Is it possible to auto-generate a GUID into an Insert statement?
Also, what type of field should I use to store this GUID?
Answer
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');
The preferred datatype for storing GUIDs is RAW(16).
As Gopinath answer:
select sys_guid() from dual
union all
select sys_guid() from dual
union all
select sys_guid() from dual
You get
88FDC68C75DDF955E040449808B55601
88FDC68C75DEF955E040449808B55601
88FDC68C75DFF955E040449808B55601
As Tony Andrews says, differs only at one character
88FDC68C75DDF955E040449808B55601
88FDC68C75DEF955E040449808B55601
88FDC68C75DFF955E040449808B55601
Maybe useful: http://feuerthoughts.blogspot.com/2006/02/watch-out-for-sequential-oracle-guids.html