Question

我有一个这样的数组：

var bookCh = {'TOMSAWYER':[Twain,50],
'PARELANDRA':[Lewis,150],
'ROGUECODE':[Russinovich,23],
'WRINKLE':[Lengle,12]};

如果我在ROGUECODE上匹配某些东西，我想打印控制台的下一个键（例如，如果我匹配ROGUECODE，则打印WRINKLE）。这是我必须做的匹配，但我不知道如何打印数组中的下一个键而不是当前的键。如果它们是数字键，那显然很容易，但它们不是......

var currentBook = "ROGUECODE";

for (var bookKey in bookCh) {
 if (bookKey == currentBook) {
  console.log("Next book: " +  ???);
 }
}

Answer 1

声明你的数组是这样的

var bookCh = [
  {
    'id': 'TOMSAWYER',
    'meta': [Twain, 50]
  },
  {
    'id': 'PARELANDRA',
    'meta': [Lewis, 150]
  },
  {
    'id': 'ROGUECODE',
    'meta': [Russinovich, 23]
  },
  {
    'id': 'WRINKLE',
    'meta': [Lengle, 12]
  }
]

现在当你迭代它时

for(var i=0; i<bookCh.length-1; i++) {
 // bookCh[i+1] (this is next array item)
}

Answer 2

var currentBook = "ROGUECODE";
var keys = Object.keys(bookCh);
for (var bookKey in bookCh) {
 if (bookKey == currentBook) {
    var curIdx = keys.indoxOf(bookKey); 
    var nextBookKey =  curIdx < keys.length-1 ?  keys[curIdx+1] : null;
    console.log("Next book: " + nextBookKey);
 }
}

Answer 3

我想到一种愚蠢的方式......如果你的钥匙很少......

＆＃13;

var bookCh = {
    'TOMSAWYER': ['Twain', 50],
    'PARELANDRA': ['Lewis', 150],
    'ROGUECODE': ['Russinovich', 23],
    'WRINKLE': ['Lengle', 12]
};

var currentBook = "ROGUECODE";
var order = {
    'TOMSAWYER': 'PARELANDRA',
    'PARELANDRA': 'ROGUECODE',
    'ROGUECODE': 'WRINKLE',
    'WRINKLE': 'TOMSAWYER'
};

for (var bookKey in bookCh) {
    if (bookKey == currentBook) {
        console.log("Next book: " + bookCh[order[bookKey]]);
    }
}

＆＃13;

获取下一个键

3 个答案: