查找对象键名并将对象添加到匹配数组

Question

我有这个数组：const array = [  {    '12345': [      { name: 'item one', numbers: ['12345', '77484'] },      { name: 'item two', numbers: ['12345', '65456'] },      { name: 'item three', numbers: ['12345', '33920'] },      { name: 'item four', numbers: ['12345', '99393'] }    ],    '67890': [      { name: 'item one b', numbers: ['67890', '33232'] },      { name: 'item two b', numbers: ['67890', '33456'] },      { name: 'item three b', numbers: ['67890', '77665'] },      { name: 'item four b', numbers: ['67890', '11234'] }    ]  }]这个对象：const newItem = { name: 'It is a new item', numbers: ['12345'] }正如您在数组中看到的，它包含一个对象，该对象包含两个对象数组。这些嵌套数组中的每一个的键都是一个数字，正如您所看到的，键号也包含在每个项目的数字数组中，例如：{ name: 'item one', numbers: ['12345', '77484'] },That 包含其数字数组中的键'12345'，因此'12345'是该对象的父级。我想获取 的编号newItem，将其与父键匹配并将其添加到与匹配键父项对应的数组中。请考虑到这是虚拟数据，并想象其中的数字数组newItem是一个动态值，嵌套数组有很多值。这里的想法是匹配这两个数字以知道应该将 newItem 推到哪里。

Answer 1

试试看：

array.map((row) => {

  Object.keys(row).map((value) => {

    if (newItem.numbers.indexOf(value) >= 0) {

      row[value].push(newItem);

    }

  });

});

Answer 2

你可以这样做：

    for (let i=0; i<array.length; i++) {

        array[i][newItem.numbers[0]] = array[i][newItem.numbers[0]] || [];

        array[i][newItem.numbers[0]].push(newItem);

    }

如果嵌套数组不存在，第一行初始化它。第二行将新项目推送到嵌套数组。

请参阅下面的代码片段。

const array = [

    {

        '12345': [

            { name: 'item one', numbers: ['12345', '77484'] },

            { name: 'item two', numbers: ['12345', '65456'] },

            { name: 'item three', numbers: ['12345', '33920'] },

            { name: 'item four', numbers: ['12345', '99393'] },

        ],

        '67890': [

            { name: 'item one b', numbers: ['67890', '33232'] },

            { name: 'item two b', numbers: ['67890', '33456'] },

            { name: 'item three b', numbers: ['67890', '77665'] },

            { name: 'item four b', numbers: ['67890', '11234'] },

        ],

    }

]

const newItem = {

    name: 'It is a new item',

    numbers: ['12345'],

}

for (let i=0; i<array.length; i++) {

    array[i][newItem.numbers[0]] = array[i][newItem.numbers[0]] || [];

    array[i][newItem.numbers[0]].push(newItem);

}

console.log(array); 

Answer 3

在解构的帮助下。

const [obj] = array;

const {

  numbers: [key],

} = newItem;

obj[key].push(newItem);

const array = [

  {

    "12345": [

      {

        name: "item one",

        numbers: ["12345", "77484"],

      },

      {

        name: "item two",

        numbers: ["12345", "65456"],

      },

      {

        name: "item three",

        numbers: ["12345", "33920"],

      },

      {

        name: "item four",

        numbers: ["12345", "99393"],

      },

    ],

    "67890": [

      {

        name: "item one b",

        numbers: ["67890", "33232"],

      },

      {

        name: "item two b",

        numbers: ["67890", "33456"],

      },

      {

        name: "item three b",

        numbers: ["67890", "77665"],

      },

      {

        name: "item four b",

        numbers: ["67890", "11234"],

      },

    ],

  },

];

const newItem = {

  name: "It is a new item",

  numbers: ["12345"],

};

const [obj] = array;

const {

  numbers: [key],

} = newItem;

obj[key].push(newItem);

console.log(array);

Answer 4

const array = [

  {

    '12345': [

      { name: 'item one', numbers: ['12345', '77484'] },

      { name: 'item two', numbers: ['12345', '65456'] },

      { name: 'item three', numbers: ['12345', '33920'] },

      { name: 'item four', numbers: ['12345', '99393'] }

    ],

    '67890': [

      { name: 'item one b', numbers: ['67890', '33232'] },

      { name: 'item two b', numbers: ['67890', '33456'] },

      { name: 'item three b', numbers: ['67890', '77665'] },

      { name: 'item four b', numbers: ['67890', '11234'] }

    ]

  }

]

const newItem = { name: 'It is a new item', numbers: ['12345'] }

newItem.numbers.forEach(n => array[0]?.[n].push(newItem));

console.log(array);