plsql中，两张表中的两个字段比较的问题

[superff7]

请教高人，我要比较两张表中两个字段“名称”异同的情况，把不同名称的记录提取出来，但是遇到一个问题。

比如说a表中“名称”字段里是“挂号费”，b表中“名称”字段里是“急诊挂号费”，比较结果显示这两条是不同记录，实际上这两条记录是一回事。我怎么样能在比较时候，能使名称类似的字段不显示？

这里是否用到%通配符？

[superff7]

为什么我这么写了
create table test
as
select distinct 虚立收费表1.*
from 虚立收费表1 ,标准收费表
where instr(标准收费表.名称,虚立收费表1.名称) = 0

还是不行？
标准收费表中有一条“单纯疱疹病毒Ⅱ型抗体测定(包括IgG 、IgM) ”，虚立收费表中有一条“单纯疱疹病毒Ⅱ型抗体测定”

执行完上面的语句后，test表中还是有“单纯疱疹病毒Ⅱ型抗体测定”，预想的是test表中没有这一项了

[华夏小卒]

数据

[dawugui]

select a.* from a , b where instr(a.col,b.col) = 0
union all
select b.* from a , b where instr(b.col,a.col) = 0

[dawugui]

1.  INSTR: returns the first-occurrence position of a character within a string   
SQL>
SQL>
SQL> -- create demo table
SQL> create table Employee(
  2    ID                 VARCHAR2(4 BYTE)         NOT NULL,
  3    First_Name         VARCHAR2(10 BYTE),
  4    Last_Name          VARCHAR2(10 BYTE),
  5    Start_Date         DATE,
  6    End_Date           DATE,
  7    Salary             Number(8,2),
  8    City               VARCHAR2(10 BYTE),
  9    Description        VARCHAR2(15 BYTE)
 10  )
 11  /

Table created.

SQL>
SQL> -- prepare data
SQL> insert into Employee(ID,  First_Name, Last_Name, Start_Date,                     End_Date,                       Salary,  City,       Description)
  2               values ('01','Jason',    'Martin',  to_date('19960725','YYYYMMDD'), to_date('20060725','YYYYMMDD'), 1234.56, 'Toronto',  'Programmer')
  3  /

1 row created.

SQL> insert into Employee(ID,  First_Name, Last_Name, Start_Date,                     End_Date,                       Salary,  City,       Description)
  2                values('02','Alison',   'Mathews', to_date('19760321','YYYYMMDD'), to_date('19860221','YYYYMMDD'), 6661.78, 'Vancouver','Tester')
  3  /

1 row created.

SQL> insert into Employee(ID,  First_Name, Last_Name, Start_Date,                     End_Date,                       Salary,  City,       Description)
  2                values('03','James',    'Smith',   to_date('19781212','YYYYMMDD'), to_date('19900315','YYYYMMDD'), 6544.78, 'Vancouver','Tester')
  3  /

1 row created.

SQL> insert into Employee(ID,  First_Name, Last_Name, Start_Date,                     End_Date,                       Salary,  City,       Description)
  2                values('04','Celia',    'Rice',    to_date('19821024','YYYYMMDD'), to_date('19990421','YYYYMMDD'), 2344.78, 'Vancouver','Manager')
  3  /

1 row created.

SQL> insert into Employee(ID,  First_Name, Last_Name, Start_Date,                     End_Date,                       Salary,  City,       Description)
  2                values('05','Robert',   'Black',   to_date('19840115','YYYYMMDD'), to_date('19980808','YYYYMMDD'), 2334.78, 'Vancouver','Tester')
  3  /

1 row created.

SQL> insert into Employee(ID,  First_Name, Last_Name, Start_Date,                     End_Date,                       Salary, City,        Description)
  2                values('06','Linda',    'Green',   to_date('19870730','YYYYMMDD'), to_date('19960104','YYYYMMDD'), 4322.78,'New York',  'Tester')
  3  /

1 row created.

SQL> insert into Employee(ID,  First_Name, Last_Name, Start_Date,                     End_Date,                       Salary, City,        Description)
  2                values('07','David',    'Larry',   to_date('19901231','YYYYMMDD'), to_date('19980212','YYYYMMDD'), 7897.78,'New York',  'Manager')
  3  /

1 row created.

SQL> insert into Employee(ID,  First_Name, Last_Name, Start_Date,                     End_Date,                       Salary, City,        Description)
  2                values('08','James',    'Cat',     to_date('19960917','YYYYMMDD'), to_date('20020415','YYYYMMDD'), 1232.78,'Vancouver', 'Tester')
  3  /

1 row created.

SQL>
SQL>
SQL>
SQL> -- display data in the table
SQL> select * from Employee
  2  /

ID   FIRST_NAME LAST_NAME  START_DAT END_DATE      SALARY CITY       DESCRIPTION
---- ---------- ---------- --------- --------- ---------- ---------- ---------------
01   Jason      Martin     25-JUL-96 25-JUL-06    1234.56 Toronto    Programmer
02   Alison     Mathews    21-MAR-76 21-FEB-86    6661.78 Vancouver  Tester
03   James      Smith      12-DEC-78 15-MAR-90    6544.78 Vancouver  Tester
04   Celia      Rice       24-OCT-82 21-APR-99    2344.78 Vancouver  Manager
05   Robert     Black      15-JAN-84 08-AUG-98    2334.78 Vancouver  Tester
06   Linda      Green      30-JUL-87 04-JAN-96    4322.78 New York   Tester
07   David      Larry      31-DEC-90 12-FEB-98    7897.78 New York   Manager
08   James      Cat        17-SEP-96 15-APR-02    1232.78 Vancouver  Tester

8 rows selected.

SQL>
SQL>
SQL>
SQL>
SQL>
SQL>
SQL> --INSTR: returns the first-occurrence position of a character within a string.
SQL> --it returns a numeric value.
SQL>
SQL> --If it does not find the character value, it returns a 0. For example:
SQL>
SQL> SELECT First_name, INSTR(First_name,'a') AS INSTR FROM Employee;

FIRST_NAME      INSTR
---------- ----------
Jason               2
Alison              0
James               2
Celia               5
Robert              0
Linda               5
David               2
James               2

8 rows selected.

SQL>
SQL>
SQL>
SQL>
SQL>
SQL>
SQL>
SQL>
SQL> -- clean the table
SQL> drop table Employee
  2  /

Table dropped.

SQL>
SQL>
SQL>
SQL>

2.  Simple demo for INSTR function: returns a location within the string where search pattern begins   
SQL> --INSTR: returns a location within the string where search pattern begins.
SQL>
SQL>
SQL> SELECT INSTR('This is a test','is') FROM dual;

INSTR('THISISATEST','IS')
-------------------------
                        3

SQL>
SQL>

3.  INSTR: Look for the second occurrence of 'is'   
SQL>
SQL>
SQL> -- Look for the second occurrence of "is,"
SQL>
SQL> SELECT INSTR('This is a test','is',1,2) FROM dual;

INSTR('THISISATEST','IS',1,2)
-----------------------------
                            6

SQL>
SQL>
SQL>

4.  If search pattern is not in the string, the INSTR function returns 0   
SQL>
SQL>
SQL> -- If search pattern is not in the string, the INSTR function returns 0:
SQL>
SQL> SELECT INSTR('This is a test','abc',1,2) FROM dual;

INSTR('THISISATEST','ABC',1,2)
------------------------------
                             0

5.  Combine INSTR and SUBSTR together   
SQL>
SQL>
SQL> SELECT SUBSTR('aaa, bb ccc', INSTR('aaa, bb ccc',', ')) FROM dual;

SUBSTR('
--------
, bb ccc

6.  If the INSTR pattern is not found, then the entire string would be returned 

SQL>
SQL> -- If the INSTR pattern is not found, then the entire string would be returned
SQL>
SQL> SELECT SUBSTR('aaa bbb c', INSTR('aaa bbb c','zonk')) FROM dual;

SUBSTR('A
---------
aaa bbb c

[--小F--]

乌龟你3楼的寒到我了

[dawugui]

你要是个女的,我就抱抱.

[SQL77]

[Quote=引用 5 楼 dawugui 的回复:]
你要是个女的,我就抱抱.
[/Quote]
哈哈,小F,紧紧的抱抱

[--小F--]

那算了 请我喝酒不是一样

[SQL77]

[Quote=引用 7 楼 fredrickhu 的回复:]
那算了 请我喝酒不是一样
[/Quote]
哈哈,喝酒暖身体

[dawugui]

闪.

[superff7]

[Quote=引用 2 楼 dawugui 的回复:]
select a.* from a , b where instr(a.col,b.col) = 0 
union all
select b.* from a , b where instr(b.col,a.col) = 0 

[/Quote]

谢谢回复
instr(a.col,b.col)中b.col必须是具体的值吧？