explain internal _id format (e.g. author_1_0000000000000019)

[gr2m]

I couldn’t tell from the README what the 1 means in author_1_0000000000000019? For example here: https://github.com/nolanlawson/relational-pouch#dbrelparsedociddocid

I understand the author_ prefix, and I understand the 0000000000000019 part is the author’s id, just not the part in between :)

Can you explain and maybe amend the README?

[jkleinsc]

@gr2m the number in the middle is the "type" of ID used.

If the id is undefined the value is 0, if the id is a number, the value is 1, if the id is a string the value is 2, and if the id is an object the value is 3. For example:

Given an object of type author with the following ids:

var authorId = 123;
var relPouchId = db.rel.makeDocID({
  id: authorId,
  type: 'author'
});
//relPouchId = 'author_1_123'

var authorId = 'onetwothree';
var relPouchId = db.rel.makeDocID({
  id: authorId,
  type: 'author'
});
//relPouchId = 'author_2_onetwothree'

var authorId = { foo: 'bar' };
var relPouchId = db.rel.makeDocID({
  id: authorId,
  type: 'author'
});
//relPouchId = 'author_3'

The reason for these numbers is sorting/collation so that ids sort as follows:
// undefined < numbers < strings < object

The only use case I have seen for the undefined and "object" id is when you are looking for a range of docs using startkey/endkey and you use a start key with an id of undefined and an end key with an object id to get the "max value", eg as used here: https://github.com/nolanlawson/relational-pouch/blob/7f3b34f70be189cabbdf0f525b64f6be464b0921/lib/index.js#L425

[gr2m]

This is a great explanation, thanks John! Maybe we can add this to or link this from the README?

[jkleinsc]

@gr2m yeah, I think it could be added here: https://github.com/nolanlawson/relational-pouch#dbrelmakedocidparsedid
@nolanlawson Did I properly explain the reasoning for the middle part of relational-pouch ids above?

[nolanlawson]

@jkleinsc Yep! :)

[broerse]

@gr2m If you like to write a PR for the readme I will commit it.

[gr2m]

@broerse #69

[broerse]

Thanks!