Instr()函数的使用计算字符串中出现某个字母或单词的个数

The following example searches the string "CORPORATE FLOOR", beginning with the third character, for the string "OR". It returns the position in CORPORATE FLOOR at which the second occurrence of "OR" begins:

SELECT INSTR('CORPORATE FLOOR','OR', 3, 2)
  "Instring" FROM DUAL;
 
  Instring
----------
        14

In the next example, Oracle counts backward from the last character to the third character from the end, which is the first "O" in "FLOOR". Oracle then searches backward for the second occurrence of OR, and finds that this second occurrence begins with the second character in the search string :

SELECT INSTR('CORPORATE FLOOR','OR', -3, 2)
"Reversed Instring"
     FROM DUAL;
 
Reversed Instring
-----------------
               2

默认查找‘o’的第一次出现的位置，默认从第一个字符开始:

SELECT INSTR('Hello World','o') FROM dual;

结果为：5

默认查找‘o’的第一次出现的位置，指定从从第6个字符开始:

SELECT INSTR('Hello World','o',6) FROM dual;

结果为:8

查找从第二次出现，从第一个位置开始的的'O'的位置:

SELECT INSTR('Hello World', 'o', 1, 2) FROM DUAL;

结果为:8
使用INSTR计算字符出现的个数：

DECLARE
  V_STRING_IN VARCHAR2(100) := 'Hello World';
  V_SUB_STR   VARCHAR2(100) := 'o';
  V_COUNTS    NUMBER;

  FUNCTION GETCOUNT(P_STRING_IN VARCHAR2, P_SUB_STR VARCHAR2) RETURN NUMBER IS
    V_COUNT      NUMBER := 1;
    V_SUBSTR_LOC NUMBER;
  BEGIN
    LOOP
      V_SUBSTR_LOC := INSTR(P_STRING_IN, P_SUB_STR, 1, V_COUNT);
      EXIT WHEN V_SUBSTR_LOC = 0 OR V_SUBSTR_LOC IS NULL;
      V_COUNT := V_COUNT + 1;
    END LOOP;
    RETURN V_COUNT - 1;
  EXCEPTION
    WHEN OTHERS THEN
      RETURN 'ERRORS';
  END;

BEGIN
  V_COUNTS := GETCOUNT(P_STRING_IN => V_STRING_IN, P_SUB_STR => V_SUB_STR);
  DBMS_OUTPUT.PUT_LINE('出现个数' || V_COUNTS);
END;