Question

假设我有一个function0: function(e) { e.preventDefault(); var valid = false; var dfd = $.Deferred(); //$.when(function1(e)).done(function(v){ // had to change .done to .then $.when(function1(e)).then(function(v){ valid = v; // -> right here valid is undefined if(!valid) { alertSomething(); return; } // had to change the scope and make it inside the .then so that it only executes when I have the value keepExecutingCode(); dfd.resolve(); }); dfd.promise(); },，我想将其映射到另一个Observable<Player>，其中Observable<Integer>等于Integer，但是有一个条件：我想映射所有玩家，但是第一个和最后一个（我们应该为他们再检查一下）。所以在迭代编程中它会像这样：

player.height

如何以Rx方式重写（你应该假设我给了Observable而不是List）？

Answer 1

假设：

Observable<Player> players;

Single<Integer> playerHeight(int playerId);

您必须使用publish(Function)将序列分为第一个，中间一个和最后一个，然后将它们组合在一起：

players
.publish(sharedPlayers -> {
    return Observable.merge(
       // work only on the very first player
       sharedPlayers.take(1)
           .filter(firstPlayer -> isGoodEnough(firstPlayer))
           .flatMapSingle(firstPlayer -> playerHeight(firstPlayer.playerId)),
       // work with not the first and not the last
       sharedPlayers.skip(1).skipLast(1)
           .flatMapSingle(midPlayers -> playerHeight(midPlayers .playerId)),
       // work with the last which shouldn't be the first again
       sharedPlayers.skip(1).takeLast(1)
           .filter(lastPlayer -> isGoodEnough(lastPlayer))
           .flatMapSingle(lastPlayer-> playerHeight(lastPlayer.playerId))
    );
})
.subscribe(/* ... */);

请根据需要调整此解决方案。

如何使用Observable中的1st和least元素？

1 个答案: