Hmm.. SQL*Plus: Release 10.2.0.1.0 - Production on Fri Sep 19 20:21:20 2008 Copyright (c) 1982, 2005, Oracle. All rights reserved. Enter password: Connected to: Oracle Database 10g Enterprise Edition Release 10.2.0.3.0 - 64bit Production With the Partitioning and Data Mining options SQL> create table ucdv_cc_summ (patient_id number(05) 2 ,study_number VARCHAR2(100) 3 ) 4 / Table created. SQL> begin 2 insert into ucdv_cc_summ VALUES ( 1 , 'UCDCC#128'); 3 insert into ucdv_cc_summ VALUES ( 2 , 'UCDCC#157'); 4 insert into ucdv_cc_summ VALUES ( 2 , 'UCDCC#165'); 5 insert into ucdv_cc_summ VALUES ( 2 , 'UCDCC#171'); 6 end; 7 / PL/SQL procedure successfully completed. SQL> column study_number FORMAT A15 wrap; SQL> SELECT patient_id , study_number FROM ucdv_cc_summ 2 / PATIENT_ID STUDY_NUMBER ---------- --------------- 1 UCDCC#128 2 UCDCC#157 2 UCDCC#165 2 UCDCC#171 SQL> CREATE VIEW ucdv_cc_summaryae AS 2 SELECT patient_id , study_number FROM ucdv_cc_summ 3 / View created. SQL> SELECT patient_id , study_number FROM ucdv_cc_summaryae 2 / PATIENT_ID STUDY_NUMBER ---------- --------------- 1 UCDCC#128 2 UCDCC#157 2 UCDCC#165 2 UCDCC#171 SQL> select patient_id, study_number 2 from ucdv_cc_summaryae 3 where study_number = 'UCDCC#128' 4 / PATIENT_ID STUDY_NUMBER ---------- --------------- 1 UCDCC#128 SQL> HTH GovindanK -- //www.freelists.org/webpage/oracle-l