How to delete an image file from GridFS by file metadata?

Question

I have an image with the following metadata:

> db.fs.files.find().pretty()
{
    "_id" : ObjectId("4576874577342672346"),
    "chunkSize" : 262144,
    "user_name" : "my name",
    "filename" : "image.jpg",
    "length" : 7103,
    "uploadDate" : ISODate("2014-01-23T13:31:48.155Z"),
    "user_email" : "[email protected]",
    "md5" : "1234567890"
}
>

I want to delete the image from Python (PyMongo).

The documentation on delete() seems to stipulate that the only accepted parameter in the delete() function is the file_id:

http://api.mongodb.org/python/current/api/gridfs/#gridfs.GridFS.delete

Programmatically, I have the following values available that can be matched in the files metadata:

• user_name
• filename
• user_email

How do I either:

• Get the file_id (through use of the above values if necessary) or
• Delete the file based on metadata fields other than file_id?

Additionally, I am only currently testing with single chunk files, if interacting with larger files in the future will deleting by file_id or other metadata remove all associated chunks as well?

Answer 1

Here's something I just tried without thinking if it's necessary or the best way to do it, but it works.

So programatically I could have the _id available from querying on the files metadata:

Python Shell:

>>> import pymongo
>>> import os
>>> hostname = os.environ['OPENSHIFT_MONGODB_DB_URL']
>>> conn = pymongo.MongoClient(host=hostname)
>>> db = conn.grid_files
>>> collection = db.fs.files
>>> result = collection.find_one({"user_email":"[email protected]","name":"my name","filename":"image.jpg"})
>>> result['_id']
ObjectId('52e119c47091447a86891d98')

# now use the _id to delete the file
>>> files_id = result['_id']
>>> import gridfs
>>> fs = gridfs.GridFS(db)
>>> fs.delete(files_id)