我有一个数组:
arr =
[{"nid":"MIA","keys":[{"sid":"sm1"},{"sid":"sm2"}]},
{"nid":"MID","keys":[{"sid":"sm1"}]},
{"nid":"MIT","keys":[{"sid":"sm1"},{"sid":"sm2"},{"sid":"sm3"},{"sid":"sm4"},{"sid":"sm5"},{"sid":"sm6"},{"sid":"sm7"},{"sid":"sm8"},{"sid":"sm9"},{"sid":"sm10"}]},
{"nid":"MIO","keys":[{"sid":"sm1"},{"sid":"sm2"},{"sid":"sm3"}]},
{"nid":"MIS","keys":[{"sid":"sm1"},{"sid":"sm2"}]},
{"nid":"MIH","keys":[{"sid":"sm1"}]}]arr由6个元素组成。这六个元素中的每一个都由另一个数组keys组成。我需要按每个键数的升序重新排列这六个元素。这意味着我需要以这种方式重新排列数组:
arr =
[
{"nid":"MID","keys":[{"sid":"sm1"}]},
{"nid":"MIH","keys":[{"sid":"sm1"}]},
{"nid":"MIA","keys":[{"sid":"sm1"},{"sid":"sm2"}]},
{"nid":"MIS","keys":[{"sid":"sm1"},{"sid":"sm2"}]},
{"nid":"MIO","keys":[{"sid":"sm1"},{"sid":"sm2"},{"sid":"sm3"}]},
{"nid":"MIT","keys":[{"sid":"sm1"},{"sid":"sm2"},{"sid":"sm3"},{"sid":"sm4"},{"sid":"sm5"},{"sid":"sm6"},{"sid":"sm7"},{"sid":"sm8"},{"sid":"sm9"},{"sid":"sm10"}]},
]我试图获取每个数组元素的keys中的元素数,如下所示:
var arrMap = [];
arr.forEach(function(array_) {
key_ = array_.keys;
var count = 0;
key_.forEach(function(arrKey) {
count++;
var keyCode = arrKey.sid;
})
arrMap.push({'nid':array_.nid, 'count': count});
})
console.log(arrMap);这给了我以下输出:
[{"nid":"MIA","count":2},{"nid":"MID","count":1},{"nid":"MIT","count":10},{"nid":"MIO","count":3},{"nid":"MIS","count":2},{"nid":"MIH","count":1}]现在,我对如何使用关键元素的count重新排列数组感到困惑。如有任何指导或帮助,将不胜感激!
发布于 2018-02-22 03:49:02
你可以试试.sort
let arr = [ {"nid":"MIA","keys":[{"sid":"sm1"},{"sid":"sm2"}]},
{"nid":"MID","keys":[{"sid":"sm1"}]},
{"nid":"MIT","keys":[{"sid":"sm1"},{"sid":"sm2"},{"sid":"sm3"},{"sid":"sm4"},{"sid":"sm5"},{"sid":"sm6"},{"sid":"sm7"},{"sid":"sm8"},{"sid":"sm9"},{"sid":"sm10"}]},
{"nid":"MIO","keys":[{"sid":"sm1"},{"sid":"sm2"},{"sid":"sm3"}]},
{"nid":"MIS","keys":[{"sid":"sm1"},{"sid":"sm2"}]},
{"nid":"MIH","keys":[{"sid":"sm1"}]} ];
arr.sort((a, b) => a.keys.length - b.keys.length);
console.log(arr);
发布于 2018-02-22 03:59:57
您可以通过使用数组的排序函数来实现这一点。
arr.sort(function(a,b){return a.keys.length - b.keys.length})https://stackoverflow.com/questions/48919285
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