ES6对象构造嵌套参数的默认值

Question

我正在使用es6对象解构来为函数提供默认参数.

function mapStateToProps({ shops: { cakeShop: {}, pieShop: {} }) {
  return {
    CakeShopName: shops.cakeShop.Name,
    PieShopName: shops.pieShop.Name
  }
}

上面的问题是,如果我打电话

mapStateToProps({})

代码抛出Cannot read property 'Name' of undefined.嵌套对象shops未设置为其默认值,并且代码具有空引用.

如何确保将嵌套对象shops设置为正确的默认值,即使shops自身已定义？

Answer 1

听起来像是在混淆默认值的混乱.您的语法会破坏参数对象,但实际上并不会引入任何参数标识符.shops您的功能范围中没有变量.

我假设您实际上想要引入cakeShop和pieShop变量,并为它们提供默认值.要做到这一点,你要写

function mapStateToProps({ shops: { cakeShop = {}, pieShop = {} }) {
// short for             { shops: { cakeShop: cakeShop = {}, pieShop: pieShop = {} }) {
// parameter names (that will be bound):      ^^^^^^^^                ^^^^^^^
  return {
    CakeShopName: cakeShop.Name,
    PieShopName: pieShop.Name
  }
}

你也可以使用

function mapStateToProps({ shops: { cakeShop: {name: CakeShopName} = {}, pieShop: {name: PieShopName} = {} }) {
  return {CakeShopName, PieShopName};
}

Answer 2

要处理嵌套对象的默认值,它应该是

function mapStateToProps({ shops: { cakeShop = {}, pieShop = {} } = {} } = {}) {
...
}