javascript - 在两个不同的json数组中合并相同的特定json对象

我想在依赖于json数据ID的两个不同的json数组上合并相同的特定json对象。

JSON数据集1

{
    "Product":[
                {
                "product_id": "123",
                "location_id": "222",
                "product_code": "abc",
                },
                {
                "product_id": "456",
                "location_id": "111",
                "product_code": "xyz",
                }
            ]
}

JSON数据集2

{
    "Location":[

            {
                "location_id": 111,
                "location_name": "alpha"
            },
            {
                "location_id": 222,
                "location_name": "tango"
            }


        ]
}

结果将是这样的

{
    "Product":[

                {
                    "product_id": "456",
                    "location_id": "111",
                    "product_code": "xyz",
                    "location_name": "alpha"
                },
                {
                    "product_id": "123",
                    "location_id": "222",
                    "product_code": "abc",
                    "location_name": "tango"
                }
            ]
}

到目前为止，这是我完成的代码。

var finalJson = {};

                    _.each(_.keys(productArray,locationArray), function(key) {
                        finalJson[key] = _.flatten(_.zip(productArray[key], locationArray[key]));
                    });

                    console.log(finalJson);

最佳答案

一个简单的算法可以使用嵌套循环遍历两个数组，如下所示：

let allProducts = [{
    "product_id": "123",
    "location_id": "222",
    "product_code": "abc",
  },
  {
    "product_id": "456",
    "location_id": "111",
    "product_code": "xyz",
  }
];


let allLocations = [

  {
    "location_id": 111,
    "location_name": "alpha"
  },
  {
    "location_id": 222,
    "location_name": "tango"
  }
];

let result = allProducts.map((product) => {
  let matchingLocation = allLocations.find((location) => {
    return location.location_id == product.location_id;
  });
  return Object.assign(product, matchingLocation);
})

console.log(result);