从对象中动态删除多个键/值对

Question

假设我有一个对象数组。我不想要某些键/值。删除一个键/值对的传统方法是使用delete，如下所示：

for (var i = 0; i < tracks.length; i++) {
  delete tracks[i]["currency"];
  ...
}

我要拉入的物品可能有 30 多对。有没有办法可以说明我想要哪些配对并删除所有其他配对？因此，例如在这个对象数组中，我只想保留trackName、kind、price

var tracks = [{
    trackNumber: "01",
    trackName: "Track 1",
    trackDuration: "5:35",
    kind: "song",
    currency: "USD",
    price: 1.29
}, {
    trackNumber: "02",
    trackName: "Track 2",
    trackDuration: "5:15",
    kind: "song",
    currency: "USD",
    price: 1.29
}, {
    trackNumber: "03",
    trackName: "Track 3",
    trackDuration: "5:07",
    kind: "song",
    currency: "USD",
    price: 1.29
}, {
    trackNumber: "04",
    trackName: "Track 4",
    trackDuration: "0:16",
    kind: "song",
    currency: "USD",
    price: 1.29
}, {
    trackNumber: "05",
    trackName: "Track 5",
    trackDuration: "5:35",
    kind: "song",
    currency: "USD",
    price: 1.29
}];

Answer 1

我能提供的最简单的解决方案：

/**
 * Returns new collection which items contains only specified properties
 * @param {Array} collection of items
 * @param {Array} properties to keep
 */

function keepCollectionProperties(collection, properties) {
    return collection.map(function(item) {
        var newItem = {}
        for(var i = 0, prop; prop = properties[i]; i++)
            if (typeof item[prop] !== 'undefined')
                newItem[prop] = item[prop]
        return newItem
    })
}

var tracks = [{
    trackNumber: "01",
    trackName: "Track 1",
    trackDuration: "5:35",
    kind: "song",
    currency: "USD",
    price: 1.29
}, {
    trackNumber: "02",
    trackName: "Track 2",
    trackDuration: "5:15",
    kind: "song",
    currency: "USD",
    price: 1.29
}, {
    trackNumber: "03",
    trackName: "Track 3",
    trackDuration: "5:07",
    kind: "song",
    currency: "USD",
    price: 1.29
}, {
    trackNumber: "04",
    trackName: "Track 4",
    trackDuration: "0:16",
    kind: "song",
    currency: "USD",
    price: 1.29
}, {
    trackNumber: "05",
    trackName: "Track 5",
    trackDuration: "5:35",
    kind: "song",
    currency: "USD",
    price: 1.29
}];

var output = keepCollectionProperties(tracks, ['trackName', 'kind', 'price'])
document.write(JSON.stringify(output))

注意：避免使用delete，它会改变对象的“形状”并损害性能。尽可能使用item.prop = null。我的函数返回新集合，所以没有必要，但如果您打算再次填充新集合，最好将不需要的道具设置为null。

Answer 2

这是我的建议，因为我觉得它更优雅：

// create new array of objects from old one
var fixed = jQuery.map(tracks, function(element, index){
    return {"trackName": element.trackName, "kind": element.kind, "price": element.price};
}); 
// delete old array   
tracks = null;

Answer 3

不确定这是否是您想要的，但它会遍历列表，并返回仅包含给定键的相同列表。 keys 是一个键数组。

function LimitArray(keys, list){
    var returnArr = [];
    for (var i in list){
       var row = {};
       for (key in list[i]){
          row[key] = list[i][key];
       }
       returnArr.push(row);
    }
    return returArr;
}

你也可以用其他几种方式来修改原始列表，但我不确定这是否是你特别想要的，所以我只是返回了一个新列表。

如果你想改变原来的，你可以这样做：

var select = ["a","b"]
list.map(function(val,index, array)
   {
   var keys = Object.keys(val);
   for (var key in keys){
      if(-1 == select.indexOf(keys[key]])  delete val[keys[key]];
   }}.bind(this));

我相信也应该做需要做的事情。它将遍历列表，并删除所有不在选择中的键。

Answer 4

一个自我描述的解决方案：

// what you want to keep:
var want_to_keep = ['trackName', 'kind', 'price'];

// loop trough every main element of tracks:
for (var i=0; i<tracks.length; i++)
    // now go trough every element of the current track:
    for (var key in tracks[i])
        // check if the current track key is wanted:
        if (want_to_keep.indexOf(key) < 0)
            delete tracks[i][el];

Answer 5

遍历数组并从每个对象中保留您想要的内容。

var keep = ['trackName', 'kind', 'price'];

for(var i = 0;i < tracks.length; i++){

    for(var key in tracks[i]){
        if(keep.indexOf(key) === -1)delete tracks[i][key];
    }

}

Answer 6

你介意使用下划线 ".pick(object, *keys) " , ".omit(object, *keys) " 并将新对象推入新数组。