為什么我的JavaScript承諾無法解決？

Question

exports.validatePasscode = (req,res) => {
    var number = req.body.number || null;
    var passcode = parseInt(req.body.passcode) || null;

    return _users_helper.validatePasscode(number, passcode).then( () =>  {
        return _users_helper.fetchUserByPhone(number)
    }).then(user => {
        console.log("abc-abc", user); 
    }).catch(err => { 
        console.log(err);
    });
}

var fetchUserByPhone = function(number){
    return new Promise(function(resolve, reject){
        return db.ref('/phones-users/' + number).once('value').then(function(snapshot){
            if(snapshot.exists() == false){ return resolve(null); };
            return fetchUserById(snapshot.child('user_id').val());
        });
    });
};
exports.fetchUserByPhone = fetchUserByPhone;


var fetchUserById = function(id){
    return new Promise(function(resolve, reject){
        return db.ref('/users/' + id).once('value').then(function(snapshot){
            if(snapshot.exists()){
                var result = snapshot.val();
                result.id = id;
                console.log("User found result", result); //this prints
                resolve(result);
            }else{
                console.log("User not found");
                resolve(null);
            }
        });
    });
}
exports.fetchUserById = fetchUserById;

當我運行此代碼時， User found result正確打印。 但是， abc-abc無法打印。 為什么？

Answer 1

這不是新承諾的工作方式，通過電話獲取用戶應為：

var fetchUserByPhone = function(number){
  return db.ref('/phones-users/' + number)
  .once('value')
  .then(function(snapshot){
    if(snapshot.exists() == false){ 
      return null; 
    };
    return  fetchUserById(snapshot.child('user_id').val());
  });
};

按ID提取用戶應為：

var fetchUserById = function(id){
  return db.ref('/users/' + id).once('value')
  .then(function(snapshot){
      if(snapshot.exists()){
          var result = snapshot.val();
          result.id = id;
          console.log("User found result", result); //this prints
          return result;
      }else{
          console.log("User not found");
          return null;
      }
  });
}

當您調用已經返回類型（例如類型）的promise的函數時，無需創建新promise。

Answer 2

由fetchUserByPhone返回的fetchUserByPhone未得到解決/拒絕。

由於您的db方法無論如何都返回一個Promise，因此沒有必要創建Promise包裝器。 可能的選擇：

 exports.validatePasscode = (req, res) => { var number = req.body.number || null; var passcode = parseInt(req.body.passcode) || null; return _users_helper.validatePasscode(number, passcode).then(() => { return _users_helper.fetchUserByPhone(number) }).then(user => { console.log("abc-abc", user); }).catch(err => { console.log(err); }); } var fetchUserByPhone = function(number) { return db.ref('/phones-users/' + number).once('value').then(function(snapshot) { if (snapshot.exists() == false) { return Promise.reject(null); }; return fetchUserById(snapshot.child('user_id').val()); }); }; exports.fetchUserByPhone = fetchUserByPhone; var fetchUserById = function(id) { return db.ref('/users/' + id).once('value').then(function(snapshot) { if (snapshot.exists()) { var result = snapshot.val(); result.id = id; console.log("User found result", result); //this prints return result; } else { console.log("User not found"); return Promise.reject(null); } }); } exports.fetchUserById = fetchUserById;