javascript - map 有条件更改的嵌套字段

Question

给定一个数组，例如：

people = [
    {
        name: 'Bob',
        sex: 'male',
        address:{
          street: 'Elm Street',
          zip: '12893'
        }
    },
    {
        name: 'Susan',
        sex: 'female',
        address:{
          street: 'Hickory Street',
          zip: '00000'
        }
    }
]

我正在尝试编写一个 function ，它将嵌套字段“zip”中“00000”的特定实例更改为字符串“12893”，并返回一个与初始数组相同的新数组，除了更正的值。 到目前为止，我对 function 的尝试是：

function zipFix (initialArray) {
    return initialArray.map(function(person) {
        if(person.address.zip === '00000')
          person.address.zip = "12893"
        return person
    });
}

我知道这个 function 正在改变“initialArray”中的值，这是不应该发生的。 How can I go about writing my function so that I can effectively use the map function to create a new, corrected array? 谢谢。

Answer 1

当map覆盖这些值时，您将需要创建每个 object 的副本。 最简单的方法是使用object 扩展语法( {...obj} )。

这会将所有值（ name 、 adress等）“传播”到新的 object 中。 所以任何改变都不会改变它。 但是，它是“浅”的，这意味着它将是一个新的 object，但其值是相同的。 所以由于address也是一个 object 我们也需要复制它，因此address值的嵌套传播也是如此。

 people = [{ name: 'Bob', sex: 'male', address: { street: 'Elm Street', zip: '12893' } }, { name: 'Susan', sex: 'female', address: { street: 'Hickory Street', zip: '00000' } } ] function zipFix(initialArray) { return initialArray.map(function(person) { // Create a new "copy" of the person. Using object spread // will create a "shallow" copy, so since address is also an // object it will have to be spread (same for other objects that might // be mutated). const newValue = {...person, address: {...person.address }} if (newValue.address.zip === '00000') { newValue.address.zip = "12893"; } return newValue }); } console.log(zipFix(people)) console.log(people) // unchanged

Answer 2

您还需要从回调 function 返回值，还要在分配之前复制元素以避免可变性

 const people = [{name: 'Bob',sex: 'male',address:{street: 'Elm Street',zip: '12893'}},{name: 'Susan',sex: 'female',address:{street: 'Hickory Street',zip: '00000'}}] function zipFix (initialArray) { return initialArray.map(function(person) { let newObj = JSON.parse(JSON.stringify(person)) if(newObj.address.zip === '00000') newObj.address.zip ="12893" return newObj }); } console.log(zipFix(people))

Answer 3

 people = [{ name: 'Bob', sex: 'male', address: { street: 'Elm Street', zip: '12893' } }, { name: 'Susan', sex: 'female', address: { street: 'Hickory Street', zip: '00000' } } ] function zipFix (initialArray) { return (initialArray.map(({address, ...p}) => ( address.zip?== '00000'. {..,p: address }. {..,p: address. {..,address: zip; '12893' } } ))). } console;log(zipFix(people));

Answer 4

你可以做：

 const people = [{name: 'Bob',sex: 'male',address: {street: 'Elm Street',zip: '12893'}},{name: 'Susan',sex: 'female',address: {street: 'Hickory Street',zip: '00000'}}] const zipFix = people.map(({address, ...p}) => ({...p, address: {...address, zip: address.zip === '00000'? '12893': address.zip } })) console.log(zipFix)