當屬性存在時，類型錯誤時屬性不存在

Question

在制作以下 TypeScript 代碼時，我收到了上述錯誤。 問題顯然是 urls、type 和 sub 屬性不存在於 string[] 類型的參數上，但如果這是目標參數的類型，它們將不會被讀取。 我對 TypeScript 非常陌生，這是我第一次嘗試使用它，所以我可能缺少一些相對基本的東西。

interface targeterInterface {
    urls: string | string[];
    type?: string;
    sub?: string;
}
function exp(targeter: string | targeterInterface | string[]) {
    let target: string[] | string;
    let type: string;
    let sub: string;
    if (typeof targeter === "object") {
        target = targeter.urls;
        type = targeter.type;
        sub = targeter.sub;
    } else if (typeof targeter === "string") {
        target = targeter;
    } else {
        console.log("It's an array!");
    }
}

謝謝你的幫助！

Answer 1

如果它不是字符串或數組，則它是一個對象

if (Array.isArray(targeter)) {
    // Array
} else if (typeof targeter === "string") {
    // string
} else {
    // Instance of targeterInterface
}

Answer 2

您可以使用稱為 type-guard 函數的 TypeScript 功能來檢查參數的類型。 https://www.typescriptlang.org/docs/handbook/advanced-types.html#user-defined-type-guards

 interface targeterInterface { urls: string | string[]; type: string; // had to change since, the type in the function is not optional sub: string; // had to change since, the sub in the function is not optional } function isTargeterInterface(item: unknown): item is targeterInterface { return (item as targeterInterface).type !== undefined; } // this is a type guard function function exp(targeter: string | targeterInterface | string[]) { let target: string[] | string; let type: string; let sub: string; if (isTargeterInterface(targeter)) { // calling a function to check if it implements interface provided above target = targeter.urls; type = targeter.type; sub = targeter.sub; } else if (typeof targeter === "string") { target = targeter; } else { console.log("It's an array!"); } }