Update a row based on the minimum value

Question

I need to update a row value in a table (table1) which is having one-to-may relationship with another table (table2) based on priority. Below is the table structure and sample data. I have a merge statement which can just do this, but it's taking too much to execute if the data in table2 is huge.
I want to reduce the execution time, can anyone suggest me another way to do this.
--Table Definition
CREATE TABLE TABLE_1 (ENAME VARCHAR2(20), ID NUMBER(20), DISPLAY_NAME VARCHAR2(20), PRIMARY KEY (ENAME));
CREATE TABLE TABLE_2 (ENAME VARCHAR2(20), DISPLAY_NAME VARCHAR2(20), PRIORITY NUMBER(1));
--Sample data
INSERT INTO TABLE_1 (ENAME,ID) VALUES ('EMP1','1000');
INSERT INTO TABLE_1 (ENAME,ID) VALUES ('EMP2','2000');
INSERT INTO TABLE_2 (ENAME,DISPLAY_NAME,PRIORITY) VALUES ('EMP1','APPLE',1);
INSERT INTO TABLE_2 (ENAME,DISPLAY_NAME,PRIORITY) VALUES ('EMP1','DELL',2);
INSERT INTO TABLE_2 (ENAME,DISPLAY_NAME,PRIORITY) VALUES ('EMP1','MS',NULL);
--Merge to perfor the update
MERGE INTO TABLE_1 T USING
(select ename, display_name from TABLE_2 where (ename,priority) in (
select distinct ename, min(priority) over (partition by ename) from TABLE_2 min_prior)) Q
ON (T.ENAME = Q.ENAME)
WHEN MATCHED THEN UPDATE SET DISPLAY_NAME = Q.DISPLAY_NAME;

Answer 1

Hi,
You can use the aggregate FIRST function, like this:
MERGE INTO     table_1          t
USING   (
         SELECT    ename
         ,            MIN (display_name) KEEP (DENSE_RANK FIRST ORDER BY priority)     
                          AS first_display_name
         FROM      table_2
         GROUP BY  ename
     )               q
ON      (t.ename = q.ename)
WHEN MATCHED THEN UPDATE
SET      t.display_name      = q.first_display_name
;If the combination (ename, priority) is not unique in table_2, then this uses the first display_name (in normal sort order) among those with the lowest priority. This probably can't happen, becuase your original statement would raise an error if it did.
For performance problems, see the forum FAQ {message:id=9360003}