Compare Dates and select the max date ?

Question

Hello,
I am trying to write a script and will compare the dates in " eff_startdt" and give me the lastest date at the outcome.
I have data that some service locations have more than one contract date and I need to get the latest dated conract dates to work on the data but I tried many things I am always getting errors. When I run the script below I get " missing expression" error. I dont' see anything missing.
also somehow Max() is keep giving me errors when I do something like [  ON service_locs = vmeterid WHERE SERVICE_LOCS = SERVICE_LOCS AND EFF_STARTDT = MAX(EFF_STARTDT)  ]
Can someone pls give me advice on this. Thanks
SELECT DISTINCT Broker, customer_name, service_locs, fee_kwh, qtr_monthly, eff_startdt, eff_enddt
FROM VMETER
INNER JOIN BROKER_DATA
ON service_locs = vmeterid WHERE SERVICE_LOCS = SERVICE_LOCS AND (SELECT MAX(EFF_STARTDT) FROM VMETER)
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Answer 1

Hi,
I will try to explain on my example. I have got a table:
DESC SOLD_ITEMS;
Name                                    Value NULL? Type
COMPONENT                                          VARCHAR2(255)
SUBCOMPONENT                                       VARCHAR2(255)
YEAR                                               NUMBER(4)
MONTH                                              NUMBER(2)
DAY                                                NUMBER(2)
DEFECTS                                            NUMBER(10)
DESCRIPTION                                        VARCHAR2(200)
SALE_DATE                                          DATE
COMP_ID                                            NUMBERI have insert example data into my table:
select component, subcomponent, sale_date,comp_id
  2  from sold_items;
COMPONENT       SUBCOMPONENT    SALE_DAT    COMP_ID                            
graph           bar             06/04/03          1                            
graph           bar             06/04/01          2                            
search          user search     06/04/02          3                            
search          user search     06/04/01          4                            
search          product search  06/03/20          5                            
search          product search  06/03/16          6                            
graph           bar             06/05/01          7                            
graph           bar             06/05/02          8                            
graph           bar             06/05/02          9
As you can see there are a few components and subcomponents duplicated with different date and comp_id value.
I want to get component and subcomponent combination with latest date.
SELECT COMPONENT, SUBCOMPONENT, MAX(SALE_DATE)
  2  FROM SOLD_ITEMS
  3* GROUP BY COMPONENT, SUBCOMPONENT;
Efect:
COMPONENT       SUBCOMPONENT    MAX(SALE                                       
graph           bar             06/05/02                                       
search          user search     06/04/02                                       
search          product search  06/03/20
For your purpose I will do it using join and subquery. Maybe it will help you resolve your problem:
SELECT COMPONENT, SUBCOMPONENT, SALE_DATE, RANK
  2  FROM (SELECT T1.COMPONENT, T1.SUBCOMPONENT, T2.SALE_DATE,
  3          ROW_NUMBER() OVER (PARTITION BY T1.COMPONENT, T2.SUBCOMPONENT ORDER BY T2.SALE_DATE DESC) AS ROW
  4          FROM SOLD_ITEMS T1, SOLD_ITEMS T2
  5          WHERE T1.COMP_ID = T2.COMP_ID)
  6* WHERE ROW = 1;
I joined the same table but it act as two different tables inside subquery. It will group values (partition by statement) and order result descending using t2.sale_date column. As you can see columns are returned from both tables. If you would like to add some conditions, you can do it after WHERE ROW=1 code.
Results:
COMPONENT       SUBCOMPONENT    SALE_DAT       RANK                            
graph           bar             06/05/02          1                            
search          product search  06/03/20          1                            
search          user search     06/04/02          1Hope this help you
Peter D.