的差異的方式返回承諾

Question

的我實現以下兩個承諾：

function getUserSportListPromise(_uid){ 
     return firebase.database().ref('users/' + user.uid + '/sports/').once('value').then(function(snap){ 
      var userSportList = []; 
      if(snap.exists()){ 
      snap.forEach(function(sport){ 
       userSportList.push(sport.val().name); 
      }); 
      } 

      return userSportList; 
     }); 
     } 

     function getRegistrationsFromSportList(sportList){ 
     var promiseHash = {}; 
     sportList.forEach(function (sport){ 
      promiseHash[sport] = firebase.database().ref('/sport_registrations/' + sport).once('value'); 
     }); 
     return $q.all(promiseHash); 
     } 

爲什麼這段代碼顯示結果：

var userSportListPromise = getUserSportListPromise(user.uid); 
     var registrationHashPromise = userSportListPromise.then(function(sportList){ 
     return getRegistrationsFromSportList(sportList); 
     }).then(function(registrationsHash){ 
     angular.forEach(registrationsHash, function(regsSnap, sport){ 
      console.info(sport + ' has the following registrations: '); 
      console.dir(regsSnap); 
     }); 
     }); 

鑑於此代碼顯示undefined：

getUserSportListPromise(user.uid) 
     .then(function(sportList){ 
      getRegistrationsFromSportList(sportList); 
     }) 
     .then(function(registrationsHash){ 
     angular.forEach(registrationsHash, function(regsSnap, sport){ 
      console.dir(registrationsHash); 
      console.info(sport + ' has the following registrations: ' + regsSnap); 
     }); 
     }); 

函數getUserSportListPromise返回一個promise的散列，爲什麼上面的代碼傳遞一個undefine d在last（）調用中註冊哈希？

Answer 1

你缺少一個return這裏：

.then(function(sportList){ 
     getRegistrationsFromSportList(sportList); 
    }) 

那個匿名函數沒有返回語句，所以它返回undefined。它應該是：

.then(function(sportList){ 
     return getRegistrationsFromSportList(sportList); 
    }) 

或者只是：

.then(getRegistrationsFromSportList)