数组:

 array1 = [{id:1, name:"raju"},{id:2, name:"ravi"},{id:4, name:"john"},{id:6, name:"jack"}];
 array2= [{id:1, degree:"b.com"},{id:3, degree:"b.a"},{id:4, degree:"c.a"},{id:5, degree:"horticulture"}];
 array3= [{id:1, age:20},{id:3, age:21},{id:6, age:27},{id:7, age:25}];

所需结果是:
resultarray = [
    {id:1, name: "raju", degree:"b.com",age:20},
    {id:2, name: "ravi"},
    {id:3, degree:"b.a", age:21},
    {id:4, name:"john", degree:"c.a"},
    {id:5, degree:"horticulture"},
    {id:6, name:"jack", age:27},
    {id:7, age:25}
 ]

我尝试了不同的功能,并尝试了两个数组,但无法合并没有ID进行比较的对象。

最佳答案

您可以使用reducedestructuring

这里的主意是

  • 首先将所有数组合并到一个数组中。
  • 现在使用reduce我们在id对象
  • 中创建op作为键
  • 如果已经有id密钥,我们将inpop[inp.id]合并
  • 如果没有id,我们将创建一个新的键,其值为inp


  • let array1 = [{id:1, name:"raju"},{id:2, name:"ravi"},{id:4, name:"john"},{id:6, name:"jack"}];
    let array2= [{id:1, degree:"b.com"},{id:3, degree:"b.a"},{id:4, degree:"c.a"},{id:5, degree:"horticulture"}];
    let array3= [{id:1, age:20},{id:3, age:21},{id:6, age:27},{id:7, age:25}];
    
    let temp = [...array1,...array2,...array3]
    
    let op = temp.reduce((op,inp)=>{
      op[inp.id] = op[inp.id] || inp
      op[inp.id] = {...op[inp.id],...inp}
      return op
    },{})
    
    console.log(Object.values(op))

    09-11 18:13
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