问题
Table Name : crm_mrdetails
id | mr_name | me_email | mr_mobile | mr_doctor|
-----------------------------------------------------
1 | John |abc@gmail.com | 1234555555 | ,1,2,3 |
Table Name : crm_mr_doctor
id | dr_name | specialization|
----------------------------------
1 | Abhishek | cordiologist |
2 | Krishnan | Physician |
3 | Krishnan | Nurse |
The concatenated values in mrdetails.mr_doctor are the foreign keys for mr_doctor.id. I need to join on them to produce output like this:
id | mr_name | me_email |Doctor_Specialization|
-------------------------------------------------
1 | John |abc@gmail.com |cordiologist,Physician,Nurse|
I'm new to Oracle, I'm using Oracle 12C. Any help much appreciated.
回答1:
First of all we must acknowledge that is a bad data model. The column mr_doctor violates First Normal Form. This is not some abstruse theoretical point. Not being in 1NF means we must write more code to lookup the meaning of the keys instead of using standard SQL join syntax. It also means we cannot depend on the column containing valid IDs: mr_doctor can contain any old nonsense and we must write a query which will can handle that. See Is storing a delimited list in a database column really that bad? for more on this.
Anyway. Here is a solution which uses regular expressions to split the mr_doctor column into IDs and then joins them to the mr_doctor table. The specialization column is concatenated to produce the required output.
select mrdet.id,
mrdet.mr_name,
mrdet.me_email,
listagg(mrdoc.specialization, ',')
within group (order by mrdoc.specialization) as doctor_specialization
from mr_details mrdet
join (
select distinct id,
regexp_substr(mr_doctor, '(,?)([0-9]+)(,?)', 1, level, null, 2) as dr_id
from mr_details
connect by level <= regexp_count(mr_doctor, '(,?)([0-9]+)')
) as mrids
on mrids.id = mrdet.id
left outer join mr_doctor mrdoc
on mrids.dr_id = mr_doc.id
group by mrdet.id,
mrdet.mr_name,
mrdet.me_email
/
This solution is reasonably resilient despite the data model being brittle. It will return results if the string has too many commas, or spaces. It will ignore values which are letters or otherwise aren't numbers. It won't hurl if the extracted number doesn't match an ID in the mr_doctor table. Obviously the results are untrustworthy for those reasons, but that's part of the price of a shonky data model.
Can you please explain the following:
(,?)([0-9]+)(,?)
The pattern matches zero or one comma followed by one or more digits followed by zero or one comma. Perhaps the (,?) in the matched patterns aren't strictly necessary. However, without them, this string 2 3 4 would match the same three IDs as this string 2,3,4. Maybe that's correct maybe it isn't. When the foreign keys are stored in a CSV column instead of being enforced through a proper constraint what does 'correct' even mean?
回答2:
You have to split data in mr_doctor column into rows, join table crm_mrdoctor and then use listagg().
How to split data? Splitting string into multiple rows in Oracle
select t.id, max(mr_name) mr_name,
listagg(specialization, ', ') within group (order by rn) specs
from (
select id, mr_name, levels.column_value rn,
trim(regexp_substr(mr_doctor, '[^,]+', 1, levels.column_value)) as did
from crm_mrdetails t,
table(cast(multiset(select level
from dual
connect by level <=
length(regexp_replace(t.mr_doctor, '[^,]+')) + 1)
as sys.odcinumberlist)) levels) t
left join crm_mr_doctor d on t.did = d.id
group by t.id
Demo and result:
with crm_mrdetails (id, mr_name, mr_doctor) as (
select 1, 'John', ',1,2,3' from dual union all
select 2, 'Anne', ',4,2,6,5' from dual union all
select 3, 'Dave', ',4' from dual),
crm_mr_doctor (id, dr_name, specialization) as (
select 1, 'Abhishek', 'cordiologist' from dual union all
select 2, 'Krishnan', 'Physician' from dual union all
select 3, 'Krishnan', 'Nurse' from dual union all
select 4, 'Krishnan', 'Onkologist' from dual union all
select 5, 'Krishnan', 'Surgeon' from dual union all
select 6, 'Krishnan', 'Nurse' from dual
)
select t.id, max(mr_name) mr_name,
listagg(specialization, ', ') within group (order by rn) specs
from (
select id, mr_name, levels.column_value rn,
trim(regexp_substr(mr_doctor, '[^,]+', 1, levels.column_value)) as did
from crm_mrdetails t,
table(cast(multiset(select level
from dual
connect by level <=
length(regexp_replace(t.mr_doctor, '[^,]+')) + 1)
as sys.odcinumberlist)) levels) t
left join crm_mr_doctor d on t.did = d.id
group by t.id
Output:
ID MR_NAME SPECS
------ ------- -------------------------------------
1 John cordiologist, Physician, Nurse
2 Anne Onkologist, Physician, Nurse, Surgeon
3 Dave Onkologist
回答3:
Please change the column names according to your requirement.
CREATE OR REPLACE Function ReplaceSpec
(String_Inside IN Varchar2)
Return Varchar2 Is
outputString Varchar2(5000);
tempOutputString crm_doc.specialization%TYPE;
Begin
FOR i in 1..(LENGTH(String_Inside)-LENGTH(REPLACE(String_Inside,',',''))+1)
LOOP
Select specialization into tempOutputString From crm_doc
Where id = PARSING_STRING(String_Inside,i);
If i != 1 Then
outputString := outputString || ',';
end if;
outputString := outputString || tempOutputString;
END LOOP;
Return outputString;
End;
/
The Parsing_String function to help split the comma separated values.
CREATE OR REPLACE Function PARSING_STRING
(String_Inside IN Varchar2, Position_No IN Number)
Return Varchar2 Is
OurEnd Number; Beginn Number;
Begin
If Position_No < 1 Then
Return Null;
End If;
OurEnd := Instr(String_Inside, ',', 1, Position_No);
If OurEnd = 0 Then
OurEnd := Length(String_Inside) + 1;
End If;
If Position_No = 1 Then
Beginn := 1;
Else
Beginn := Instr(String_Inside, ',', 1, Position_No-1) + 1;
End If;
Return Substr(String_Inside, Beginn, OurEnd-Beginn);
End;
/
Please note that I have given only a basic function to get your output. You might need to add some exceptions etc.
Eg. When the doc_id [mr_doctor] is empty, what to do.
Usage
select t1.*,ReplaceSpec(doc_id) from crm_details t1
if your mr_doctor data always starts with a comma use:
Select t1.*,ReplaceSpec(Substr(doc_id,2)) from crm_details t1
回答4:
You can use a recursive sub-query and simple string functions (which may be faster than using regular expressions and a correlated hierarchical query):
Oracle Setup:
CREATE TABLE crm_mrdetails (id, mr_name, mr_doctor) as
select 1, 'John', ',1,2,3' from dual union all
select 2, 'Anne', ',4,2,6,5' from dual union all
select 3, 'Dave', ',4' from dual;
CREATE TABLE crm_mr_doctor (id, dr_name, specialization) as
select 1, 'Abhishek', 'cordiologist' from dual union all
select 2, 'Krishnan', 'Physician' from dual union all
select 3, 'Krishnan', 'Nurse' from dual union all
select 4, 'Krishnan', 'Onkologist' from dual union all
select 5, 'Krishnan', 'Surgeon' from dual union all
select 6, 'Krishnan', 'Nurse' from dual;
Query:
WITH crm_mrdetails_bounds ( id, mr_name, mr_doctor, start_pos, end_pos ) AS (
SELECT id,
mr_name,
mr_doctor,
2,
INSTR( mr_doctor, ',', 2 )
FROM crm_mrdetails
UNION ALL
SELECT id,
mr_name,
mr_doctor,
end_pos + 1,
INSTR( mr_doctor, ',', end_pos + 1 )
FROM crm_mrdetails_bounds
WHERE end_pos > 0
),
crm_mrdetails_specs ( id, mr_name, start_pos, specialization_id ) AS (
SELECT id,
mr_name,
start_pos,
TO_NUMBER(
CASE end_pos
WHEN 0
THEN SUBSTR( mr_doctor, start_pos )
ELSE SUBSTR( mr_doctor, start_pos, end_pos - start_pos )
END
)
FROM crm_mrdetails_bounds
)
SELECT s.id,
MAX( s.mr_name ) AS mr_name,
LISTAGG( d.specialization, ',' )
WITHIN GROUP ( ORDER BY s.start_pos )
AS doctor_specialization
FROM crm_mrdetails_specs s
INNER JOIN crm_mr_doctor d
ON ( s.specialization_id = d.id )
GROUP BY s.id
Output:
ID | MR_NAME | DOCTOR_SPECIALIZATION -: | :------ | :--------------------------------- 1 | John | cordiologist,Physician,Nurse 2 | Anne | Onkologist,Physician,Nurse,Surgeon 3 | Dave | Onkologist
db<>fiddle here
回答5:
Please go through https://oracle-base.com/articles/misc/string-aggregation-techniques String Aggregation Techniques
or
SELECT deptno, LTRIM(MAX(SYS_CONNECT_BY_PATH(ename,',')) KEEP (DENSE_RANK LAST ORDER BY curr),',') AS employees FROM (SELECT deptno, ename, ROW_NUMBER() OVER (PARTITION BY deptno ORDER BY ename) AS curr, ROW_NUMBER() OVER (PARTITION BY deptno ORDER BY ename) -1 AS prev FROM emp) GROUP BY deptno CONNECT BY prev = PRIOR curr AND deptno = PRIOR deptno START WITH curr = 1
or
listagg and wm_concat an also be used as other people have used it
回答6:
How about this one? I have not tested it, so there could be any syntax error.
select id,mr_name,me_email,listagg(specialization,',') within group (order by specialization) as Doctor_Specialization
from
(select dtls.id,dtls.mr_name,dtls.me_email,dr.specialization
from crm_mrdetails dtls,
crm_mr_doctor dr
where INSTR(','||dtls.mr_doctor||',' , ','||dr.id||',') > 0
) group by id,mr_name,me_email;
来源：https://stackoverflow.com/questions/45278653/oracle-query-join-with-comma-separated-data