uuid

I found that jackson comes equipped with a UUID serizlizer/deserializer that can be used like this: @Data @NoArgsConstructor public class MyClass { @JsonSerialize(using=UUIDSerializer.class) @JsonDeserialize(using=UUDIDeserializer.class) private UUID myUUID; } And then using ObjectMapper on MyClass will correctly serialize/deserialize the myUUID field. However, my class has a set of UUIDs that I want to serialize. I tried annotating the field the same way as above, but it complains that Set cannot be converted to UUID (as I half expected). I know I can create my own serializer/deserializers by

I came across a blog of using UUID with Hibernate and MySql. Now the problem is, whenever I take a look at the database the ID's will be non-readable format (binary-16). How can I store UUID as a readable format like 7feb24af-fc38-44de-bc38-04defc3804fe instead of ¡7ôáßEN¹º}ÅÑs I was using this code @Id @GeneratedValue( generator = "uuid2" ) @GenericGenerator( name = "uuid2", strategy = "uuid2" ) @Column( name = "id", columnDefinition = "BINARY(16)" ) private UUID id; And the result is ¡7ôáßEN¹º}ÅÑs . But I want it as readable UUID so I used the following code which didn't help me @Id

Apple since iOS7 ha deprecated and unavailable the use of the property -uniqueIdentifier . The other properties -identifierForVendor and -advertisingIdentifier have the big problem that they change the value after uninstalling and reinstalling the app. I need that value to identify uniquely the device connected to the server. The app will be distributed only internally using an enterprise account, so no problem for the review process. Is there a private a method to get it? [UPDATE WITH SOME TEST] As David said I used identifier for vendor on iOS7.1 device here are some results from my tests.

问题描述： @RequestMapping( value = "/auth1/{uuid}/xxx" , method = RequestMethod.GET) public void imageCode1 (@ PathVariable ( value = "uuid" ) String uuid) { logger.info(uuid); } 见以上代码，url中的uuid如何解析成为参数传递进来。 解析过程：(接收请求：如/auth1/xxxx-xxx-xxx/xxx) 1. 将/auth1/{uuid}/xxx根据/拆成 auth1、{uuid}、xxx 2. 将{uuid}替换成(.*),并纪录key为uuid 3. 同样将/auth1/xxxx-xxx-xxx/xxx拆成auth1、xxxx-xxx-xxx、xxx 4. 进行正则匹配,并根据group得到uuid=xxxx-xxx-xxx. 5. 将uuid=xxxx-xxx-xxx放入request的一个attribute中。 6. 根据反射和标注得到pathvariable名为uuid 7. 去request得到这个uuid,然后进行方法调用。 下面是测试springmvc的解析代码。 public static void main (String[] args) { AntPathMatcher matcher

I am looking at a piece of code, and i can see that the following code has been written which generates a string, and that string has been set as the primary key of the table. return new BigInteger(UUID.randomUUID().toString().replaceAll("-", ""), 16).toString(36); Till now, there are lots of records and they seem to be unique as the primary key constraint is not violated. I understand that the number will be random but will it be unique? thanks Tuco There is no guarantee that the numbers will be unique, however it is extremely improbable that a duplicate will ever be generated because there

安装一个 KVM服务器 案例 1 ：安装一个 KVM 服务器 案例 2 ： KVM 平台构建及简单管理 案例 3 ： virsh 基本管理操作 案例 4 ： xml 配置文件的应用 案例 5 ：为虚拟机制作快照备份 案例 6 ：快建新虚拟机 1 案例 1 ：安装一个 KVM 服务器 1.1 问题 本例要求准备一台 RHEL7.2 服务器，将其搭建为 KVM 平台，主要完成下列操作： 关闭本机的 SELinux 保护、防火墙服务 挂载 RHEL7 光盘到 /mnt/dvd ，将其配置为本机 YUM 源（ baseurl=file:///mnt/dvd ） 安装 KVM 相关包组，确保已启用 libvirtd 服务 1.2 方案 RHEL7 中的虚拟化服务软件组： 虚拟化平台 ——"Virtualization Platform" 虚拟化主机 ——"Virtualization Host" 虚拟化客户端 ——"Virtualization Client" 另外需要注意， yum 命令的软件组管理操作与 RHEL6 有些小变化： yum groups list[hidden][ 组名 ]... yum groups info[ 组名 ]... yum groups install[ 组名 ]... yum groups remove[ 组名 ]... 1.3 步骤

I have stored data that has a UUID v1 as key and I now want to know when the UUID was generated. How do I do that the nice way? This works with irb: require 'simple_uuid' uuid = SimpleUUID::UUID.new uuid.to_guid # actually this is what I have stored in my database Time.at(SimpleUUID::UUID.new(uuid.to_guid).seconds) # returns the time and date of the UUID Thanks to my friend KillerFox, he showed me the simple solution. 来源： https://stackoverflow.com/questions/6696518/extract-time-from-a-uuid-v1-in-ruby

I have a simple C++ function that calls "uuid_generate_random". It seems the generated UUID is not standard compliant (UUID Version 4). Any Ideas? The code I am using is (uuid_generate_random & uuid_unparse functions are available in libuuid): char uuidBuff[36]; uuid_t uuidGenerated; uuid_generate_random(uuidGenerated); uuid_unparse(uuidGenerated, uuidBuff); I understand that "uuid_generate_random" generates the UUID version 4, however, the string that is being returned is like: Run1) 2e9dc3c5-426a-eea0-8186-f6b65b5dc361 Run2) 112c6a78-51bc-cbbb-ae9c-92a33e1fd95f Run3) 5abd3d6d-0259-ce5c-8428

环境： CentOS Linux release 7.7.1908 (Core) PHP 7.3.11 UUID Extention 1.0.4 感觉上PHP对UUID的支持似乎不是很上心，PECL中的UUID扩展仅仅是对libuuid的打包，并没有像其他语言那样直接提供完整的UUID实现。由于libuuid仅仅提供了版本1和版本4的UUID，所以……，开心一点，这两个就够用了。 uuid_create( [int $uuid_type = UUID_TYPE_DEFAULT] ) : string $uuid_type 　　如下常量之一： 　　　　 UUID_TYPE_DEFAULT // 0 　　　　 UUID_TYPE_TIME // 1 　　　　 UUID_TYPE_DCE // 4 　　　　 UUID_TYPE_NAME // 1 　　　　 UUID_TYPE_RANDOM // 4 　　libuuid提供了UUID版本1和版本4的实现，也只定义了 UUID_TYPE_DCE_TIME 和 UUID_TYPE_DCE_RANDOM 两个常量。PHP的UUID扩展中，常量是通过映射libuuid的常量定义的。所以，虽然扩展中根据UUID标准定义了4个常量名，但对于当前版本的libuuid，也只能映射到两个常量。 UUID_TYPE_TIME 映射为libuuid的常量