I need to resolve a math equation/function in pl/sql.
What kind of math operations/functions are available in oracle pl/sql that could help me resolve a math function like t
Alas the Oracle database is not a mathematical tool. It has lots of arithmetical and statistical functions but it doesn't have built-in functionality capable of interpreting equations. Sorry.
By sheer coincidence Marc (AKA Odie_63) has recently published a Reverse Polish Notation calculator which he has written in PL/SQL. It doesn't do precisely what you want but I'm including a link for the benefit of any seekers who may stumble upon this thread in the future. Find out more.
As APC said, there is no built-in functionality to do this. But you can use the WolframAlpha API from PL/SQL:
declare
v_equation varchar2(32767) :=
'(3.5/(1+x))+(3.5/(1+x)^2)+(3.5/(1+x)^3)+(3.5/(1+x)^4)+(100/(1+x)^4)=101.55';
v_escaped_url varchar2(32767);
v_uri httpuritype;
v_xml xmltype;
v_count number := 1;
begin
--Escape the URL.
--I used chr(38) for ampersand, in case your IDE think it's a substitution variable
v_escaped_url :=
'http://api.wolframalpha.com/v2/query?appid=EQGHLV-UYUEYY9ARU'||chr(38)||'input='
||utl_url.escape(v_equation, escape_reserved_chars => true)
||chr(38)||'format=plaintext';
--Create an HTTPURIType, and get the XML
v_uri := httpuritype.createUri(v_escaped_url);
v_xml := v_uri.getXML;
--Complex solutions
while v_xml.existsNode('/queryresult/pod[@title="Complex solutions"]/subpod['||v_count||']') = 1 loop
dbms_output.put_line(v_xml.extract('/queryresult/pod[@title="Complex solutions"]/subpod['||v_count||']/plaintext/text()').getStringVal());
v_count := v_count + 1;
end loop;
--Real solutions
v_count := 1;
while v_xml.existsNode('/queryresult/pod[@title="Real solutions"]/subpod['||v_count||']') = 1 loop
dbms_output.put_line(v_xml.extract('/queryresult/pod[@title="Real solutions"]/subpod['||v_count||']/plaintext/text()').getStringVal());
v_count := v_count + 1;
end loop;
end;
/
Results:
x = -1.00006-0.996229 i
x = -1.00006+0.996229 i
x = -1.99623
x = 0.0308219
There are a lot of potential downsides to this approach. It will be very slow, and the API is not free. My example works because I used my free developer appid, but it's only good for a small number of calls.