Need help with a JOIN query

Question

Hi,
I am looking for the following result:
PROVIDER_ID     SPECIALTY     BUCKET     CODE     RATING     FEE
     1     FP                 EM     100     9     20
     1     FP                 SP     300     15     0
     1     INFUS                 EM     100     3     20
     1     INFUS                 EM     200     6     15The base tables are provided below in the with clause. What I am trying to do is, where "code" matches show the fee from t1 else show 0.
I tried a few variotions but just can't seem to get it right.
with t1 as
select 1 as provider_id, 100 as code, 20 as fee, 10 as default_multiplier from dual union all
select 1 as provider_id, 200 as code, 15 as fee, 30 as default_multiplier from dual
, t2 as
select 100 as code, 'INFUS' AS specialty, 'EM' AS bucket, 3 as rating from dual union all
select 200 as code, 'INFUS' AS specialty, 'EM' AS bucket, 6 as rating from dual union all
select 100 as code, 'FP' AS specialty, 'EM' AS bucket, 9 as rating from dual union all
select 300 as code, 'FP' AS specialty, 'SP' AS bucket, 15 as rating from dual
SELECT   t1.provider_id
       , t2.specialty
       , t2.bucket
       , t2.code
       , t2.rating
       , t1.fee
FROM     t1, t2
ORDER BY 1, 2, 3Any help will be appreciated.
Thanks.

Answer 1

Could I possibly add one more twist to this.
The current result:
PROVIDER_ID SPECI BU       CODE     RATING        FEE
          1 FP    EM        100          9         20
          1 FP    SP        300         15          0
          1 INFUS EM        100          3         20
          1 INFUS EM        200          6         75
          2 FP    EM        100          9         40
          2 FP    SP        300         15          0
          2 INFUS EM        100          3         40
          2 INFUS EM        200          6          0
          3 FP    EM        100          9          0
          3 FP    SP        300         15          0
          3 INFUS EM        100          3          0
          3 INFUS EM        200          6         60The current code:
with t1 as
select 1 as provider_id, 100 as code, 20 as fee, 10 as default_multiplier from dual union all
select 1 as provider_id, 200 as code, 75 as fee, 10 as default_multiplier from dual union all
select 1 as provider_id, 500 as code, 75 as fee, 10 as default_multiplier from dual union all
select 2 as provider_id, 100 as code, 40 as fee, 20 as default_multiplier from dual union all
select 3 as provider_id, 200 as code, 60 as fee, 30 as default_multiplier from dual
, t2 as
select 100 as code, 'INFUS' AS specialty, 'EM' AS bucket, 3 as rating from dual union all
select 200 as code, 'INFUS' AS specialty, 'EM' AS bucket, 6 as rating from dual union all
select 100 as code, 'FP' AS specialty, 'EM' AS bucket, 9 as rating from dual union all
select 300 as code, 'FP' AS specialty, 'SP' AS bucket, 15 as rating from dual
, t3 as
SELECT   t1.provider_id
       , t2.specialty
       , t2.bucket
       , t2.code
       , t2.rating
       , CASE t1.code
            WHEN t2.code
               THEN t1.fee
            ELSE 0
         END AS fee
       , RANK () OVER (PARTITION BY t1.provider_id, t2.specialty, t2.bucket, t2.code ORDER BY CASE t1.code
             WHEN t2.code
                THEN t1.fee
             ELSE 0
          END DESC) AS the_rank
    FROM t1
       , t2
SELECT DISTINCT provider_id
              , specialty
              , bucket
              , code
              , rating
              , fee
           FROM t3
          WHERE the_rank = 1
       ORDER BY 1
              , 2
              , 3
              , 4I added the code 500 to t1. Howevere, I don't want this code to show up in the result becuase it is does not exist in t2.
I also added code 200 to t1, to see if that shows properly.
I had to do the rank, becuase I need the proper count of rows.
In reality the row counts are:
t1: 20 Million
t2: 10 Thousand.
So, I would like to avoid doing DISTINCT and ANALYTIC functions on this large set, so any better way of achiving the same result, with a better statement.
Thanks.