How do I design the path to get the following data

[leeyisoft]

Uid1 = 1, Uid2 = 2, DType1 = <<"ios">>,  DType2 = <<"andriod">>.
DID1 = <<"did1">>, DID2 = <<"did2">>.
Pid1 = self(). Pid2 = self().

Uid1Bin = integer_to_binary(Uid1).

Path1 = <<"/:s/", Uid1Bin/binary, "/", DType1/binary>>.
khepri:put(Path1, {Uid1, DType1, Pid1, DID1}).

Path2 = <<"/:s/", Uid1Bin/binary, "/", DType2/binary>>.
khepri:put(Path2, {Uid1, DType2, Pid2, DID2}).

(imboy@127.0.0.1)171> khepri:get(Path1).
{ok,{1,<<"ios">>,<0.11996.0>,<<"did11">>}}
(imboy@127.0.0.1)172> khepri:get(Path2).
{ok,{1,<<"andriod">>,<0.11820.1>,<<"did2">>}}

How do I design the path to get the following data：

khepri:get({s, Uid})
{ok, [
  {1,<<"ios">>,<0.11996.0>,<<"did11">>}, 
  {1,<<"andriod">>,<0.11820.1>,<<"did2">>}
]}

[the-mikedavis]

You can lookup many paths in the store using khepri:get_many/3 with path patterns:

> khepri:get_many("/:s/1/*"). 
{ok,#{[s,<<"1">>,<<"andriod">>] =>
          {1,<<"andriod">>,<0.249.0>,<<"did2">>},
      [s,<<"1">>,<<"ios">>] =>
          {1,<<"ios">>,<0.249.0>,<<"did1">>}}}
> khepri:get_many("/:s/**").
{ok,#{[s,<<"1">>,<<"andriod">>] =>
          {1,<<"andriod">>,<0.249.0>,<<"did2">>},
      [s,<<"1">>,<<"ios">>] =>
          {1,<<"ios">>,<0.249.0>,<<"did1">>}}}

* matches any tree node and ** matches any path in the tree node. Also see the docs for khepri:get_many/3 and khepri_path:unix_pattern() in the documentation (https://rabbitmq.github.io/khepri/).

[leeyisoft]

In the example data above, there are two Uids and two device ids (did).

May I ask, how to write the method to count the total number of user ids and the number of device ids?

The way I know how to do this is to query all the data and then iterate through the list, but that's obviously not efficient (it's a problem if you have a million users online at the same time).以上翻译结果来自有道神经网络翻译（YNMT）· 通用场景

在上面的示例数据中，有两个uid，两个设备ID（did）。

请问，统计用户ID总数量，统计设备ID中数量的 方法如何编写？

我知道的方法是把所有数据查询出来在遍历列表，但是这样显然效率不高（有一百万用户同时在线就麻烦了）

[michaelklishin]

There are no functions that counts keys, you have to use khepri:fold/5. See #194.

[leeyisoft]

#194

Thank you for your help.
Here's how I did it:

    {ok, Li} = khepri:fold("/:s/*/*", fun([s, U, _DType] , _NP, Acc) ->
        [U|Acc]
    end, []),
    length(lists:uniq(Li)).

[dumbbell]

A khepri:count() function was added later in the API. It uses the fold function underneath, but it's a common enough use case that it warranted a dedicated function.

[leeyisoft]

It would be even better if khepri could retain mnesia's syntax, or support SQL syntax!

如果 khepri 能够保留 mnesia 的语法，或者支持SQL语法，那就更完美了！

[michaelklishin]

There are no plans to support anything from Mnesia.

Khepri's data model is not relational. If someone wants to come up with a SQL-like querying language
for Khepri, they are welcome to submit a draft PR. Khepri is open source software after all.