在从同一个类继承的一个数组中收集各种不同的对象时,如何在TypeScript中设置一个优等的类,以便TypeScript不显示错误?
我正在尝试这样:
interface IVehicle{
modelName: string
}
interface ICar extends IVehicle{
numberOfDoors: number,
isDropTop: boolean
}
interface IBike extends IVehicle{
hasDynamo: boolean
}
var vehicles: IVehicle[] =
[
{
modelName: "carModelName", // Error
numberOfDoors: 4,
isDropTop: true
},
{
modelName: "bikeModelName",
hasDynamo: true
}
]
这样做,我遇到了错误.
如果我不想显示任何错误,我只能添加上级接口IVehicle的对象.
解决方法:
修复语法错误后,您可以指定数组中每个条目的类型.
interface IVehicle {
modelName: string
}
interface ICar extends IVehicle {
numberOfDoors: number,
isDropTop: boolean
}
interface IBike extends IVehicle {
hasDynamo: boolean
}
let vehicles: IVehicle[] =
[
{
modelName: "carModelName",
numberOfDoors: 4,
isDropTop: true,
} as ICar,
{
modelName: "bikeModelName",
hasDynamo: true
} as IBike
]
或者只是将数组的类型更改为车辆,汽车或自行车的数组,如下所示:
let vehicles: Array<IVehicle | ICar | IBike> =
[
{
modelName: "carModelName",
numberOfDoors: 4,
isDropTop: true,
},
{
modelName: "bikeModelName",
hasDynamo: true
}
]
如果以后要确定IVehicle是IBike还是ICar,您可以使用用户定义的类型保护来执行此操作.
function isBike(vehicle: IVehicle): vehicle is IBike {
return (<IBike>vehicle).hasDynamo !== undefined;
}
function isCar(vehicle: IVehicle): vehicle is ICar {
return (<ICar>vehicle).numberOfDoors !== undefined;
}
function log(vehicle: IVehicle) {
if (isBike(vehicle)) {
// tsc knows vehicle is IBike
console.log(vehicle.hasDynamo);
} else if (isCar(vehicle)) {
// tsc knows vehicle is ICar
console.log(vehicle.numberOfDoors);
} else {
console.log(vehicle.modelName);
}
}
您可以在本手册的Advanced types部分中阅读有关它们的更多信息.
您还可以在here操场上找到整个代码的工作示例.