pymongo : how to copy database ?

[Nicolas Clairon]

Hi,

I want to copy a database to another name into the same server with
pymongo. The mongodb documentation says:

> db.copyDatabase(<from_dbname>, <to_dbname>, <from_host>);

but there's no such method on pymongo (no Connection.copy_database).
So I tried with the `command` method without success :

>>> con = pymongo.Connection()
>>> con.test.command({'copydb':True, 'fromdb': 'test', 'todb': 'test2'})
<class 'pymongo.errors.OperationFailure'>: command {'fromdb': 'test',
'todb': 'test2', 'copydb': True} failed: no such cmd

>>> con.test.command({'copydb':True})
<class 'pymongo.errors.OperationFailure'>: command {'copydb': True}
failed: access denied

Any ideas ?

N.

[Dwight Merriman]

you can use runCommand to do it.  evaling copyDatabase() in the sheell without the parens will show what is does, hopefully you can figure out the right command from that as i don't know pymongo:

> db.copyDatabase
function (fromdb, todb, fromhost, username, password) {
    assert(isString(fromdb) && fromdb.length);
    assert(isString(todb) && todb.length);
    fromhost = fromhost || "";
    if (username && password) {
        var n = this._adminCommand({copydbgetnonce:1, fromhost:fromhost});
        return this._adminCommand({copydb:1, fromhost:fromhost, fromdb:fromdb, todb:todb, username:username, nonce:n.nonce, key:this.__pwHash(n.nonce, username, password)});
    } else {
        return this._adminCommand({copydb:1, fromhost:fromhost, fromdb:fromdb, todb:todb});
    }
}

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[Nicolas Clairon]

Ok, I succeed to copy a database via the mongo shell :

> use admin
> db.runCommand({'copydb':1, 'fromdb': 'test', 'todb': 'test2'})

but when I run the same command with pymongo, I doesn't work:

>>> import pymongo
>>>> con = pymongo.Connection()

>>> con.admin.command({'copydb':1, 'fromdb': 'test', 'todb': 'test2'})

<class 'pymongo.errors.OperationFailure'>: command {'fromdb': 'test',

'todb': 'test2', 'copydb': 1} failed: no such cmd

What append ?

# pymongo 1.4
# mongodb 1.3.2

[Eliot Horowitz]

I think you need to use SON or something like that in pymongo since
dicts aren't ordered

[Nicolas Clairon]

OK, I was just mis-use SON :

>>> from pymongo.son import SON
>>> con.admin.command(SON(copydb=1, fromdb='test', todb='test2'))
<class 'pymongo.errors.OperationFailure'>: command SON([('fromdb',
'test'), ('todb', 'test2'), ('copydb', 1)]) failed: no such cmd

I also tried :

>>> con.admin.command(SON(copydb=1, fromdb='test', todb='test2'))
<class 'pymongo.errors.OperationFailure'>: command SON([('fromdb',
'test'), ('todb', 'test2'), ('copydb', 1)]) failed: no such cmd

Which is weird as arguments are ordered...

Anyway, after digging, I tried :

>>> con.admin.command(SON([("copydb", 1), ("fromdb", 'test'), ("todb", 'test2')]))
{u'ok': 1.0}

Thanks !

[Dwight Merriman]

the command name has to be the first BSON field.  i don't know how to do that with PyMongo will wait for Mike to answer.