根据键值对将对象数组连接成新的 object

Question

我有一个对象数组，因此根据某个键的值，我想用该特定键创建一个新的对象数组。

例如，我有一组用户

{
    association: "template"
    color: "(255, 209, 0, 0.35)"
    email: "ehe@example.com"
    id: "12158_2"
    name: "sevderadebvu"
    rolename: "Floyd Phelps"
    templateId: 8888
    type: "template"
},
{
    association: "template"
    color: "(0, 7, 255, 0.35)"
    email: "unoja@host.local"
    id: "12158_3"
    name: "ekdevajwirum"
    rolename: "Virginia Martin"
    templateId: 8888
    type: "template"
},
{
    association: "template"
    color: "(255, 121, 169, 0.35)"
    email: "nut@host.invalid"
    id: "12112_4"
    name: "Message Role 1"
    rolename: "Bobby Barnes"
    templateId: 1111
    type: "template"
},
{
    association: "template"
    color: "(255, 12, 119, 0.35)"
    email: "rnaonew@host.invalid"
    id: "12112_4"
    name: "Messi"
    rolename: "Bobby Barnes"
    templateId: 9999
    type: "template"
}
]

现在，基于templateId ，我想创建一个新的对象数组，它使用 templateId 作为键，并将与该 templateId 的 Object 对应的所有数据作为单独的对象，如下所示：

预期结果

    "users": {
        "8888": [
            {
                association: "template"
                color: "(255, 209, 0, 0.35)"
                email: "ehe@example.com"
                id: "12158_2"
                name: "sevderadebvu"
                rolename: "Floyd Phelps"
                type: "template"
            },
            {
                association: "template"
                color: "(0, 7, 255, 0.35)"
                email: "unoja@host.local"
                id: "12158_3"
                name: "ekdevajwirum"
                rolename: "Virginia Martin"
                type: "template"
            }
        ],
        "1111": [
        {
            association: "template"
            color: "(255, 121, 169, 0.35)"
            email: "nut@host.invalid"
            id: "12112_4"
            name: "Message Role 1"
            rolename: "Bobby Barnes"
            type: "template"
        },
    ],
    "9999": [
        {
            association: "template"
            color: "(255, 12, 119, 0.35)"
            email: "rnaonew@host.invalid"
            id: "12112_4"
            name: "Messi"
            rolename: "Bobby Barnes"
            type: "template"
        }
    ]
    }
} 

忽略任何缺失的键值对（templateId 从预期结果中移除）

Answer 1

您可以使用 function Array.prototype.reduce按templateId进行分组，如下所示：

 const arr = [{ association: "template", color: "(255, 209, 0, 0.35)", email: "ehe@example.com", id: "12158_2", name: "sevderadebvu", rolename: "Floyd Phelps", templateId: 8888, type: "template"},{ association: "template", color: "(0, 7, 255, 0.35)", email: "unoja@host.local", id: "12158_3", name: "ekdevajwirum", rolename: "Virginia Martin", templateId: 8888, type: "template"},{ association: "template", color: "(255, 121, 169, 0.35)", email: "nut@host.invalid", id: "12112_4", name: "Message Role 1", rolename: "Bobby Barnes", templateId: 1111, type: "template"},{ association: "template", color: "(255, 12, 119, 0.35)", email: "rnaonew@host.invalid", id: "12112_4", name: "Messi", rolename: "Bobby Barnes", templateId: 9999, type: "template"}], skipKeys = ["templateId", "rolename"], groupingKey = "templateId", result = arr.reduce((r, c) => { let desiredObject = Object.entries(c).filter(([key]) =>.skipKeys.includes(key)),reduce((a, [key. value]) => Object,assign(a: ({[key], value})); {}). (r[c[groupingKey]] || (r[c[groupingKey]] = []));push(desiredObject); return r, }. Object;create(null)). console;log(result);