我有以下用于发布plsql json请求的基本代码。被执行的web服务没有任何响应,因为它只是用于执行某项任务。 但每次执行块时,我都会从Apache Tomcat获取状态码400。 我哪里错了?
declare
http_resp utl_http.resp;
http_req utl_http.req;
json_msg VARCHAR2(500);
begin
http_req := utl_http.begin_request('http://192.168.1.194:8080/NotificationApp/sendNotification.rest', 'POST');
utl_http.set_body_charset(http_req, 'UTF-8');
utl_http.set_header(http_req, 'Content-Type', 'application/json');
json_msg := '{"code":100,"id": "APA91bFSmD_gBsUwP_hraRZL20mt8p4ejGn5fC7tlciINT50Ad8oIod2T-64GVk_8rrjoqXGEpYuRcoQogG0L7aOKIjeeisTcmHiUUONbnZzn4_u0ED7QD_iNeVkh2RU8Pa-HBHwgJUgOT-TyvlM9hB4Yn9fvWER","data": "alert alert"}';
utl_http.write_text(http_req, dbms_lob.substr(json_msg,dbms_lob.getLength(json_msg),1));
http_resp := utl_http.get_response(http_req);
if (http_resp.status_code >= 400) and
(http_resp.status_code <= 499)
then
dbms_output.put_line(http_resp.status_code);
end if;
utl_http.end_response(http_resp);
end;
提前致谢
答案 0 :(得分:9)
经过大量搜索后,我从blog获得了以下代码。对我来说很好。
declare
req utl_http.req;
res utl_http.resp;
url varchar2(4000) := 'http://192.168.1.194:8080/NotificationApp/sendNotification.rest';
name varchar2(4000);
buffer varchar2(4000);
content varchar2(4000) := '{"code":100,"id": "APA91bFSmD_gBsUwO_hraRZL20mt8p4ejGn5fC7tlciINT50Ad8oIod2T-64GVk_8rProqXGEpYuDcoQogG0L7a0TuyeeisTcmHiUUONbnZzn4_u0ED7QD_iNeVkh1ZgU8Pa-HRtfgJUgOT-TyvlM9hB4Yn9fvOPud","data": "alert alert"}';
begin
req := utl_http.begin_request(url, 'POST',' HTTP/1.1');
utl_http.set_header(req, 'user-agent', 'mozilla/4.0');
utl_http.set_header(req, 'content-type', 'application/json');
utl_http.set_header(req, 'Content-Length', length(content));
utl_http.write_text(req, content);
res := utl_http.get_response(req);
begin
loop
utl_http.read_line(res, buffer);
dbms_output.put_line(buffer);
end loop;
utl_http.end_response(res);
exception
when utl_http.end_of_body then
utl_http.end_response(res);
end;
end;
答案 1 :(得分:0)
现在,我需要更改行req:= utl_http.begin_request(url,&#39; POST&#39;,&#39; HTTP / 1.1&#39;);
我做过并且工作过:
req := utl_http.begin_request(url, 'POST');