1

我正在尝试过滤重复值并将唯一值作为对象数组获取。我不知道如何根据颜色获得唯一值。所以下面是我的数据：

[
    {
        "code": "xxxx1",
     
        "priceData": {
            "currencyIso": "USD",
            "value": 649.99
        },
 
        "variants": [
            {
                "color": "#212028 |Black",
            }
        ]
    },
    {
        "code": "xx2",
        "priceData": {
            "currencyIso": "USD",
            "value": 999.99
        },

        "variants": [
            {
                "color": "#212028 |Black",
            },
        ]
    },
    {
        "code": "xx3",
        "priceData": {
            "currencyIso": "USD",
            "value": 549.99
        },
        "variants": [
            {
                "color": "#D3CCC1 |Silver",
            },
           
        ]
    },
    {
        "code": "xxx-4",

        "priceData": {
            "currencyIso": "USD",
            "value": 649.99
        },
   
        "variants": [
            {
                "color": "#D3CCC1 |Silver",
               
            }
        ]
    }
]

期望值为：

[
    {
        "code": "xxxx1",
     
        "priceData": {
            "currencyIso": "USD",
            "value": 649.99
        },
 
        "variants": [
            {
                "color": "#212028 |Black",
            }
        ]
    },
    {
        "code": "xx3",
        "priceData": {
            "currencyIso": "USD",
            "value": 549.99
        },
        "variants": [
            {
                "color": "#D3CCC1 |Silver",
            },
           
        ]
    },
   
]

下面的代码仅返回对象的变量数组。但我希望作为我上面提到的预期结果

  let variants2 = Array.from(
            new Set(
              variants.map(
                (a) => a.variants
              )
            )
          ).map((variants) => {
            return variants.find((a) => a.color=== a.color);
          });
          [
            ...new Map(variants2.map((item) => [item.value, item])).values(),
          ];

有人可以帮我解决这个问题吗？

4

1 回答 1

3

许多可能的方法之一（输入数组称为items）：

Array.from(new Map(items.reverse().map(item => [item.variants[0].color, item])).values()).reverse();

测试：

const items = [
    {
        "code": "xxxx1",
     
        "priceData": {
            "currencyIso": "USD",
            "value": 649.99
        },
 
        "variants": [
            {
                "color": "#212028 |Black",
            }
        ]
    },
    {
        "code": "xx2",
        "priceData": {
            "currencyIso": "USD",
            "value": 999.99
        },

        "variants": [
            {
                "color": "#212028 |Black",
            },
        ]
    },
    {
        "code": "xx3",
        "priceData": {
            "currencyIso": "USD",
            "value": 549.99
        },
        "variants": [
            {
                "color": "#D3CCC1 |Silver",
            },
           
        ]
    },
    {
        "code": "xxx-4",

        "priceData": {
            "currencyIso": "USD",
            "value": 649.99
        },
   
        "variants": [
            {
                "color": "#D3CCC1 |Silver",
               
            }
        ]
    }
];
const items2 = Array.from(new Map(items.reverse().map(item => [item.variants[0].color, item])).values()).reverse();
console.log(items2);

请注意，items.reverse()将原始数组反转到位。如果您需要其原始订单，则必须items.reverse()再次调用（创建后items2）。

对于每种独特的颜色，这种方法会将第一个对象放入结果中，如您的示例所示。如果获取最后一个对象也可以，则可以.reverse()从上述代码中删除这两个调用：

Array.from(new Map(items.map(item => [item.variants[0].color, item])).values());
于 2021-08-03T19:07:29.823 回答