我正在使用TypeScript库(角度),该库具有将Type作为参数的函数:
curl -X POST https://api.sandbox.paypal.com/v1/payments/payouts \
-H "content-type: application/json" \
-H "Authorization: Bearer Access-Token" \
-d '{
"sender_batch_header": {
"sender_batch_id": "1524086406556",
"email_subject": "This email is related to simulation"
},
"items": [
{
"recipient_type": "EMAIL",
"receiver": "payouts-simulator-receiver@paypal.com",
"note": "ERRPYO002",
"sender_item_id": "15240864065560",
"amount": {
"currency": "USD",
"value": "1.00"
}
}]
}'
但是我不知道如何在我的Scala代码中定义它。具体来说,如果还有[],则Type [any]对应的scala类型是什么:
ComponentFactoryResolover.resolveComponent(component: Type<any>)
答案 0 :(得分:0)
Scala.js对于运行时类值(或构造函数)没有非常特定的类型。由于我们传递给此类参数的实际值通常是为某些js.constructorOf[Foo]
调用Foo
并返回js.Dynamic
的结果,因此我建议翻译TypeScript的{{1} }放入Scala.js中的Type<T>
。