如何在Oracle中生成GUID? [英] How to generate a GUID in Oracle?

问题描述

是否可以将GUID自动生成为Insert语句?

Is it possible to auto-generate a GUID into an Insert statement?

我还应该使用哪种类型的字段来存储此GUID?

Also, what type of field should I use to store this GUID?

推荐答案

您可以使用SYS_GUID()函数在插入语句中生成GUID:

You can use the SYS_GUID() function to generate a GUID in your insert statement:

insert into mytable (guid_col, data) values (sys_guid(), 'xxx');

用于存储GUID的首选数据类型是RAW(16).

The preferred datatype for storing GUIDs is RAW(16).

作为Gopinath的答案:

As Gopinath answer:

 select sys_guid() from dual
 union all
 select sys_guid() from dual
 union all 
 select sys_guid() from dual

你得到

88FDC68C75DDF955E040449808B55601
88FDC68C75DEF955E040449808B55601
88FDC68C75DFF955E040449808B55601

88FDC68C75DDF955E040449808B55601
88FDC68C75DEF955E040449808B55601
88FDC68C75DFF955E040449808B55601

正如托尼·安德鲁斯(Tony Andrews)所说， 仅在一个字符处不同

As Tony Andrews says, differs only at one character

88FDC68C75D D F955E040449808B55601
88FDC68C75D E F955E040449808B55601
88FDC68C75D F F955E040449808B55601

88FDC68C75DDF955E040449808B55601
88FDC68C75DEF955E040449808B55601
88FDC68C75DFF955E040449808B55601

可能有用: http://feuerthoughts .blogspot.com/2006/02/watch-out-for-sequential-oracle-guids.html

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