typescript 打字稿：传播类型只能从对象类型创建
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Typescript: Spread types may only be created from object types
提问by Morten Poulsen
function foo<T extends object>(t: T): T {
return {
...t // Error: [ts] Spread types may only be created from object types.
}
}
I am aware that there are issues on github, but I can't figure out what is fixed and what is not and they have 2695 open issues. So I am posting here. I am using latest Typescript 2.9.2.
我知道 github 上存在问题，但我无法弄清楚哪些已修复，哪些未修复，并且他们有 2695 个未解决的问题。所以我在这里发帖。我正在使用最新的 Typescript 2.9.2。
Should the above code not work? And how can I fix it if possible?
上面的代码应该不起作用吗？如果可能，我该如何解决？
回答by jmattheis
This is fixed in TypeScript Version 3.2. See Release Notes.
这在 TypeScript 3.2 版中得到了修复。请参阅发行说明。
Looks like spread with a generic type isn't supported yet, but there is a GitHub issue about it: Microsoft/TypeScript#10727.
看起来尚不支持具有泛型类型的传播，但有一个关于它的 GitHub 问题：Microsoft/TypeScript#10727。
For now you can either use type assertionlike @Jevgenicommented:
function foo<T extends object>(t: T): T {
return { ...(t as object) } as T;
}
or you can use Object.assignwhich has proper type definitions.
或者您可以使用Object.assignwhich 具有正确的类型定义。
function foo<T extends object>(t: T): T {
return Object.assign({}, t);
}
回答by Muhammad Mabrouk
You can use blank curly brackets {} or an interface like below examples:
您可以使用空白大括号 {} 或类似以下示例的接口：
goodsArray.map(good => {
return {
id: good.payload.doc.id,
...good.payload.doc.data() as {}
};
});
or
或者
goodsArray.map(good => {
return {
id: good.payload.doc.id,
...good.payload.doc.data() as Goods // 'Goods' is my interface name
};
});
回答by Titian Cernicova-Dragomir
Version 3.2 of Typescript fixed this. The two PRs that improve handling of spread and rest parameters are:
Typescript 3.2 版修复了这个问题。改善传播和休息参数处理的两个 PR 是：
You can try it out now using npm install [email protected].
您现在可以使用npm install [email protected].
With 3.2 your code works as is.
使用 3.2，您的代码按原样运行。
Version 3.2 has been released on November 29th, 2018, you can read more about it here.
3.2 版本已于 2018 年 11 月 29 日发布，您可以在此处阅读更多相关信息。
回答by Aathil Ahamed
This answer may help to fix the same problem while using angular and firestoe.
这个答案可能有助于在使用 angular 和 firestoe 时解决同样的问题。
return { id: item.payload.doc.id,
...item.payload.doc.data()}
as Employee
return { id: item.payload.doc.id,
...item.payload.doc.data()}
as Employee
this on will throw he same error. for fixing you have to change like this one.
这将引发他同样的错误。为了修复，你必须像这样改变。
return { id: item.payload.doc.id,
...item.payload.doc.data() as Employee }
return { id: item.payload.doc.id,
...item.payload.doc.data() as Employee }
回答by Simon_Weaver
If you are still getting this error post version 3.2 then you probably have a type error 'upstream' resulting in an object being unknowninstead of an actual object. For instance if you do { ...getData() }and your getDatafunction has a compile error you can get this error.
如果您在 3.2 版后仍然收到此错误，那么您可能有一个类型错误“上游”，导致对象unknown不是实际对象。例如，如果您这样做{ ...getData() }并且您的getData函数有编译错误，您可能会收到此错误。
So check for all compiler errors because it's probably a red herring.
所以检查所有编译器错误，因为它可能是一个红鲱鱼。
