[英]Unable to access parameters from union types (TypeScript)
為什么無法訪問像這樣的聯合類型中的屬性:
export interface ICondition {
field: string
operator: string
value: string
}
export interface IConditionGroup {
conditions: ICondition[]
group_operator: string
}
function foo(item: ICondition | IConditionGroup) {
if(typeof item.conditions === "undefined") { // does not work
let field = item.field; // does not work
///.. do something
} else {
let conditions = item.conditions; // does not work
/// .. do something else
}
}
我得到這些錯誤:
error TS2339: Property 'conditions' does not exist on type 'ICondition | IConditionGroup'.
error TS2339: Property 'conditions' does not exist on type 'ICondition | IConditionGroup'.
error TS2339: Property 'field' does not exist on type 'ICondition | IConditionGroup'.
但是我必須強制轉換類型才能使其正常工作-像這樣:
function foo2(inputItem: ICondition | IConditionGroup) {
if(typeof (<IConditionGroup>inputItem).conditions === "undefined") {
let item= (<ICondition>inputItem);
let field = item.field;
///.. do something
} else {
let item= (<IConditionGroup>inputItem);
let conditions = item.conditions;
/// .. do something else
}
}
我知道類型信息在JS中丟失了,為什么我必須在TS中顯式轉換它?
Typescript使用Type Guards處理此問題,通常很簡單:
if (typeof item === "string") { ... } else { ... }
要么
if (item instanceof MyClass) { ... } else { ... }
但是對於您來說,由於您無法使用接口,因此您需要創建自己的用戶定義類型保護 :
function isConditionGroup(item: ICondition | IConditionGroup): item is IConditionGroup {
return (item as IConditionGroup).conditions !== undefined;
}
function foo(item: ICondition | IConditionGroup) {
if (isConditionGroup(item)) {
let conditions = item.conditions;
// do something
} else {
let field = item.field;
// do something else
}
}
( 操場上的代碼 )
您也可以在沒有類型保護的情況下執行此操作:
function foo(item: ICondition | IConditionGroup) {
if ((item as IConditionGroup).conditions !== undefined) {
let conditions = (item as IConditionGroup).conditions;
// do something
} else {
let field = (item as ICondition).field;
// do something else
}
}
但這是冗長的方式,因為您需要輸入3次而不是一次的斷言item
。
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