如何在Java Rest API中获取真实路径InputStream文件

Question

我有一个Java Rest API Post方法，该方法发送图像（InputStream）作为参数，并且必须将其保存在Oracle的blob列中。

我需要获取此InputStream的完整路径（真实路径）以将该图像保存在数据库中。我的代码在下面。

@POST
@Path("/upload")
@Consumes(MediaType.MULTIPART_FORM_DATA)
public Response uploadFile(
                @FormDataParam("file") InputStream uploadedInputStream,
                @FormDataParam("file") FormDataContentDisposition fileDetail) {
String UPLOAD_FOLDER = "c:/uploadedFiles/"; // my rest api does not have this file, how to get at runtime?    
String uploadedFileLocation = UPLOAD_FOLDER + fileDetail.getFileName(); // this line is ok

我想做这样的事情：

String UPLOAD_FOLDER = uploadedInputStream.getRealPathName();

或

String UPLOAD_FOLDER = fileDetail.getRealPathName();

Answer 1

如何在客户端上获取路径并将其与上传请求一起发送到服务器？

Answer 2

我通过将输入流转换为字节数组解决了这个问题。我将byte []转发到了数据库持久性方法。我的代码如下：

public byte[] toByteArray(InputStream is) throws IOException{
    ByteArrayOutputStream baus = new ByteArrayOutputStream();
    byte[] buffer = new byte[1024];
    int len;
    while((len= is.read(buffer)) != -1){
        baus.write(buffer, 0, len);
    }
    return baus.toByteArray();
}

@POST
@Path("/upload")
@Consumes(MediaType.MULTIPART_FORM_DATA)
public Response uploadFile(
                @FormDataParam("file") InputStream uploadedInputStream,
                @FormDataParam("file") FormDataContentDisposition fileDetail) {
    ...
    byte[] b = toByteArray(uploadedInputStream);
    business.saveUploadedFileInDatabase(b);
    ...
}

public boolean uploadFile(byte[] b) throws SQLException, IOException{
    ...
    PreparedStatement ps = conn.prepareStatement("INSERT INTO TABLE_IMAGE_TEST (CODE, IMAGE) VALUES (?, ?)");
    pstmt.setLong(1, 1L);
    pstmt.setBytes(2, b);
    pstmt.execute();
    ...
}