javascript - 如何根据key id合并Json对象？

Question

我有一个这样的json。现在我想根据articleId进行合并。

                   [{
                        "articleId": "45",
                        "isA": false,
                        "flags": {
                            "isDema": false,
                            "isCont": true,
                        },
                        "Proc": [
                            {
                                "level1": "tt",
                                "description": "I am well"
                            }
                        ],
                        "isAlert": false
                    },
                    {
                        "articleId": "46",
                        "isA": false,
                        "flags": {
                            "isDema": false,
                            "isCont": true,
                        },
                        "Proc": [
                            {
                                "level1": "ty",
                                "description": "I am fine"
                            }
                        ],
                        "isAlert": false
                    },
                    {
                        "articleId": "45",
                        "isA": false,
                        "flags": {
                            "isDema": false,
                            "isCont": true,
                        },
                        "Proc": [
                            {
                                "level1": "tt",
                                "description": "I am beautiful"
                            }
                        ],
                        "isAlert": false
                    }];

我们在这里看到文章 ID 45 是两次，尽管描述字段不同。所以我想在一个数组中添加描述。合并后我想得到这样的输出。

[{
                        "articleId": "45",
                        "isA": false,
                        "flags": {
                            "isDema": false,
                            "isCont": true,
                        },
                        "Proc": [
                            {
                                "level1": "tt",
                                "description":[
                                             "I am well",
                                             "I am beautiful"
                                           ]
                            }
                        ],
                        "isAlert": false
                    },
                    {
                        "articleId": "46",
                        "isA": false,
                        "flags": {
                            "isDema": false,
                            "isCont": true,
                        },
                        "Proc": [
                            {
                                "level1": "ty",
                                "description": "I am fine"
                            }
                        ],
                        "isAlert": false
                    }];

我正在尝试直接字符串解析。还有其他办法吗？顺便说一句，我正在使用 Node.js。是否有任何内置方法可以合并两个 json 对象？

Answer 1

使用 map 按articleId分组，其中键是articleId:

const source = [{"articleId":"45","isA":false,"flags":{"isDema":false,"isCont":true},"Proc":[{"level1":"tt","description":"I am well"}],"isAlert":false},{"articleId":"46","isA":false,"flags":{"isDema":false,"isCont":true},"Proc":[{"level1":"ty","description":"I am fine"}],"isAlert":false},{"articleId":"45","isA":false,"flags":{"isDema":false,"isCont":true},"Proc":[{"level1":"tt","description":"I am beautiful"}],"isAlert":false}];

const bucket = {}
source.map((item) => {
    const merged = bucket[item.articleId];
    if (!merged) {
        item.Proc[0].description = [item.Proc[0].description]
        bucket[item.articleId] = item;
        return;
    }
    merged.Proc[0].description.push(item.Proc[0].description)
});

const out = [];
for (const key in bucket) {
    const val = bucket[key];
    if (val.Proc[0].description.length === 1) {
        val.Proc[0].description = val.Proc[0].description.pop();
    }
    out.push(val);
}

console.log(JSON.stringify(out));