小编Dam*_*mon的帖子

我有一个对象数组

people = [
    {id: "1", name: "abc", gender: "m", age:"15" },
    {id: "2", name: "a", gender: "m", age:"25" },
    {id: "3", name: "efg", gender: "f", age:"5" },
    {id: "4", name: "hjk", gender: "m", age:"35" },
    {id: "5", name: "ikly", gender: "m", age:"41" },
    {id: "6", name: "ert", gender: "f", age:" 30" },
    {id: "7", name: "qwe", gender: "f", age:" 31" },
    {id: "8", name: "bdd", gender: "m", age:" 78" },
]

我有另一个数组:

id_filter = [1,4,5,8]

如果id匹配数组中的任何元素,我想过滤人的id_filter …

我希望在调用第二个“then”时不执行第三个“then”。然而，即使承诺被拒绝（第二个“then”被调用）并且代码返回“rejected”然后“undefined”，它仍然调用第三个“then”。如何不运行第三个“then”，这样“undefined”就不会出现？

var FirstPromise = function() {

  let promiseme = new Promise(function(res, rej) {

    if ('a' == 'b') {
      res(1);
    } else {
      rej('rejected');
    }
  })

  return promiseme;
};

function succeddcallback(msg) {
  return msg * 2;
}

function errorcallback(msg) {
  console.log(msg);
}

FirstPromise()
  .then(succeddcallback, null)
  .then(null, errorcallback)
  .then(function(succedded) {
    console.log(succedded);
  }, function(failed) {
    console.log(failed);
  })

我想找到一个字符串的所有组合，保持顺序，但任何长度。例如：

string_combinations("wxyz")
# => ['w', 'wx', 'wxy', 'wxyz', 'wxz', 'wy', 'wyz', 'wz', 'x', 'xy', 'xyz', 'xz', 'y', 'yz', 'z']

我希望您只能使用循环并避免使用 ruby​​ 方法，就像#combination我试图找到最干净的方法来实现它，如果我在另一种语言中遇到它。

有没有办法在小于 O(n^3) 的时间内做到这一点？我最初的想法是这样的：

def string_combinations(str)
    result = []

    (0...str.length).each do |i|
        result << str[i]
        ((i+1)...str.length).each do |j|
            result << str[i] + str[j]
            ((j+1)...str.length).each do |k|
                result << str[i] + str[j..k]
                # Still not covering everything.
            end
        end
    end

    result
end