如何在InputStream中接收多个文件并相应地处理它？

Question

我想收到从客户端上传的多个文件.我上传了多个文件并使用JAX-RS(Jersey)请求我的服务器端(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);
            }
}

如何在客户端上传的服务器端单独编写文件.

例如.我上传了诸如的文件My_File.txt, My_File.PNG, My_File.doc.我需要My_File.txt, My_File.PNG, My_File.doc在服务器端编写与上面相同的内容.

我怎样才能做到这一点？

Answer 1

你可以尝试这样的事情:

@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()
        }
    }
}