update DRV_TEMP_MID set ZTBM=p_ZTBM,SHSJ=sysdate where INSTR(','||p_id||',',','||id||',')>0;这个语句是什么意思,重点解析instr在这里是什么意思,谢谢!!!
解决方案 »
- 关于不用UNION问题.。。急在线等
- 如何能将文本文件更快的写入表中
- 配置客户端时出错
- 请教一个关于更新的问题
- 监听程序无法启动专用服务器进程
- ORA-12535:TNS:操作超时,只要连到外网就不能访问oracle了。(看了好多解决方案但是还是不行,求救!)
- sql server 中的print在oracle中对应的是什么?
- oracale数据库问题,请大家指教!!
- 请教高手:ORA-03113: end-of-file on communication channel错误发生的原因,如何解决?
- ORACLE8.1.5安装问题!!!——高手救命,高分相赠
- 急求各位高手帮忙解决
- 请教一个oracle备份的问题
类似于sql server的chaindex你这句的意思就是查询在ID中第一次存在p_id的位置.由于你的ID是以逗号分隔的,所有需要前后都加上逗号.类似于:id = ',1,2,3,4,5,6,10,12'
p_id = '1'如果不加逗号,则会把1,12,10搞混.
加上逗号后,则准确了.
类似于sql server的chaindex你这句的意思就是查询在ID中第一次存在p_id的位置.由于你的ID是以逗号分隔的,所有需要前后都加上逗号.类似于:id = '1,2,3,4,5,6,10,12'
p_id = '1'如果不加逗号,则会把1,12,10搞混.
加上逗号后,则准确了.
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 Tester8 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 28 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')
-------------------------
3SQL>
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)
-----------------------------
6SQL>
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)
------------------------------
05. Combine INSTR and SUBSTR together
SQL>
SQL>
SQL> SELECT SUBSTR('aaa, bb ccc', INSTR('aaa, bb ccc',', ')) FROM dual;SUBSTR('
--------
, bb ccc6. 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