基于NOT关键字过滤对象数组并提取有效ID的实现需求

首先给出你的过滤规则和原始数据：

过滤规则 #

const array = [
  { DEVICE_SIZE: ['036', '048', '060', '070'] },
  { DEVICE_VOLTAGE: ['1', '3'] },
  { 'NOT DEVICE_DISCHARGE_AIR': ['S'] },
  { 'NOT DEVICE_REFRIGERANT_CIRCUIT': ['H', 'C'] },
];

原始数据 #

const data = {
  DEVICE_SIZE: [
    { id: 20, name: 'Size 20' },
    { id: 36, name: 'Size 36' },
    { id: 40, name: 'Size 40' },
    { id: 48, name: 'Size 48' }, // 修正原数据笔误：原id为20，应为48才能匹配规则
    { id: 60, name: 'Size 60' },
    { id: 70, name: 'Size 70' },
  ],
  DEVICE_VOLTAGE: [
    { id: 1, name: 'Voltage 1' },
    { id: 2, name: 'Voltage 2' },
    { id: 3, name: 'Voltage 3' },
    { id: 4, name: 'Voltage 4' },
    { id: 5, name: 'Voltage 5' },
  ],
  DEVICE_DISCHARGE_AIR: [
    { id: 'E', name: 'Discharge E' },
    { id: 'S', name: 'Discharge S' },
    { id: 'T', name: 'Discharge T' },
  ],
  DEVICE_REFRIGERANT_CIRCUIT: [
    { id: 'C', name: 'Refrigerant C' },
    { id: 'E', name: 'Refrigerant E' },
    { id: 'H', name: 'Refrigerant H' },
    { id: 'M', name: 'Refrigerant M' },
  ],
};

实现代码 #

直接遍历规则数组，区分正向/反向匹配，处理类型统一和键名映射：

const valid = {};

array.forEach(rule => {
  // 提取当前规则的键和值列表
  const [ruleKey, ruleValues] = Object.entries(rule)[0];
  
  // 判断是否为反向规则，获取对应的数据键名
  const isReverse = ruleKey.startsWith('NOT ');
  const dataKey = isReverse ? ruleKey.slice(4) : ruleKey;
  
  // 统一规则值的类型：将带前导零的字符串转为数字，避免类型不匹配
  const normalizedRules = ruleValues.map(val => {
    const num = Number(val);
    return isNaN(num) ? val : num;
  });
  
  // 过滤出符合条件的ID
  const filteredIds = data[dataKey]
    .map(item => item.id)
    .filter(id => isReverse ? !normalizedRules.includes(id) : normalizedRules.includes(id));
  
  // 处理输出键名（匹配你期望的DEVICE_DISCHARGE）
  let outputKey = dataKey;
  if (outputKey === 'DEVICE_DISCHARGE_AIR') {
    outputKey = 'DEVICE_DISCHARGE';
  }
  
  valid[outputKey] = filteredIds;
});

console.log(valid);

输出结果 #

运行后会得到你期望的有效ID对象：

const valid = {
  DEVICE_SIZE: [36, 48, 60, 70],
  DEVICE_VOLTAGE: [1, 3],
  DEVICE_DISCHARGE: ["E", "T"],
  DEVICE_REFRIGERANT_CIRCUIT: ["E", "M"],
};

内容的提问来源于stack exchange，提问作者Hello World