我是新流程,目前正在开发一个流程严格的项目。我有一个类型别名传递给一个类。可以找到代码段here
// @flow strict
export type CustomType = {
a: string,
b: boolean,
c: boolean,
d: boolean,
e: boolean
};
let d = true;
let b = true;
let customType:CustomType = {
d,
b,
c: true,
}
class Custom{
constructor(customType = {}) {}
}
let custom = new Custom(customType);
传递对象时,不存在所有属性customType。这里最好的解决方法是什么?
答案 0 :(得分:3)
您可以将CustomType
对象的键输入为optional:
(Try)
// @flow strict
export type CustomType = {
a?: string,
b?: boolean,
c?: boolean,
d?: boolean,
e?: boolean
};
let d = true;
let b = true;
let customType: CustomType = {
d,
b,
c: true,
}
class Custom{
constructor(customType: CustomType = {}) {}
}
let custom = new Custom(customType);
或者,您可以使用$Shape<T>
。它只允许您传递您感兴趣的键:
(Try)
// @flow strict
export type CustomType = {
a: string,
b: boolean,
c: boolean,
d: boolean,
e: boolean
};
let d = true;
let b = true;
let customType: $Shape<CustomType> = {
d,
b,
c: true,
}
class Custom{
constructor(customType: $Shape<CustomType> = {}) {}
}
let custom = new Custom(customType);