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" FROM DUAL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" FROM DUAL
--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 || ',';
    WHILE LENGTH(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);
    END LOOP;
  END IF;
end;

来源：oschina

链接：https://my.oschina.net/u/2862827/blog/3058484

标签

substr

instr

Oracle

instr函数

oracle函数