Question

我遇到一个错误，只能通过添加any作为返回值来解决。

export const dbConnections: any = {};

export const connectDb: Promise<void> = async () => {
    if (dbConnections.isConnected) {
        return;
    }

    try {
        const db = await mongoose.connect(config.get('mongoURI'), {
            useNewUrlParser: true,
            useUnifiedTopology: true,
            useFindAndModify: false,
            useCreateIndex: true,
        });

        dbConnections.isConnected = db.connections[0].readyState;
    } catch (err) {
        createError('Error caught connecting to db!', err);
    }
};

这会引发错误，

export const connectDb: Promise<void> = async () => {
                                        ^^^^^^^^^^^^^
Type '() => Promise<void>' is missing the following properties
  from type 'Promise<void>': then, catch, [Symbol.toStringTag], finally

如果我使用any而不是Promise<void>，则错误消失了，但这显然不是我要解决的方法。如何解决此皮棉错误？

Answer 1

问题在函数声明中。您需要将返回类型指定为Promise<void>。

export const connectDb = async (): Promise<void> => {
    if (dbConnections.isConnected) {
        return;
    }

    try {
        const db = await mongoose.connect(config.get('mongoURI'), {
            useNewUrlParser: true,
            useUnifiedTopology: true,
            useFindAndModify: false,
            useCreateIndex: true,
        });

        dbConnections.isConnected = db.connections[0].readyState;
    } catch (err) {
        createError('Error caught connecting to db!', err);
    }
};

Answer 2

打字稿中的异步函数是返回承诺值的函数。您可以这样定义函数类型：

export const dbConnections: any = {};

export const connectDb: () => Promise<void> = async () => {
    ...
};

ESlint错误，类型'（）=> Promise <void>'从类型'Promise <void>'中缺少以下属性：然后，最后捕获[Symbol.toStringTag]

2 个答案: