't have permission to create the folder
     */
    private void createFolderIfNotExists(String dirName)
            throws SecurityException {
        File theDir = new File(dirName);
        if (!theDir.exists()) {
            theDir.mkdir();
        }
}
}

创建Rest Web服务以接收映像

2 个答案:

示例REST Web服务Java类以接收图像