Maths formula parser

Question

Hi,
Does oracle provide any function that parses and evaluets a math formula/expression by resolving the operator precedence in it.
e.g. I have a math formula in the form of a string variable '(2+3)*(4+5)'
Does oracle have a function that will resolve this and return answer 45.
OR
Does anyone have a stored proc/func for this that I can use.
Any help is appreciated.
Thanks in advance.

Answer 1

I created this a long time ago in response to a post on this forum. Someone wanted an IIF() construct.
CREATE FUNCTION iif (p_expr IN VARCHAR2,
                     p_true_result IN VARCHAR2,
                     p_false_result IN VARCHAR2) RETURN VARCHAR2 IS
   l_sqlstr VARCHAR2(4000);
   l_result PLS_INTEGER;
BEGIN
   l_sqlstr := 'SELECT 1 FROM dual WHERE '||REPLACE(p_expr,'"','''');
   BEGIN
      EXECUTE IMMEDIATE l_sqlstr INTO l_result;
      RETURN p_true_result;
   EXCEPTION WHEN NO_DATA_FOUND THEN
      RETURN p_false_result;
   END;
END;{code}
or, for a strictky PL/SQL version (i.e. you can't call it in sql)
{code}CREATE FUNCTION iif (p_expr IN VARCHAR2) RETURN BOOLEAN IS
   l_sqlstr VARCHAR2(4000);
   l_result PLS_INTEGER;
BEGIN
   l_sqlstr := 'SELECT 1 FROM dual WHERE '||REPLACE(p_expr,'"','''');
   BEGIN
      EXECUTE IMMEDIATE l_sqlstr INTO l_result;
      RETURN TRUE;
   EXCEPTION WHEN NO_DATA_FOUND THEN
      RETURN FALSE;
   END;
END;The first version would be called like:
SELECT IIF('"A" in ("A","B","C") and 3 > 2 and 4 between 2 and 5','TRUE','FALSE') FROM dual;The second something like:
IF IIF('"A" in ("A","B","C") and 3 > 2 and 4 between 2 and 5') THEN
END IF;Just keep in mind Brian's comments about SQL injection from earlier in this thread
John