ruby 在 Rails 中使用 uuidtools 生成短的 UUID 字符串
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Generating a short UUID string using uuidtools In Rails
提问by Siddharth
I have to generate a unique and random string which is to be stored in database. For doing this I have used the "uuidtools" gem. Then in my controller I have added the following line:
我必须生成一个唯一的随机字符串，该字符串将存储在数据库中。为此，我使用了“uuidtools” gem。然后在我的控制器中，我添加了以下行：
require "uuidtools"
and then in my controllers create method I have declared a 'temp' variable and generating a unique and random 'uuid' string like this:
然后在我的控制器 create 方法中，我声明了一个 'temp' 变量并生成一个唯一且随机的 'uuid' 字符串，如下所示：
temp=UUIDTools::UUID.random_create
which is creating a string like this one:
它正在创建一个这样的字符串：
f58b1019-77b0-4d44-a389-b402bb3e6d50
Now my problem is I have to make it short, preferably within 8-10 character. Now how do I do it?? Is it possible to pass any argument to make it a desirable length string??
现在我的问题是我必须让它简短，最好在 8-10 个字符内。现在我该怎么做？？是否可以传递任何参数使其成为理想的长度字符串？？
Thanks in Advance...
提前致谢...
回答by Dogbert
You don't need uuidtools for this. You can use Secure Randomfor this.
为此，您不需要 uuidtools。您可以为此使用安全随机。
[1] pry(main)> require "securerandom"
=> true
[2] pry(main)> SecureRandom.hex(20)
=> "82db4d707c4c5db3ebfc349da09c991b7ca0faa1"
[3] pry(main)> SecureRandom.base64(20)
=> "CECjUqNvPBaq0o4OuPy8RvsEoCY="
Passing 4and 5to hexwill generate 8 and 10 character hex strings respectively.
传递4和5tohex将分别生成 8 个和 10 个字符的十六进制字符串。
[5] pry(main)> SecureRandom.hex(4)
=> "a937ec91"
[6] pry(main)> SecureRandom.hex(5)
=> "98605bb20a"
回答by Rameshwar Vyevhare
Please see in detail, How I used securerandom in one of my project recently, definitely help you!
请详细查看，我最近如何在我的一个项目中使用securerandom，绝对对您有帮助！
create usesguid.rb file in your lib/usesguid.rb and paste below code in that -
在你的 lib/usesguid.rb 中创建 usesguid.rb 文件并粘贴下面的代码 -
require 'securerandom'
module ActiveRecord
module Usesguid #:nodoc:
def self.append_features(base)
super
base.extend(ClassMethods)
end
module ClassMethods
def usesguid(options = {})
class_eval do
self.primary_key = options[:column] if options[:column]
after_initialize :create_id
def create_id
self.id ||= SecureRandom.uuid
end
end
end
end
end
end
ActiveRecord::Base.class_eval do
include ActiveRecord::Usesguid
end
add following line in your config/application.rb to load file -
在 config/application.rb 中添加以下行以加载文件 -
require File.dirname(__FILE__) + '/../lib/usesguid'
Create migration script for UUID function as mentioned below to -
为 UUID 函数创建迁移脚本，如下所述 -
class CreateUuidFunction < ActiveRecord::Migration
def self.up
execute "create or replace function uuid() returns uuid as 'uuid-ossp', 'uuid_generate_v1' volatile strict language C;"
end
def self.down
execute "drop function uuid();"
end
end
Here is example for contact migration, how we can use it -
这是联系人迁移的示例，我们如何使用它 -
class CreateContacts < ActiveRecord::Migration
def change
create_table :contacts, id: false do |t|
t.column :id, :uuid, null:false
t.string :name
t.string :mobile_no
t.timestamps
end
end
end
Final how to use into your model
最终如何使用到您的模型中
class Contact < ActiveRecord::Base
usesguid
end
This will help you to configure UUID for your rails application.
这将帮助您为 Rails 应用程序配置 UUID。
This can be useful for Rails 3.0, 3.1, 3.2 and 4.0 as well.
这对 Rails 3.0、3.1、3.2 和 4.0 也很有用。
Please let me know If you have any issue while using it, so simple!
如果您在使用时遇到任何问题，请告诉我，就这么简单！
