The first big question when diving in to this is "how do you want to store changesets"?
- Diffs?
- Whole record copies?
My personal approach would be to store diffs. Because the display of these diffs is really a special action, I would put the diffs in a different "history" collection.
I would use the different collection to save memory space. You generally don't want a full history for a simple query. So by keeping the history out of the object you can also keep it out of the commonly accessed memory when that data is queried.
To make my life easy, I would make a history document contain a dictionary of time-stamped diffs. Something like this:
{
_id : "id of address book record",
changes : {
1234567 : { "city" : "Omaha", "state" : "Nebraska" },
1234568 : { "city" : "Kansas City", "state" : "Missouri" }
}
}
To make my life really easy, I would make this part of my DataObjects (EntityWrapper, whatever) that I use to access my data. Generally these objects have some form of history, so that you can easily override the save()
method to make this change at the same time.
UPDATE: 2015-10
It looks like there is now a spec for handling JSON diffs. This seems like a more robust way to store the diffs / changes.
Short Answer
Just to add a direct response to your initial question: YES, if you use BSON Object ID generation, then for most drivers the IDs are almost certainly going to be unique across collections. See below for what "almost certainly" means.
Long Answer
The BSON Object ID's generated by Mongo DB drivers are highly likely to be unique across collections. This is mainly because of the last 3 bytes of the ID, which for most drivers is generated via a static incrementing counter. That counter is collection-independent; it's global. The Java driver, for example, uses a randomly initialized, static AtomicInteger.
So why, in the Mongo docs, do they say that the IDs are "highly likely" to be unique, instead of outright saying that they WILL be unique? Three possibilities can occur where you won't get a unique ID (please let me know if there are more):
Before this discussion, recall that the BSON Object ID consists of:
[4 bytes seconds since epoch, 3 bytes machine hash, 2 bytes process ID, 3 bytes counter]
Here are the three possibilities, so you judge for yourself how likely it is to get a dupe:
1) Counter overflow: there are 3 bytes in the counter. If you happen to insert over 16,777,216 (2^24) documents in a single second, on the same machine, in the same process, then you may overflow the incrementing counter bytes and end up with two Object IDs that share the same time, machine, process, and counter values.
2) Counter non-incrementing: some Mongo drivers use random numbers instead of incrementing numbers for the counter bytes. In these cases, there is a 1/16,777,216 chance of generating a non-unique ID, but only if those two IDs are generated in the same second (i.e. before the time section of the ID updates to the next second), on the same machine, in the same process.
3) Machine and process hash to the same values. The machine ID and process ID values may, in some highly unlikely scenario, map to the same values for two different machines. If this occurs, and at the same time the two counters on the two different machines, during the same second, generate the same value, then you'll end up with a duplicate ID.
These are the three scenarios to watch out for. Scenario 1 and 3 seem highly unlikely, and scenario 2 is totally avoidable if you're using the right driver. You'll have to check the source of the driver to know for sure.
Best Answer
For question #1, let's break it into two parts. First, increment any document that has "items.item_name" equal to "my_item_two". For this you'll have to use the positional "$" operator. Something like:
Note that this will only increment the first matched subdocument in any array (so if you have another document in the array with "item_name" equal to "my_item_two", it won't get incremented). But this might be what you want.
The second part is trickier. We can push a new item to an array without a "my_item_two" as follows:
For your question #2, the answer is easier. To increment the total and the price of item_three in any document that contains "my_item_three," you can use the $inc operator on multiple fields at the same time. Something like: