Question

聚合物1.0组件是否有办法注册页面上发生的所有按键操作，无论重点在哪里？

在预组件世界中，我会加入文档正文中的关键事件。什么是干净的，组件驱动的方法？

Answer 1

我能够通过实施Polymer.IronA11yKeysBehavior behavior来实现这一目标。

我基本上只是阅读为该演示定义的组件的源代码，is here

我在对原始问题的评论中提到的错误只是让我keyEventTarget properties properties对象放在<dom-module id="x-component"> <template> Type shift+a to see feedback. <div id="feedback"></div> </template> </dom-module> <script> Polymer({ is: 'x-component', behaviors: [Polymer.IronA11yKeysBehavior], keyBindings: { 'shift+a': '_handleKey' }, _handleKey: function(e) { this.$['feedback'].innerHTML = this.$['feedback'].innerHTML + "<p>again</p>"; }, properties: { keyEventTarget: { type: Object, value: function() { return document.body; } } } }); </script>对象上方的一两个屏幕上; d已经放入页面，优先。

这是一个非常简单的组件，可以显示它的实际效果：

def play_game(initial_funds, price_to_play, iterations):
    prob_tree = {}
    for iteration in range(1, iterations + 1):
        prob_tree[iteration] = []
        # first number always represents money in hand, second represents money in pot.
    prob_tree[1].append([initial_funds - price_to_play, 1])
    for iteration in range(2, iterations + 1):
        # for each possible outcome listed in the line above in the probability tree
        for possibility in prob_tree[iteration - 1]:
            # if there isn't any money in the pot AND not enough current funds, just repeat it twice in the next line for counting purposes.
            if possibility[0] < price_to_play and possibility[1] == 0:
                prob_tree[iteration].append(possibility)
                prob_tree[iteration].append(possibility)
                # add the current possibility combo twice to the current iteration.
            # elif there is money in pot, append a heads and tails possibility to the current row
            elif possibility[0] < price_to_play and possibility[1] > 0:
                case2_old_pot = possibility[1]
                case2_win_new_pot = possibility[1] * 2
                case2_lose_new_pot = 0
                case2_player_funds = possibility[0]
                # if successful, the money in the pot will double, and the player's money will remain constant
                prob_tree[iteration].append([case2_player_funds, case2_win_new_pot])
                # if not successful, the money in the pot will be added into the player's money, and the pot will return to 0
                prob_tree[iteration].append([case2_player_funds + case2_old_pot, case2_lose_new_pot])
            # elif there is no money in the pot but there is sufficient money in the player's pocket to play again1
            elif possibility[0] >= price_to_play and possibility[1] == 0:
                # first, the money in the player's pocket goes down by the amount price_to_play, and the pot amount goes up to one.
                case3_funds_after_buyin = possibility[0] - price_to_play
                case3_pot_after_buyin = 1
                case3_pot_if_successful = 2
                case3_pot_if_unsuccessful = 0
                case3_funds_if_successful = case3_funds_after_buyin
                case3_funds_if_unsuccessful = case3_funds_after_buyin + case3_pot_after_buyin
                # then, either the player gets back the pot and it goes to 0
                prob_tree[iteration].append([case3_funds_if_unsuccessful, case3_pot_if_unsuccessful])
                prob_tree[iteration].append([case3_funds_if_successful, case3_pot_if_successful])
    counter = 0
    for outcome in prob_tree[iterations]:
        if outcome[0] < price_to_play and outcome[1] == 0:
            counter += 1
    print float(counter)/(2**iterations)

play_game(2, 2, 25)

制作可以收听页面上所有按键的聚合物组件

1 个答案: