uuid | 易学教程

linux 的uuid码

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在提到这个之前，有个概念，就是什么是uuid呢？ UUID码全称是通用唯一识别码 (Universally Unique Identifier, UUID),它是一个软件建构的标准，亦为自由软件基金会 (Open Software Foundation, OSF) 的组织在分布式计算环境 (Distributed Computing Environment, DCE) 领域的一部份。 UUID 的目的，是让分布式系统中的所有元素，都能有唯一的辨识信息，而不需要通过中央控制端来做辨识信息的指定。如此一来，每个人都可以创建不与其它人冲突的 UUID。在这样的情况下，就不需考虑数据库创建时的名称重复问题。它会让网络任何一台计算机所生成的uuid码，都是互联网整个服务器网络中唯一的。它的原信息会加入硬件，时间，机器当前运行信息等等。 UUID格式是：包含32个16进位数字，以“-”连接号分为五段，形式为8-4-4-4-12的32个字符。范例；550e8400-e29b-41d4-a716-446655440000 ,所以：UUID理论上的总数为216 x 8=2128，约等于3.4 x 1038。也就是说若每奈秒产生1兆个UUID，要花100亿年才会将所有UUID用完。其实，大家做数据库设计时候，肯定听说过，guid(全局唯一标识符)码，它其实是与uuid类似，由微软支持。这里编码

shell实现随机数多种方法（data，random，uuid）

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在日常生活中，随机数实际上经常遇到，想丢骰子，抓阄，还有抽签。呵呵，非常简单就可以实现。那么在做程序设计，真的要通过自己程序设计出随机数那还真的不简单了。现在很多都是操作系统内核会提供相应的api，这些原始参数是获取一些计算机运行原始信息，如内存，电压，物理信号等等，它的值在一个时间段可以保证是唯一的了。好了，废话我就不说了。呵呵。 shell脚本程序我们有那些获得随机数方法呢？一、通过时间获得随机数（date) 这个也是我们经常用到的，可以说时间是唯一的，也不会重复的，从这个里面获得同一时间的唯一值。适应所有程序里面了。例子： ? 1 2 3 4 5 6 7 8 9 10 11 12 13 [chengmo@centos5 shell]$ date +%s 1287764773 #获得时间戳，当前到：1970-01-01 00:00:00 相隔的秒数 #如果用它做随机数，相同一秒的数据是一样的。在做循环处理，多线程里面基本不能满足要求了。 [chengmo@centos5 shell]$ date +%N 738710457 #获得当前时间的纳秒数据，精确到亿分之一秒。 #这个相当精确了，就算在多cpu，大量循环里面，同一秒里面，也很难出现相同结果，不过不同时间里面还会有大量重复碰撞 [chengmo@centos5 shell]$ date +%s%N

Implementing Unique ID as UUID and saving it in Keychain

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问题 I need unique ID in my app and i know, we cannot use UDID anymore so as per my research, using UUID as device unique ID and saving it in keychain will ensure the unique id remain same even if user reinstalls my app. I found the below code from one of the answer to similar question here on stackoverflow however, i need to know how to access keychainUtils and IDManager? Appreciate your help in advance. + (NSString *) getUniqueUUID { NSError * error; NSString * uuid = [KeychainUtils

mysql_insert_id or something like that to return last mysql UUID()

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问题 How do you return the last generated UUID() (primary key) - is there something like mysql_insert_id for that? Table uuidtable : primary key: uuid uuid() id_u (index): integer multiple id_u matched with a primary key uuid() insert: insert into uuidtable (uuid,id_u) values (uuid(),id) where id is a number, of course, and uuid is escaped with uuid 回答1: Write yourself a trigger like so: CREATE TRIGGER ai_uuidtable AFTER INSERT ON uuidtable FOR EACH ROW SET @last_uuid = NEW.uuid; Following an

How to create a 8 digit Unique ID in Python?

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问题 I am making a web app for practice, it is a job portal. I want to create 8 digit unique IDs for job which will be visible to the end user. Ids can have numbers and alphabets. Pattern needs to be 8 digits without hyphens or dashes like XXXXXXXX . I know there is UUID thing in python. But they seem to generate the id in their specific format. Is there a way to get the id in my required format? If not can someone please guide me on how to go about it? I know there is a similar thread How can I

Implementing UUID as primary key

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问题 I need to implement UUID as primary key but I'm not sure how to do it in Django. My code class LinkRenewAd(models.Model): # This model will generate the uuid for the ad renew link def make_uuid(self): return str(uuid.uuid1().int>>64) uuid = models.CharField(max_length=36, primary_key=True, default=make_uuid, editable=False) main = models.ForeignKey(Main) expiration_date = models.DateTimeField() date_inserted = models.DateTimeField(auto_now_add=True) date_last_update = models.DateTimeField

db4o, Linq, and UUID's

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问题 Apparently the db4o website was recently redone, and now old urls are giving 404 errors. Everytime I think I've found the answer, I get a 404 error. I have a simple db4o database I've setup to store people. public class Person { public string Firstname { get; set;} public string Lastname {get;set;} } I've been able to run Linq queries on the database, and everything works wonderfully. I'm programming in a web environment, so I will need someone to identify and fetch unique objects from the

java递归和反向递归

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1. 递归查询树tree结构有两种做法：第一种，递归查询数据库结构，第二种，一次性将数据库表中的所有数据查出来，然后再递归查出来的list集合，第一种做法适合数据量较少的tree结构，因为要一直查询数据库数据量大时速度回相对较慢，所以数据量大时建议使用第二种方法，如图1 所示是一个常见的树tree结构图1 2. 反向递归（逆向递归）查询树tree结构根据关键字过滤数据大家有么有遇到过这个问题：我想要根据关键字过滤查询出相关数据和它的上级结构，得到图1所示结果，可往往不知道怎么做，查不出上级结构总是得到图3 类似的结构，要解决这个比较常见的问题就要用到反向递归的算法，网上我那个网搜不到类似的解决方案，本人一时兴趣来了，做了一套递归和反递归的解决方案，简单易懂，大家可以相互交流一下图2 图3 3.示例代码 [java] view plain copy /** * 说明方法描述：将list转为树tree结构 * * @param allRrecords * @return * @time 2016年5月10日下午6:00:35 * @author yangdong */ public List<Record> useListRecordToTree(List<Record> allRrecords) { List<Record> listParentRecord

UUID madness with mssql

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问题 My database entry has a UUID with the value (extracted using the Microsoft SQL Server Management Studio) CDF86F27-AFF4-2E47-BABB-2F46B079E98B After this is loaded into my Scala application, the toString method yields this value 276ff8cd-f4af-472e-babb-2f46b079e98b How does this happen? And how can I programmatically create a UUID instance, when I have only the bare string CDF86F27-AFF4-2E47-BABB-2F46B079E98B at hand? Relevant Slick code (former: table definition, latter: database access

How to get unchanged device id from ios7 device programmatically

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问题 I need to get the unique device id to populate unique user id in the database. I have used the following code to get that. NSString *strApplicationUUID= [[[UIDevice currentDevice] identifierForVendor] UUIDString]; But when I am reinstalling the app, the UUID is changing all the time. But as far as I know the device id never changes. For that I used below code with the third party SSKeychain to save and retrieve the old UDID: NSString *Appname = [[[NSBundle mainBundle] infoDictionary]

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