问题
I have a query requirement from ----. Trying to solve it with CONNECT BY, but can't seem to get the results I need.
Table (simplified):
create table CSS.USER_DESC (
USER_ID VARCHAR2(30) not null,
NEW_USER_ID VARCHAR2(30),
GLOBAL_HR_ID CHAR(8)
)
-- USER_ID is the primary key
-- NEW_USER_ID is a self-referencing key
-- GLOBAL_HR_ID is an ID field from another system
There are two sources of user data (datafeeds)... I have to watch for mistakes in either of them when updating information.
Scenarios:
- A user is given a new User ID... The old record is set accordingly and deactivated (typically a rename for contractors who become fulltime)
- A user leaves and returns sometime later. HR fails to send us the old user ID so we can connect the accounts.
- The system screwed up and didn't set the new User ID on the old record.
- The data can be bad in a hundred other ways
I need to know the following are the same user, and I can't rely on name or other fields... they differ among matching records:
ROOTUSER NUMROOTS NODELEVEL ISLEAF USER_ID NEW_USER_ID GLOBAL_HR_ID USERTYPE LAST_NAME FIRST_NAME
-----------------------------------------------------------------------------------------------------------------------------
EX0T1100 2 1 0 EX0T1100 EX000005 CONTRACTOR VON DER HAAVEN VERONICA
EX0T1100 2 2 1 EX000005 00126121 EMPLOYEE HAAVEN, VON DER VERONICA
GL110456 1 1 1 GL110456 00126121 EMPLOYEE VONDERHAAVEN VERONICA
EXOT1100 and EX000005 are connected properly by the NEW_USER_ID field. The rename occurred before there were global HR IDs, so EX0T1100 doesn't have one. EX000005 was given a new user ID, 'GL110456', and the two are only connected by having the same global HR ID.
Cleaning up the data isn't an option.
The query so far:
select connect_by_root cud.user_id RootUser,
count(connect_by_root cud.user_id) over (partition by connect_by_root cud.user_id) NumRoots,
level NodeLevel, connect_by_isleaf IsLeaf, --connect_by_iscycle IsCycle,
cud.user_id, cud.new_user_id, cud.global_hr_id,
cud.user_type_code UserType, ccud.last_name, cud.first_name
from css.user_desc cud
where cud.user_id in ('EX000005','EX0T1100','GL110456')
-- Using this so I don't get sub-users in my list of root users...
-- It complicates the matches with GLOBAL_HR_ID, however
start with cud.user_id not in (select cudsub.new_user_id
from css.user_desc cudsub
where cudsub.new_user_id is not null)
connect by nocycle (prior new_user_id = user_id);
I've tried various CONNECT BY clauses, but none of them are quite right:
-- As a multiple CONNECT BY
connect by nocycle (prior global_hr_id = global_hr_id)
connect by nocycle (prior new_user_id = user_id)
-- As a compound CONNECT BY
connect by nocycle ((prior new_user_id = user_id)
or (prior global_hr_id = global_hr_id
and user_id != prior user_Id))
UNIONing two CONNECT BY queries doesn't work... I don't get the leveling.
Here is what I would like to see... I'm okay with a resultset that I have to distinct and use as a subquery. I'm also okay with any of the three user IDs in the ROOTUSER column... I just need to know they're the same users.
ROOTUSER NUMROOTS NODELEVEL ISLEAF USER_ID NEW_USER_ID GLOBAL_HR_ID USERTYPE LAST_NAME FIRST_NAME
-----------------------------------------------------------------------------------------------------------------------------
EX0T1100 3 1 0 EX0T1100 EX000005 CONTRACTOR VON DER HAAVEN VERONICA
EX0T1100 3 2 1 EX000005 00126121 EMPLOYEE HAAVEN, VON DER VERONICA
EX0T1100 3 (2 or 3) 1 GL110456 00126121 EMPLOYEE VONDERHAAVEN VERONICA
Ideas?
Update
Nicholas, your code looks very much like the right track... at the moment, the lead(user_id) over (partition by global_hr_id) gets false hits when the global_hr_id is null. For example:
USER_ID NEW_USER_ID CHAINNEWUSER GLOBAL_HR_ID LAST_NAME FIRST_NAME
FP004468 FP004469 AARON TIMOTHY
FP004469 FOONG KOK WAH
I've often wanted to treat nulls as separate records in a partition, but I've never found a way to make ignore nulls work. This did what I wanted:
decode(global_hr_id,null,null,lead(cud.user_id ignore nulls) over (partition by global_hr_id order by user_id)
... but there's got to be a better way. I haven't been able to get the query to finish yet on the full-blown user data (about 40,000 users). Both global_hr_id and new_user_id are indexed.
Update
The query returns after about 750 seconds... long, but manageable. It returns 93k records, because I don't have a good way of filtering level 2 hits out of the root - you have start with global_hr_id is null, but unfortunately, that isn't always the case. I'll have to think some more about how to filter those out.
I've tried adding more complex start with clauses before, but I find that separately, they run < 1 second... together, they take 90 minutes >.<
Thanks again for you help... plodding away at this.
回答1:
You have provided sample of data for only one user. Would be better to have a little bit more. Anyway, lets look at something like this.
SQL> with user_desc(USER_ID, NEW_USER_ID, GLOBAL_HR_ID)as(
2 select 'EX0T1100', 'EX000005', null from dual union all
3 select 'EX000005', null, 00126121 from dual union all
4 select 'GL110456', null, 00126121 from dual
5 )
6 select connect_by_root(user_id) rootuser
7 , count(connect_by_root(user_id)) over(partition by connect_by_root(user_id)) numroot
8 , level nodlevel
9 , connect_by_isleaf
10 , user_id
11 , new_user_id
12 , global_hr_id
13 from (select user_id
14 , coalesce(new_user_id, usr) new_user_id1
15 , new_user_id
16 , global_hr_id
17 from ( select user_id
18 , new_user_id
19 , global_hr_id
20 , decode(global_hr_id,null,null,lead(user_id) over (partition by global_hr_id order by user_id)) usr
21 from user_desc
22 )
23 )
24 start with global_hr_id is null
25 connect by prior new_user_id1 = user_id
26 ;
Result:
ROOTUSER NUMROOT NODLEVEL CONNECT_BY_ISLEAF USER_ID NEW_USER_ID GLOBAL_HR_ID
-------- ---------- ---------- ----------------- -------- ----------- ------------
EX0T1100 3 1 0 EX0T1100 EX000005
EX0T1100 3 2 0 EX000005 126121
EX0T1100 3 3 1 GL110456 126121
来源：https://stackoverflow.com/questions/12886371/oracle-self-join-on-multiple-possible-column-matches-connect-by