JavaScript删除对象数组中的重复对象&对属性求和

JavaScript to remove duplicate objects in an Object Array & sum the properties

本文关键字：对象 属性 求和 删除 数组 JavaScript      更新时间：2023-09-26

我已经尝试了几天，现在想出一个解决我的问题，我只是不能，想象我有以下JSON数组(我们将其称为jsonData: 

[
  { "id": 118748, "price":"", "stocklevel": 100,   "instock": false, "pname": "Apple TV" },
  { "id": 118805291, "price":"", "stocklevel": 432,   "instock": true, "pname": "Hitachi TV"},
  { "id": 118801891, "price":"", "stocklevel": 0,   "instock": false, "pname": "Sony TV" },
  { "id": 118748, "price":"", "stocklevel": 2345,   "instock": true, "pname": "Apple TV"},
...

现在我的JSON数组中可能有超过100个项目，我想删除具有重复id的项目，但要求和库存水平并保留数组中的顺序，因此一行应该取代该id最近出现的位置。在上面的JSON中，具有"id":118748的对象的第一个实例被删除，但它的库存水平值传递/添加了具有相同id的对象的下一个实例，因此JSON数组看起来像这样:

[
  { "id": 118805291, "price":"", "stocklevel": 432,   "instock": true, "pname": "Hitachi TV"},
  { "id": 118801891, "price":"", "stocklevel": 0,   "instock": false, "pname": "Sony TV" },
  { "id": 118748, "price":"", "stocklevel": 2445,   "instock": true, "pname": "Apple TV"},
...

我生成了以下代码来删除重复项，但我无法对库存水平总数求和，下面是我的代码:

function idsAreEqual(obj1, obj2) {
    return obj1.id === obj2.id;
}
function arrayContains(arr, val, equals) {
    var i = arr.length;
    while (i--) {
        if (equals(arr[i], val)) {
            return true;
        }
    }
    return false;
}
function removeDups(arr, equals) {
    var originalArr = arr.slice(0);
    var i, k, len, val;
    arr.length = 0;
    for (i = originalArr.length - 1, len = originalArr.length, k  = originalArr.length - 1 ; i > 0; --i) {
        val = originalArr[i];
        if (!arrayContains(arr, val, equals)) {
            arr.push(val);
        }
    }
}

removeDups(jsonData, idsAreEqual);
jsonData.reverse();
有人能帮我解决这个问题吗?请注意，我不能使用下划线，jQuery或任何其他库。

提前致谢

您可以做如下操作，我认为这比您的实现要简单一些。我也使用forEach和reduce，这是ES5，但应该在现代浏览器上工作。如果你使用的是旧的、非es5兼容的浏览器，你可以抓取一个polyfil:

var data = [
  { "id": 118748, "price":"", "stocklevel": 0,   "instock": false, "pname": "Apple TV" },
  { "id": 118805291, "price":"", "stocklevel": 432,   "instock": true, "pname": "Hitachi TV"},
  { "id": 118801891, "price":"", "stocklevel": 0,   "instock": false, "pname": "Sony TV" },
  { "id": 118748, "price":"", "stocklevel": 2345,   "instock": true, "pname": "Apple TV"}
];
function dedup_and_sum(arr, prop) {
    var seen = {},
        order = [];
    arr.forEach(function(o) {
        var id = o[prop];
        if (id in seen) {
            // keep running sum of stocklevel
            var stocklevel = seen[id].stocklevel + o.stocklevel
            // keep this newest record's values
            seen[id] = o;  
            // upid[118805291], stocklevel=432, instock=truedate stocklevel to our running total
            seen[id].stocklevel = stocklevel;
            // keep track of ordering, having seen again, push to end
            order.push(order.splice(order.indexOf(id), 1));
        }
        else {
            seen[id] = o;
            order.push(id);
        }
    });
    return order.map(function(k) { return seen[k]; });
}
// Get unique records, keeping last record of dups
// and summing stocklevel as we go
var unique = dedup_and_sum(data, 'id');
unique.forEach(function(o) {
    console.log("id[%d], stocklevel=%d, instock=%s", o.id, o.stocklevel, o.instock);
});
// output =>
// id[118805291], stocklevel=432, instock=true
// id[118801891], stocklevel=0, instock=false 
// id[118748], stocklevel=2445, instock=true

EDIT已更新以符合我们在注释中讨论的要求。

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