Cannot read property 'then' of undefined with JavaScript promises

问题

I understand at first glance this may look like a duplicate but I have seen all the answers for that tell me to put a return in but that is not working.

this is my function:

function removePastUsersFromArray(){
  pullAllUsersFromDB().then(function(users_array){
  var cookie_value = document.cookie.split('=') [1];
  const promises = []
    for (var i = 0; i < _USERS.length; i++) {
      if (_USERS[i].useruid == cookie_value){
      var logged_in_user = _USERS[i].useruid;
        promises.push(
          onChildValue(rootRef, 'users/' + logged_in_user + '/disliked_users/').then(formatUsers)
        )
        promises.push(
          onChildValue(rootRef, 'users/' + logged_in_user + '/liked_users/').then(formatUsers)
        )
      }
    }
    return Promise.all(promises);
  })
};

and I get the error at this function:

function displayRemovedPastUsersFromArray(){
  removePastUsersFromArray().then(function(promises){

basically saying that my removePastUsersFromArray is undefined. but it isn't as it clearly exists above and returns promises??

回答1:

basically saying that my removePastUsersFromArray is undefined

No, it says that the removePastUsersFromArray() call returned undefined, as that's what you're trying to call then upon.

it clearly exists above and returns promises?

It exists, yes, but it doesn't return anything. The return you have is inside the then callback, but the function itself does not have a return statement. return the promise that results from the chaining:

function removePastUsersFromArray() {
  return pullAllUsersFromDB().then(function(users_array) {
//^^^^^^
    var cookie_value = document.cookie.split('=') [1];
    const promises = []
    for (var i = 0; i < _USERS.length; i++) {
      if (_USERS[i].useruid == cookie_value){
        var logged_in_user = _USERS[i].useruid;
        promises.push(
          onChildValue(rootRef, 'users/' + logged_in_user + '/disliked_users/').then(formatUsers)
        );
        promises.push(
          onChildValue(rootRef, 'users/' + logged_in_user + '/liked_users/').then(formatUsers)
        );
      }
    }
    return Promise.all(promises);
  })
};

来源：https://stackoverflow.com/questions/39509589/cannot-read-property-then-of-undefined-with-javascript-promises