uuid

This blog explains, that the output of sys_guid() is not random for every system: http://feuerthoughts.blogspot.de/2006/02/watch-out-for-sequential-oracle-guids.html Unfortunately I have to use such a system. How to ensure to get a random UUID? Is it possible with sys_guid() ? If not how to reliably get a random UUID on Oracle? Here's a complete example, based on @Pablo Santa Cruz's answer and the code you posted. I'm not sure why you got an error message. It's probably an issue with SQL Developer. Everything works fine when you run it in SQL*Plus, and add a function: create or replace and

Why are there dashes in a .NET GUID? Are there dashes in most implementations of a GUID, or is it just a Microsoft thing? Signed, 741ecf77-9c92-4435-8e6b-85975bd13452 casperOne Technically, there are no "dashes" in a GUID . A GUID is a 128-bit value which is usually stored in the following manner (using C# here to represent the structure): public struct Guid { public ulong Data1; public ushort Data2; public ushort Data3; public fixed byte Data4[8]; } The dashes are in the string representation of a GUID. The dashes are optional and are not required in a string representation of a GUID. That

第四周知识小总结 时间过得飞快，转眼间就到第四周了，我们又学了哪些新知识呢，下面让我们一起来看下。 一 源码编译安装 1 程序包编译 程序包编译安装： Application-VERSION-release.src.rpm --> 安装后，使用rpmbuild命令制作成二进制格式的rpm包，然后再安装 源代码-->预处理-->编译-->汇编-->链接-->执行 源代码组织格式： 多文件：文件中的代码之间，很可能存在跨文件依赖关系 C、C++：make 项目管理器 configure脚本 --> Makefile.in --> Makefile java: maven 编译安装 C语言源代码编译安装三步骤： 1、./configure (1) 通过选项传递参数，指定启用特性、安装路径等；执行时会参考用户的 指定以及Makefile.in文件生成Makefile (2) 检查依赖到的外部环境，如依赖的软件包 2、make 根据Makefile文件，构建应用程序 3、make install 复制文件到相应路径 开发工具： autoconf: 生成configure脚本 automake：生成Makefile.in 注意：安装前查看README，INSTALL 编译安装 编译C源代码： 准备：提供开发工具及开发环境 开发工具：make, gcc等 开发环境：开发库，头文件 glibc

I am creating an .idl file programmatically. How do I create UUIDs for the interfaces and Methods Programmatically. Can I generate the UUID programmatically? You are probably looking for System.Guid.NewGuid() . Ben Mosher Be careful: while the string representations for .NET Guid and (RFC4122) UUID are identical, the storage format is not. .NET trades in little-endian bytes for the first three Guid parts. If you are transmitting the bytes (for example, as base64), you can't just use Guid.ToByteArray() and encode it. You'll need to Array.Reverse the first three parts (Data1-3). I do it this way

I'm using my uuid as following: @Id @GeneratedValue(generator = "uuid") @GenericGenerator(name = "uuid", strategy = "uuid") @Column(name = "uuid", unique = true) private String uuid; but I'm getting a smart Hibernate warning: Using org.hibernate.id.UUIDHexGenerator which does not generate IETF RFC 4122 compliant UUID values; consider using org.hibernate.id.UUIDGenerator instead So I want to switch to org.hibernate.id.UUIDGenerator , now my question is how should I tell it to Hibernate's generator. I saw some guy used it as a "hibernate-uuid" - so this is what I've tried, but with negative

This is the best way I could come up with to convert a MySQL GUID/UUID generated by UUID() to a binary(16): UNHEX(REPLACE(UUID(),'-','')) And then storing it in a BINARY(16) Are there any implications of doing it this way that I should know of? Quassnoi Not many implications. It will slow down the queries a little, but you will hardly notice it. UNIQUEIDENTIFIER is stored as 16-byte binary internally anyway. If you are going to load the binary into a client and parse it there, note the bit order , it may have other string representation than the initial NEWID() . Oracle 's SYS_GUID() function

警告信息如下: 3.1 WARN [org.hibernate.id.UUIDHexGenerator] (ServerService Thread Pool -- 48) HHH000409:Using org.hibernate.id.UUIDHexGenerator which does not generate IETF RFC 4122 compliant UUID values; consider using org.hibernate.id.UUIDGenerator instead 修改 @GenericGenerator(name = "system-uuid", strategy = "uuid") 为@GenericGenerator(name = "system-uuid", strategy = "uuid2") 下面是在CSDN上看到的一篇文章,保存在我的txt中,现在也贴出来. 介绍hibernate主键生成策略的文章网上比比皆是。但是如何选择一个适合于自己项目的主键生成策略缺没有什么好的指导性文章。在此希望与大家议论。 hibernate的主键生成策略主要包括了"uuid2","guid","uuid","uuid.hex","hilo","assigned","identity","select","sequence","seqhilo",

I want to generate just random UUID's, as it is just important for instances in my program to have unique identifiers. I looked into Boost UUID , but I can't manage to generate the UUID because I don't understand which class and method to use. I would appreciate if someone could give me any example of how to achieve this. A basic example: #include <boost/uuid/uuid.hpp> // uuid class #include <boost/uuid/uuid_generators.hpp> // generators #include <boost/uuid/uuid_io.hpp> // streaming operators etc. int main() { boost::uuids::uuid uuid = boost::uuids::random_generator()(); std::cout << uuid <<