I can't figure out how to stream a binary file from GridFS with spring-data-mongodb and its GridFSTemplate
when I already have the right ObjectId
.
GridFSTemplate returns either GridFSResource
(getResource()
) or GridFSFile
(findX()
).
I can get the GridFSFile
by ID:
// no way to get the InputStream?
GridFSFile file = gridFsTemplate.findOne(Query.query(Criteria.where("_id").is(id)))
but there is no obvious way how to get an InputStream
for that GridFSFile
.
Only GridFSResource
allows me to get hold of the corresonding InputStream
with InputStreamResource#getInputstream
. But the only way to get a GridFSResource
is by its filename
.
// no way to get GridFSResource by ID?
GridFSResource resource = gridFsTemplate.getResource("test.jpeg");
return resource.getInputStream();
Somehow the GridFsTemplate
API implies that filenames are unique - which they are not. The GridFsTemplate
implementation just returns the first element.
Now I'm using the native MongoDB API and everything makes sense again:
GridFS gridFs = new GridFs(mongo);
GridFSDBFile nativeFile = gridFs.find(blobId);
return nativeFile.getInputStream();
It looks like I'm misunderstanding the fundamental concepts behind the Spring Data Mongo GridFS abstraction. I'd expect (at least) one of the following things to be possible/true:
GridFSResource
by its IDGridFSResource
or InputStream
for a GridFsFile
I already haveAm I wrong or is there something odd with this particular piece of the Spring Data MongoDB API?
I stumbled upon this, too. And I am actually pretty shocked that the GridFsTemplate has been designed like this... Anyway, my ugly "solution" to this so far:
public GridFsResource download(String fileId) {
GridFSFile file = gridFsTemplate.findOne(Query.query(Criteria.where("_id").is(fileId)));
return new GridFsResource(file, getGridFs().openDownloadStream(file.getObjectId()));
}
private GridFSBucket getGridFs() {
MongoDatabase db = mongoDbFactory.getDb();
return GridFSBuckets.create(db);
}
Note: You have to inject the MongoDbFactory for this to work...