How do I get my data back with no duplicates

Question

I need no duplicate EncounterNo even if I do this it still returns dupes. Any Suggestions would be deeply appreciated!
SELECT Distinct
PWLDBA.ENTORDER.ENTORDERNO,
PWLDBA.TESTREQ.ACCESSION,
PWLDBA.ENCOUNTER.ENCOUNTERNO,
PWLDBA.PATIENT.NAMELAST,
PWLDBA.PATIENT.NAMEFIRST,
PWLDBA.PATIENT.NAMEINITIAL,
PWLDBA.PATIENT.NAMESUFFIX,
'F' AS PATIENT_GENDER,
PWLDBA.PATIENT.SSN,
PWLDBA.PATIENT.DOB,
' ' As AGE,
PWLDBA.RC.DESCRIPTION AS ETHNICITY,
'NA' AS PATIENT_OCCUPATION,
PWLDBA.PATIENT.ADDR1,
PWLDBA.PATIENT.ADDR2,
PWLDBA.PATIENT.CITY,
PWLDBA.PATIENT.STATE,
PWLDBA.PATIENT.COUNTRY,
PWLDBA.PATIENT.PHONE,
PWLDBA.SF.SRCFAC,
PWLDBA.SF.SRCFACNAME,
PWLDBA.PATIENT.PID,
PWLDBA.RD.NAMELAST AS ORDERING_PROVLNAME,
PWLDBA.RD.NAMEFIRST AS ORDERING_PROVFNAME,
PWLDBA.RD.NAMEINITIAL AS ORDERING_PROVINITIAL,
PWLDBA.RD.NAMEDEGREE AS TITLE,
PWLDBA.RD.NAMESUFFIX,
PWLDBA.RD.PMRY_SFKEY,
PWLDBA.RD.ADDR1,
PWLDBA.RD.ADDR2,
PWLDBA.RD.CITY,
PWLDBA.RD.STATE,
PWLDBA.RD.ZIP,
PWLDBA.RD.CONTACTPHONE,
PWLDBA.RD.UPIN,
PWLDBA.RD.NPI,
PWLDBA.PATIENT.ITKEY_0,
PWLDBA.PATIENT.GROUP_0,
PWLDBA.TESTREQ.COLLECTED AS PRIMARY_COL_DT,
PWLDBA.TESTREQ.RECEIVED AS PRIMARY_REC_DT,
PWLDBA.PTRESULT.FINALSTAMP AS FINAL_COMP_DT,
'NA' AS CASECURRENTSTEPDESC,
'DIGENE' AS CASE_TYPE,
PWLDBA.TEST.SHORTNAME,
PWLDBA.TEST.TESTCODE,
'NA' AS LOCALSTANDARDCODE,
PWLDBA.PWSP.DESCRIPTION AS PRIMSPECDESC,
PWLDBA.TESTREQ.SOURCE,
'NA' AS SNOMED_CODES,
PWLDBA.TESTREQ.ICD9KEY_0,
PWLDBA.BILLING.DESCRIPTION,
PWLDBA.BILLING_CPT.CPTBILLCODE AS CPTCODE,
'NA' AS LOINCCODES,
'NA' AS UMLCODES,
PWLDBA.TEST.TESTCODE,
PWLDBA.ANALYT.DESCRIPTION AS RESULTNAME,
PWLDBA.PTRESULT.VALUE AS RESULT_VALUE,
PWLDBA.PTRESULT.VALUE AS HPV_DETECTED,
PWLDBA.TESTCOMMENT.D_COMMENT,
PWLDBA.PTRESULT.ABNRANGE,
'NA' AS CLINICALHIST,
'NA' AS SPECIMENSOURCE,
PWLDBA.TEST.NAME,
'DIGENE' AS HPVTEST_MFG,
'NA' AS ADDENDUM,
'SED MEDICAL LAB' AS LABCODE
FROM
PWLDBA.TESTREQ
INNER JOIN PWLDBA.PATIENT
     ON PWLDBA.PATIENT.SFKEY=PWLDBA.TESTREQ.SFKEY
INNER JOIN PWLDBA.ENCOUNTER
ON PWLDBA.ENCOUNTER.PATIENTKEY=PWLDBA.TESTREQ.PATIENTKEY
INNER JOIN PWLDBA.SF
ON PWLDBA.SF.SFKEY=PWLDBA.PATIENT.SFKEY
INNER JOIN PWLDBA.TEST
ON PWLDBA.TEST.TESTKEY=PWLDBA.TESTREQ.TESTKEY
INNER JOIN PWLDBA.PWSP
ON PWLDBA.PWSP.PWSPKEY=PWLDBA.TESTREQ.PWSPKEY
INNER JOIN PWLDBA.PTRESULT
ON PWLDBA.PTRESULT.TESTREQKEY=PWLDBA.TESTREQ.TESTREQKEY
INNER JOIN PWLDBA.PT
ON PWLDBA.PT.PTKEY=PWLDBA.ENCOUNTER.PTKEY
INNER JOIN PWLDBA.RD
ON PWLDBA.TESTREQ.RDKEY_0=PWLDBA.RD.RDKEY
INNER JOIN PWLDBA.ANALYT
ON PWLDBA.ANALYT.ANALYTKEY=PWLDBA.PTRESULT.ANALYTKEY
INNER JOIN PWLDBA.TESTCOMMENT
ON PWLDBA.TESTCOMMENT.TESTREQKEY=PWLDBA.TESTREQ.TESTREQKEY
INNER JOIN PWLDBA.BILLING
ON PWLDBA.BILLING.PATIENTKEY=PWLDBA.TESTREQ.PATIENTKEY
INNER JOIN PWLDBA.BILLING_CPT
ON PWLDBA.BILLING_CPT.BILLINGKEY=PWLDBA.BILLING.BILLINGKEY
INNER JOIN PWLDBA.RC
ON PWLDBA.RC.RCKEY=PWLDBA.PATIENT.RCKEY
INNER JOIN PWLDBA.ENTORDER
ON PWLDBA.ENTORDER.ENTORDERKEY=PWLDBA.TESTREQ.ENTORDERKEY
WHERE PWLDBA.TEST.TESTCODE in ('8191','721L') AND PWLDBA.TESTREQ.STATUS ='D'
ORDER BY PWLDBA.ENCOUNTER.ENCOUNTERNO ASC

Answer 1

Hi,
Whenever you have a question, please post a little sample data (CREATE TABLE and INSERT statements) and the results you want from that data. Explain how you get those results from that data.
If you can show your problem using commonly available tables (liek those in the scott schema) then you don't have to post any sample data; just the results and explanation.
Always say which version of Oracle you're using.
I believe your problem is very much like this:
Say you want to display one row for every ditinct job in the scott.emp table.
If you run this query:
SELECT DISTINCT
          job, ename, hiredate
FROM       scott.emp
ORDER BY  job
,            ename
;you'll get these results:
JOB       ENAME      HIREDATE
ANALYST   FORD       03-Dec-1981
ANALYST   SCOTT      19-Apr-1987
CLERK     ADAMS      23-May-1987
CLERK     JAMES      03-Dec-1981
CLERK     MILLER     23-Jan-1982
CLERK     SMITH      17-Dec-1980
MANAGER   BLAKE      01-May-1981
MANAGER   CLARK      09-Jun-1981
MANAGER   JONES      02-Apr-1981
PRESIDENT KING       17-Nov-1981
SALESMAN  ALLEN      20-Feb-1981
SALESMAN  MARTIN     28-Sep-1981
SALESMAN  TURNER     08-Sep-1981
SALESMAN  WARD       22-Feb-1981because SELECT DISTINCT means that no row of output will be identical to any other row of output; it says nothing about individual columns. For example, there are two rows in the output above with job='ANALYST'; these rows are distinct, because even though the job column is identical, the ename column is different.
If you want only one row of output per job, like this:
JOB       ENAME      HIREDATE
ANALYST   FORD       03-Dec-1981
CLERK     ADAMS      23-May-1987
MANAGER   BLAKE      01-May-1981
PRESIDENT KING       17-Nov-1981
SALESMAN  ALLEN      20-Feb-1981then you can run a query like this:
WITH     got_r_num     AS
     SELECT     job, ename, hiredate
     ,     ROW_NUMBER () OVER ( PARTITION BY  job
                               ORDER BY          ename
                             )        AS r_num
     FROM    scott.emp
SELECT       job, ename, hiredate
FROM       got_r_num
WHERE       r_num          = 1
;Notice that, although the table contains two rows where job='ANALYST' this query displays only one of them, the one where ename='FORD'. Why didn't the query display the other row from the table, the one with ename='SCOTT'? Because of the analytic ORDER BY clause in the ROW_NUMBER function. "ROW_NUMBER () OVER (PARTITION BY job ORDER BY ename)" means that the ROWN_NUMBER function will return distinct numbers (1, 2, 3, ...) for each alue of job, with the first row in order by ename getting #1, the 2nd row getting 2, and so on. If there happened to be 2 or more rows with the same job and the same ename, then ROW_NUMBER would still assign distinct numbers to each one, but there's no way to foretell which of the rows would get the lower number.