4
如何使用bugzilla rest api報告bug?以下文檔指出bug對象或其一些字段必須包含在POST主體中。我試圖添加字段作爲POST方法參數,但我得到這個錯誤「沒有數據提供創建」狀態代碼400.我的問題是,我怎麼能包括一個錯誤對象或其一些字段在POST方法體?如何使用Bugzilla REST API發佈bug
https://wiki.mozilla.org/Bugzilla:REST_API:Methods#Create_new_bug_.28.2Fbug_POST.29
String serverURL = "https://api-dev.bugzilla.mozilla.org/test/latest";
String product = "FoodReplicator";
HttpClient client = new HttpClient();
PostMethod method = new PostMethod(serverURL + "/[email protected]&password=123456);
method.addParameter("product", "FoodReplicator");
method.addParameter("component", "Salt");
method.addParameter("summary", "testing");
method.addParameter("version", "1.0");
client.executeMethod(method);
return method.getStatusCode() + " " + method.getResponseBodyAsString();