如何在InputStream中接收多个文件并相应地处理它

Question

我想接收从我的客户端上传的多个文件。我上传了多个文件，并使用JAX-RS (Jersey)请求服务器端(Java)。

我有以下代码，

@POST
    @Consumes(MediaType.MULTIPART_FORM_DATA)
    public void upload(@Context UriInfo uriInfo,
            @FormDataParam("file") final InputStream is,
            @FormDataParam("file") final FormDataContentDisposition detail) {
FileOutputStream os = new FileOutputStream("Path/to/save/" + appropriatefileName);
    byte[] buffer = new byte[1024];
            int length;
            while ((length = is.read(buffer)) > 0) {
                os.write(buffer, 0, length);
            }
}

如何在服务器端单独写入在客户端上传的文件

例如

。上传了My_File.txt, My_File.PNG, My_File.doc等文件。我需要在服务器端写入与上述My_File.txt, My_File.PNG, My_File.doc相同的内容

我怎样才能做到这一点?

Answer 1

你可以尝试这样做:

@POST
@Consumes(MediaType.MULTIPART_FORM_DATA)
public void upload(FormDataMultiPart formParams)
{
    Map<String, List<FormDataBodyPart>> fieldsByName = formParams.getFields();
    // Usually each value in fieldsByName will be a list of length 1.
    // Assuming each field in the form is a file, just loop through them.
    for (List<FormDataBodyPart> fields : fieldsByName.values())
    {
        for (FormDataBodyPart field : fields)
        {
            InputStream is = field.getEntityAs(InputStream.class);
            String fileName = field.getName();
            // TODO: SAVE FILE HERE
            // if you want media type for validation, it's field.getMediaType()
        }
    }
}

Answer 2

有一个您正在寻找的场景的博客。希望这对你有所帮助http://opensourzesupport.wordpress.com/2012/10/27/multiple-file-upload-along-with-form-data-in-jax-rs/