如何在Javascript数组中合并/推送重复值[关闭]答案

【问题标题】：How to merge/push duplicate value in Javascript Array [closed]如何在Javascript数组中合并/推送重复值[关闭]
【发布时间】：2018-07-01 23:20:11
【问题描述】：

我有一个这样的数组对象：

[
  {"item_id":1,"name":"DOTA 2 Backpack","image":"XXX","qty":1,"original_price":1450000,"total_price":1450000}, 
  {"item_id":2,"name":"Mobile Legend Backpack","image":"XXX","qty":1,"original_price":1350000,"total_price":1350000}
]

当我添加这个具有相同值的数组时，会是这样的：

[
  {"item_id":1,"name":"DOTA 2 Backpack","image":"XXX","qty":1,"original_price":1450000,"total_price":1450000},
  {"item_id":2,"name":"Mobile Legend Backpack","image":"XXX","qty":1,"original_price":1350000,"total_price":1350000},
  {"item_id":2,"name":"Mobile Legend Backpack","image":"XXX","qty":1,"original_price":1350000,"total_price":1350000}
]

我的目标想要相同的值会这样合并：

    [
  {"item_id":1,"name":"DOTA 2 Backpack","image":"XXX","qty":1,"original_price":1450000,"total_price":1450000},
  {"item_id":2,"name":"Mobile Legend Backpack","image":"XXX","qty":2,"original_price":1350000,"total_price":2700000}
]

我很困惑并坚持下去。请有人帮助我。

【问题讨论】：

你在使用 lodash 吗？ @Axax
不，顺便说一句，我是 Javascript 的新手。
不，Item_id 不同。示例：Dota 2 -> item_id : 1 Mobile Legend -> item_id : 2
不，X 是假的。
你的“合并”代码是什么？

标签： javascript arrays object vue.js duplicates

【解决方案1】：

您可以使用Array.prototype.reduce().. 在代码示例中，我正在与元素的item_id进行比较。

代码：

const array = [{"item_id": 1,"name":"DOTA 2 Backpack","image":"XXX","qty":1,"original_price":1450000,"total_price":1450000}, {"item_id":2,"name":"Mobile Legend Backpack","image":"XXX","qty":1,"original_price":1350000,"total_price":1350000}];
const newVal = {"item_id": 2,"name":"Mobile Legend Backpack","image":"XXX","qty":1,"original_price":1350000,"total_price":1350000};

const addElement = (arr, val) => {
  return arr.reduce((a, c, i) => {
    if (c['item_id'] === val['item_id']) {
      a.found = true;
      c.qty += val.qty;
    }
    
    a.result.push(c);
    
    return !a.found && i + 1 === arr.length
      ? [].concat(arr.result, val)
      : a;
  }, {result: [], found: false});
};


console.log(addElement(array, newVal).result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

【讨论】：

非常感谢。解决了！ :)

【解决方案2】：

看看这个。或许能帮到你。

您可以使用Array Iteration methods

var array1 = [
  {"item_id":X,"name":"DOTA 2 Backpack","image":"XXX","qty":1,"original_price":1450000,"total_price":1450000}, 
  {"item_id":X,"name":"Mobile Legend Backpack","image":"XXX","qty":1,"original_price":1350000,"total_price":1350000}
];

var array2 = [
  {"item_id":X,"name":"DOTA 2 Backpack","image":"XXX","qty":1,"original_price":1450000,"total_price":1450000},
  {"item_id":X,"name":"Mobile Legend Backpack","image":"XXX","qty":1,"original_price":1350000,"total_price":1350000},
  {"item_id":X,"name":"Mobile Legend Backpack","image":"XXX","qty":1,"original_price":1350000,"total_price":1350000}
]

var props = ['name', 'item_id'];

var result = array1.filter(function(o1){

    return !array2.some(function(o2){
        return o1.item_id=== o2.item_id;         
    });
}).map(function(o){

    return props.reduce(function(newo, name){
        newo[name] = o[name];
        return newo;
    }, {});
});

输出：

[
  {"item_id":X,"name":"DOTA 2 Backpack","image":"XXX","qty":1,"original_price":1450000,"total_price":1450000},
  {"item_id":X,"name":"Mobile Legend Backpack","image":"XXX","qty":1,"original_price":1350000,"total_price":1350000}

]

希望对您有所帮助..！

【讨论】：

你读过这个问题吗？这不能回答问题，也不能解决 OP 问题。
非常感谢。解决了！ :)
@Axaxaxaxax 如果解决了。标记一下。谢谢

【解决方案3】：

如果您想将新的object 推送到array，如果它不存在并更新它，如果它存在：

您需要先使用Array#some() method 检查它是否存在。
然后使用found 索引更新其qty 和total_price。
或者如果它不存在，只需推送它。

这应该是你的代码：

var found = -1;
if (arr.some(function(el, i) {
    if (el.name === obj.name)
      found = i;
    return el.name === obj.name;
  })) {
  arr[found]["qty"] = arr[found]["qty"] + obj["qty"];
  arr[found]["total_price"] = arr[found]["total_price"] + obj["total_price"];
} else {
  arr.push(obj);
}

演示：

var arr = [{
    "item_id": 'X',
    "name": "DOTA 2 Backpack",
    "image": "XXX",
    "qty": 1,
    "original_price": 1450000,
    "total_price": 1450000
  },
  {
    "item_id": 'X',
    "name": "Mobile Legend Backpack",
    "image": "XXX",
    "qty": 1,
    "original_price": 1350000,
    "total_price": 1350000
  }
];

var obj = {
  "item_id": 'X',
  "name": "Mobile Legend Backpack",
  "image": "XXX",
  "qty": 1,
  "original_price": 1350000,
  "total_price": 1350000
};
var found = -1;
if (arr.some(function(el, i) {
    if (el.name === obj.name)
      found = i;
    return el.name === obj.name;
  })) {
  arr[found]["qty"] = arr[found]["qty"] + obj["qty"];
  arr[found]["total_price"] = arr[found]["total_price"] + obj["total_price"];
} else {
  arr.push(obj);
}

console.log(arr);

【讨论】：

非常感谢。解决了！ :)
@Axaxaxaxax 欢迎您，很高兴它有帮助。如果对您有帮助，您可以接受。

【解决方案4】：

if (items.indexOf(toAdd) != -1)
   items.push(toAdd)

或者你可以使用 jQuery grep，它返回一个包含请求值的数组列表

if  (($.grep(items, function(e){ return e.id == toAdd.id; })) == 0)
   items.push(toAdd)

【讨论】：

你知道我的问题吗？