oracle instr函数用法和(oracle 用instr 来代替 like)

浏览数：53 / 时间：2015年06月12日

最近项目中出现oracle instr函数，于是在网上找了一下关于它的资料。

Oracle中INSTR和SUBSTR的用法

Oracle中INSTR的用法：
INSTR方法的格式为
INSTR(源字符串, 要查找的字符串, 从第几个字符开始, 要找到第几个匹配的序号)
返回找到的位置，如果找不到则返回0.
例如：INSTR(‘CORPORATE FLOOR‘,‘OR‘, 3, 2)中，源字符串为‘CORPORATE FLOOR‘,在字符串中查找‘OR‘，从第三个字符位置开始查找"OR"，取第三个字后第2个匹配项的位置。

默认查找顺序为从左到右。当起始位置为负数的时候，从右边开始查找。

所以SELECT INSTR(‘CORPORATE FLOOR‘, ‘OR‘, -1, 1) "aaa" FROMDUAL的显示结果是

Instring
——————
14

oracle的substr函数的用法：
取得字符串中指定起始位置和长度的字符串  substr( string, start_position, [ length ] )
如:
    substr(‘This is a test‘, 6,2)    would return ‘is‘
    substr(‘This is a test‘,6)    would return ‘is a test‘
    substr(‘TechOnTheNet‘, -3,3)    would return ‘Net‘
    substr(‘TechOnTheNet‘, -6,3)    would return ‘The‘

select substr(‘Thisisatest‘, -4, 2) value from dual

综合应用：
SELECT INSTR(‘CORPORATE FLOOR‘, ‘OR‘, -1, 1) "Instring" FROMDUAL
--INSTR(源字符串, 目标字符串, 起始位置, 匹配序号)
SELECT INSTR(‘CORPORATE FLOOR‘,‘OR‘, 3, 2) "Instring" FROM DUAL

SELECT INSTR(‘32.8,63.5‘,‘,‘, 1, 1) "Instring" FROM DUAL

SELECT SUBSTR(‘32.8,63.5‘,INSTR(‘32.8,63.5‘,‘,‘, 1, 1)+1)"INSTRING" FROM DUAL
SELECT SUBSTR(‘32.8,63.5‘,1,INSTR(‘32.8,63.5‘,‘,‘, 1, 1)-1)"INSTRING" FROM DUAL

-- CREATED ON 2008-9-26 BY ADMINISTRATOR
DECLARE
-- LOCAL VARIABLES HERE
T  VARCHAR2(2000);
S  VARCHAR2(2000);
NUM INTEGER;
I  INTEGER;
POS INTEGER;
BEGIN
-- TEST STATEMENTS HERE
T := ‘12.3,23.0;45.6,54.2;32.8,63.5;‘;
SELECT LENGTH(T) - LENGTH(REPLACE(T, ‘;‘, ‘‘))INTO NUM FROM DUAL;
DBMS_OUTPUT.PUT_LINE(‘NUM:‘ || NUM);
POS := 0;
FOR I IN 1 .. NUM LOOP
   DBMS_OUTPUT.PUT_LINE(‘I:‘ || I);
   DBMS_OUTPUT.PUT_LINE(‘POS:‘ || POS);
   DBMS_OUTPUT.PUT_LINE(‘==:‘ || INSTR(T, ‘;‘, 1, I));
   DBMS_OUTPUT.PUT_LINE(‘INSTR:‘ || SUBSTR(T, POS + 1, INSTR(T, ‘;‘,1, I) - 1));
    POS :=INSTR(T, ‘;‘, 1, I);
END LOOP;
END;

-- Created on 2008-9-26 by ADMINISTRATOR
declare
-- Local variables here
i integer;
T     VARCHAR2(2000);
S     VARCHAR2(2000);
begin
-- Test statements here
    --历史状态
T := ‘12.3,23.0;45.6,54.2;32.8,63.5;‘;
IF (T IS NOT NULL) AND (LENGTH(T)> 0) THEN
    --T := T ||‘,‘;
    WHILELENGTH(T) > 0 LOOP
     --ISTATUSID := 0;
     S        := TRIM(SUBSTR(T, 1, INSTR(T, ‘;‘) - 1));
     IF LENGTH(S) > 0 THEN
        DBMS_OUTPUT.PUT_LINE(‘LAT:‘||SUBSTR(‘32.8,63.5‘,1,INSTR(‘32.8,63.5‘,‘,‘,1, 1)-1));
        DBMS_OUTPUT.PUT_LINE(‘LON:‘||SUBSTR(‘32.8,63.5‘,INSTR(‘32.8,63.5‘,‘,‘,1, 1)+1));
       -- COMMIT;
     END IF;
     T := SUBSTR(T, INSTR(T, ‘;‘) + 1);
    ENDLOOP;
END IF;
end;