Question

在XPages / Java中，有没有更好的方法来将ref的基本URL获取到当前应用程序？

Database.database().reference().child("questionPosts").queryOrderedByKey().observe(.childAdded) { (snapshot) in
            if let dict = snapshot.value as? NSDictionary {
                //var questionName = dict["name"] as! String
                //var created_by = dict["email"] as! String
                let questionTitle = dict["name"] as? String
                let created_by = dict["email"] as? String

                let question = Question(questionName: questionTitle!, created_by: created_by!)

                self.questions.append(question)



                print(self.questions.count)

            }
        }

Answer 1

也许更好一点：

String serverPathNsf = url.getAddress().replace(url.getSiteRelativeAddress(context), "")

url.getAddress()为您提供nsf + xsp的服务器+路径

    http://yourServer/pathTo.nsf/your.xsp

url.getSiteRelativeAddress(context)为您提供了xsp

    /your.xsp

和url.getAddress().replace(url.getSiteRelativeAddress(context), "")返回nsf的服务器+路径

    http://yourServer/pathTo.nsf

获取Java中应用程序的基本URL

1 个答案: