设置Oracle 12c R1以连接到Web服务

问题描述 投票:0回答:1

我已经使用DBMS_NETWORK_ACL_ADMIN软件包设置了ACL。在这里

enter image description here

但是当我尝试使用GET连接到上述Web服务时,出现错误

ORA-29273: HTTP request failed
ORA-24247: network access denied by access control list (ACL)
ORA-06512: at "SYS.UTL_HTTP", line 368
ORA-06512: at "SYS.UTL_HTTP", line 1118
ORA-06512: at "APEX_040200.WWV_FLOW_WEB_SERVICES", line 550
ORA-06512: at "APEX_040200.WWV_FLOW_WEBSERVICES_API", line 197
ORA-06512: at line 7
29273. 00000 -  "HTTP request failed"
*Cause:    The UTL_HTTP package failed to execute the HTTP request.
*Action:   Use get_detailed_sqlerrm to check the detailed error message.
           Fix the error and retry the HTTP request.

我也要提到主机oracle-base.com运行良好。下面的代码工作正常。

declare
l_response CLOB;
l_url varchar2(4000) := 'http://oracle-base.com/webservices/add-numbers.php'; --web service URL
l_result NUMBER;
begin
---get XML from 
l_response := apex_web_service.make_rest_request(p_url => l_url,
                                                 p_http_method => 'GET',
                                                 p_parm_name  => apex_util.string_to_table('p_int_1:p_int_2'),
                                                 p_parm_value => apex_util.string_to_table('12:12')
                                                );


dbms_output.put_line(l_response);
--parse the response
l_result :=apex_web_service.parse_xml(p_xml => XMLTYPE(l_response), --convert CLOB to XML
                           p_xpath => '//number/text()'
                        );
dbms_output.put_line(l_result);
end;

但是当我尝试对reqres.re进行相同操作时,它会失败。我如何添加reqres.in

DECLARE
  l_principal VARCHAR2(20) := 'APEX_040200';
  --l_principal VARCHAR2(20) := 'APEX_050000';
  --l_principal VARCHAR2(20) := 'APEX_050100';
  --l_principal VARCHAR2(20) := 'APEX_180200';
BEGIN
  DBMS_NETWORK_ACL_ADMIN.append_host_ace (
    host       => 'reqres.in', 
    lower_port => 80,
    upper_port => 80,
    ace        => xs$ace_type(privilege_list => xs$name_list('http'),
                              principal_name => l_principal,
                              principal_type => xs_acl.ptype_db)); 
END;
/

我正在尝试执行的命令:

declare
l_response CLOB;
l_url varchar2(4000) := 'https://reqres.in/api/users?page=2'; --web service URL
l_result NUMBER;
begin
---get XML from 
l_response := apex_web_service.make_rest_request(p_url => l_url,
                                                 p_http_method => 'GET'
--                                                 p_parm_name  => apex_util.string_to_table('p_int_1:p_int_2'),
--                                                 p_parm_value => apex_util.string_to_table('12:12')
                                                );


dbms_output.put_line(l_response);
----parse the response
--l_result :=apex_web_service.para(p_xml => XMLTYPE(l_response), --convert CLOB to XML
--                           p_xpath => '//number/text()'
--                        );
dbms_output.put_line(l_result);
--APEX_JSON.parse(l_result); -- APEX_JSON is not available

end;
oracle web-services plsql
1个回答
1
投票

要发出HTTPS请求,您需要做的第一件事是指定正确的端口。 HTTPS使用443,而HTTP使用80:

     DBMS_NETWORK_ACL_ADMIN.append_host_ace (
       host       => 'reqres.in', 
       lower_port => 443,
       upper_port => 443,
       ace        => xs$ace_type(....

这应该消除ORA-24247: network access denied by access control list (ACL)错误。但是,这样做之后,您可能会收到ORA-29024: Certificate validation failure错误。

如果发生这种情况,您需要创建一个Oracle电子钱包来存储https://reqres.in/使用的根证书(即Sectigo证书),确保数据库有权读取该钱包,并将钱包详细信息添加到呼叫apex_web_service.make_web_request

apex_web_service.make_web_request
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