javascript - 为什么云函数返回 ‘Optional()’ 的值

Question

我有一个检查电子邮件是否正在使用的函数，下面是该函数的代码:

import * as functions from 'firebase-functions'
import * as admin from 'firebase-admin'
admin.initializeApp()

exports.uniqueEmail = functions.https.onCall((data) => {

  const email = data.email;
  if (!email) {
    console.log('missing email')
    throw new functions.https.HttpsError('invalid-argument', 'Missing email parameter');
  }
  return admin.auth().getUserByEmail(email).then(function(userRecord) {
    console.log('Successfully fetched user data:', userRecord.toJSON());
    return "true"
  }).catch(function(error) {
   console.log('Error fetching user data:', error);
   console.log('Email: ', email)
   return "false"
  });
});

如您所见，该函数确实有效，它知道电子邮件是否已被接收，但它返回我的应用程序“可选(假)”而不是“假”或“真”。

这是我的 Xcode 代码:

IBAction func nextPressed(_ sender: Any) {
        var result = ""
        if(emailInputField.text != ""){
            Functions.functions().httpsCallable("uniqueEmail").call(["email": emailInputField.text]) { (result, error) in
                if let error = error as NSError? {
                    if error.domain == FunctionsErrorDomain {
                        let code = FunctionsErrorCode(rawValue: error.code)
                        let message = error.localizedDescription
                        let details = error.userInfo[FunctionsErrorDetailsKey]
                    }
                    // ...
                }
                    print(result?.data)
            }

提前致谢

Answer 1

你得到 opcional(true/false 因为)你正在打印一个可选值 '?'

你的代码:

print(result?.data)

例子:

let value = false

print("this is a false value: \(value?)")

每次你用'?'打印一个值你会先得到可选的