這裏有一種方法來創建開始於1000000列,並自動分配一個唯一的值,它是大致相當於前值+ 1(大概是因爲一個ID可能是因爲失敗的交易,可能還有一些其他的情況下「跳過」 )。
這個答案特定於PostgreSQL,因爲它使用PostgreSQL專用的setval
和pg_get_serial_sequence
函數。
遷移:
defmodule MyApp.Repo.Migrations.CreateCompany do
use Ecto.Migration
def up do
create table(:companies, primary_key: false) do
add :number, :serial, primary_key: true
timestamps()
end
execute "select setval(pg_get_serial_sequence('companies', 'number'), 999999)"
end
def down do
drop table(:companies)
end
end
型號:
defmodule MyApp.Company do
use MyApp.Web, :model
@primary_key false
schema "companies" do
field :number, :integer, primary_key: true, read_after_writes: true
timestamps()
end
end
演示:
iex(1)> Repo.insert! %Company{}
[debug] QUERY OK db=2.7ms queue=0.1ms
INSERT INTO "companies" ("inserted_at","updated_at") VALUES ($1,$2) RETURNING "number" [{{2016, 12, 5}, {15, 57, 44, 0}}, {{2016, 12, 5}, {15, 57, 44, 0}}]
%MyApp.Company{__meta__: #Ecto.Schema.Metadata<:loaded, "companies">,
inserted_at: #Ecto.DateTime<2016-12-05 15:57:44>, number: 1000000,
updated_at: #Ecto.DateTime<2016-12-05 15:57:44>}
iex(2)> Repo.insert! %Company{}
[debug] QUERY OK db=4.5ms
INSERT INTO "companies" ("inserted_at","updated_at") VALUES ($1,$2) RETURNING "number" [{{2016, 12, 5}, {15, 57, 44, 0}}, {{2016, 12, 5}, {15, 57, 44, 0}}]
%MyApp.Company{__meta__: #Ecto.Schema.Metadata<:loaded, "companies">,
inserted_at: #Ecto.DateTime<2016-12-05 15:57:44>, number: 1000001,
updated_at: #Ecto.DateTime<2016-12-05 15:57:44>}
iex(3)> Repo.insert! %Company{}
[debug] QUERY OK db=3.4ms queue=0.1ms
INSERT INTO "companies" ("inserted_at","updated_at") VALUES ($1,$2) RETURNING "number" [{{2016, 12, 5}, {15, 57, 45, 0}}, {{2016, 12, 5}, {15, 57, 45, 0}}]
%MyApp.Company{__meta__: #Ecto.Schema.Metadata<:loaded, "companies">,
inserted_at: #Ecto.DateTime<2016-12-05 15:57:45>, number: 1000002,
updated_at: #Ecto.DateTime<2016-12-05 15:57:45>}
一些注意事項:
你不使用這個自動增量鍵的原因嗎? – Dogbert
是的,因爲我想使用數字作爲ID。這不是一個好習慣嗎? –
那麼你是什麼意思「生成一個數字」呢?有自動遞增的整數鍵將創建具有值0,1,2條記錄,... – Dogbert