Oracle：获取用户的功能列表
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Oracle: Get list of functions for user
提问by David Oneill
How do I get a list of all the functions for a particular user?
如何获取特定用户的所有功能列表？
EDIT for question clarification:
编辑问题澄清：
When (as USER1) I run
当（作为 USER1）我运行
select * from all_objects
where owner = 'USER2'
and object_type = 'FUNCTION';
it doesn't return all the functions that I know USER2 owns. I suspect that it is only returning those functions that USER1 is allowed to view/execute.
它不会返回我知道 USER2 拥有的所有功能。我怀疑它只返回那些允许 USER1 查看/执行的函数。
Is that suspicion correct?
这种怀疑是否正确？
Also, if that is true, is there a way to get around this?
另外，如果这是真的，有没有办法解决这个问题？
采纳答案by Dave Costa
Yes, your suspicion is correct. The ALL_OBJECTS view will only list those items that the current user has access to.
是的，你的怀疑是正确的。ALL_OBJECTS 视图将只列出当前用户有权访问的那些项目。
If you can log in as USER2, then you can query USER_OBJECTS as that user to see all objects owned by that user.
如果您可以作为 USER2 登录，那么您可以作为该用户查询 USER_OBJECTS 以查看该用户拥有的所有对象。
If you can log in as SYSTEM, then you would have access to all objects regardless of owner, so the list provided by ALL_OBJECTS (or DBA_OBJECTS) would be complete.
如果您可以以 SYSTEM 身份登录，那么无论所有者如何，您都可以访问所有对象，因此 ALL_OBJECTS（或 DBA_OBJECTS）提供的列表将是完整的。
If you can't log in as a user that has access to all of USER2's objects, then you can't list all of USER2's objects.
如果您无法以有权访问 USER2 的所有对象的用户身份登录，则您无法列出 USER2 的所有对象。
回答by Tony Andrews
If you mean a list of functions the belong to a particular user then:
如果您的意思是属于特定用户的功能列表，则：
select object_name
from all_objects
where owner = 'WHOEVER'
and object_type = 'FUNCTION';
This will return only stand-alone functions, not procedures or function in packages, that belong to the schema 'WHOEVER'.
这将仅返回属于架构“WHOEVER”的独立函数，而不是包中的过程或函数。
To obtain a list of allfunctions that the current user can access:
获取当前用户可以访问的所有功能的列表：
select object_name
from all_objects
where object_type = 'FUNCTION';
回答by ashutosh5111
select * from dba_objects where owner = 'USER2' and object_type = 'FUNCTION';
select * from dba_objects where owner = 'USER2' and object_type = 'FUNCTION';
