javascript - $q.all - 只取那些已解决的

Question

如果是这样的话:

getCol = (colId)->
   dfrd = $q.defer()
   if colId == "bacon"
       dfrd.reject()
   else
       dfrd.resolve colId
   dfrd.promise


getCols = (columns)->
   $q.all(_.map(columns, (cs)-> getCol(cs)))


getCols(['eggs','juice']).then (cols)->
   console.log cols                    # works



getCols(['eggs','juice','bacon']).then (cols)->
   console.log cols                  # not even getting here

那么，在 getCols() 中，我怎样才能只返回那些已解决的 promise ？

Answer 1

$q.all 仅在 all of the promises you pass it 时解析已解决。例如，它非常适合仅在加载所有 4 个小部件后才显示仪表板。

如果您不希望出现这种行为，也就是说，您想尝试显示尽可能多的可以成功解析的列，则必须使用不同的方法。

loadedColumns = []

getCols = (columns) ->
  for col in columns
    willAdd = addColumn(col) # add column needs to store columns in the "loadedColumns" area, then resolve
    willAdd.then buildUI
    willAdd.catch logError


# Because this method is debounced, it'll fire the first time there is 50 ms of idleness
buildUI = _.debounce ->
  // Construct your UI out of "loadedColumns"
, 50