still get the same error.
I tried to create a procedure as below, (this is almost same as the one you had)
CREATE OR REPLACE PROCEDURE tag_owner.show_html_from_url (p_url IN VARCHAR2) AS
l_http_request UTL_HTTP.req;
l_http_response UTL_HTTP.resp;
l_text VARCHAR2(32767);
BEGIN
-- Make a HTTP request and get the response.
l_http_request := UTL_HTTP.begin_request(p_url);
l_http_response := UTL_HTTP.get_response(l_http_request);
-- Loop through the response.
BEGIN
LOOP
UTL_HTTP.read_text(l_http_response, l_text, 32766);
DBMS_OUTPUT.put_line (l_text);
END LOOP;
EXCEPTION
WHEN UTL_HTTP.end_of_body THEN
UTL_HTTP.end_response(l_http_response);
END;
EXCEPTION
WHEN OTHERS THEN
UTL_HTTP.end_response(l_http_response);
RAISE;
END show_html_from_url;
/
and then from SQL prompt, did the following
SQL> exec utl_http.set_proxy('sdev-prelive-litle.hctra.pri:443');
PL/SQL procedure successfully completed.
SQL> exec utl_http.set_wallet('file: /app/oracle/product/11204/dbv1_JANPSU2016/wallet', '***');
PL/SQL procedure successfully completed.
Then, executed the procedue
SQL> exec show_html_from_url('
https://prelive.litle.com' );
BEGIN show_html_from_url('
https://prelive.litle.com' ); END;
*
ERROR at line 1:
ORA-29273: HTTP request failed
ORA-06512: at "SYS.UTL_HTTP", line 1525
ORA-29261: bad argument
ORA-06512: at "TAG_OWNER.SHOW_HTML_FROM_URL", line 22
ORA-29273: HTTP request failed
ORA-06512: at "SYS.UTL_HTTP", line 1130
ORA-29259: end-of-input reached
ORA-06512: at line 1
Any idea what could be the issue ?