Angular8中如何根据匹配条件替换数组对象值

云上老周

2026-8-25

合并数组：匹配双字段替换对象，保留原数组无匹配项

需求说明

将oldArray与newArray合并，规则如下：

当两个数组中的对象**同时匹配makeLineName和makeProcessTypeId**时，用newArray的对应对象替换oldArray中的对象；
若newArray中没有匹配makeLineName（结合makeProcessTypeId）的对象，则保留oldArray的原对象。

示例数据

const oldArray = [
  {       
    makeLineName: "TestDemo1",
    makeProcessTypeId: "type1",
    avtBCT: 80,
    MaxBCT: 80
  },
  {      
    makeLineName: "Test565",
    makeProcessTypeId: "type2",
    avtBCT: '',
    MaxBCT: ''
  },
  {      
    makeLineName: "Luck", 
    makeProcessTypeId: "type3",
    avtBCT: 60,
    MaxBCT: 60
  }
];

const newArray = [
  {       
    makeLineName: "TestDemo1",
    makeProcessTypeId: "type1",
    avtBCT: 500,
    MaxBCT: 500
  },
  {      
    makeLineName: "Test565",
    makeProcessTypeId: "type2",
    avtBCT: 600,
    MaxBCT: 600
  }
];

解决方案

通过Map构建高效查找表，再遍历原数组完成替换，时间复杂度为O(n)，适合处理大数据量：

// 构建newArray的映射表，用双字段组合作为唯一匹配键
const newItemMap = new Map();
newArray.forEach(item => {
  const matchKey = `${item.makeLineName}-${item.makeProcessTypeId}`;
  newItemMap.set(matchKey, item);
});

// 遍历oldArray，替换匹配项，保留无匹配元素
const filteredData = oldArray.map(oldItem => {
  const matchKey = `${oldItem.makeLineName}-${oldItem.makeProcessTypeId}`;
  return newItemMap.get(matchKey) || oldItem;
});

预期输出

const filteredData = [
  {       
    makeLineName: "TestDemo1",
    makeProcessTypeId: "type1",
    avtBCT: 500,
    MaxBCT: 500
  },
  {      
    makeLineName: "Test565",
    makeProcessTypeId: "type2",
    avtBCT: 600,
    MaxBCT: 600
  },
  {      
    makeLineName: "Luck", 
    makeProcessTypeId: "type3",
    avtBCT: 60,
    MaxBCT: 60
  }
];