typescript - 使用TypeScript，如何强烈键入mysql查询结果

Question

我是Typescript的新手，无法确定我是否在正确输入查询结果。这是我的代码的本质...

import mysql2, {Pool} from "mysql2";
const pool: Pool = mysql2.createPool({...}).promise();

interface IUser {
    uid   : number;
    uname : string;
}

class UserController {

    public async getUser(id: number): Promise<IUser> {
        const [rows]: Array<{rows: IUser}> = await pool.query(
            "SELECT * FROM `users` WHERE `email` = ?",["me@me.org"]);        
        return rows;
    }
}

TypeScript编译器(3.3.1)提示我的 return rows语句。

TS2740: Type '{rows: IUser;}' is missing the following properties from type 'IUser': uid and uname.

如果我忽略 // @ts-ignore的返回，那么一切都很好。我把我的物体放回原处，没有任何错误。

难道我做错了什么？

我做了一些更改，但是对于TypeScript为什么不提示，我实在感到困惑。似乎根本不对。

    class UserController {

        public async getUser(id: number): Promise<{rows: IUser}> {
            const [rows]: Array<{rows: IUser}> = await pool.query(
                "SELECT * FROM `users` LIMIT 3");        
            return rows;
        }
    }

这是正确的吗？？？

所以，没有，这全都错了。

当我想到 query返回 [rows, fields]时，它也开始变得更有意义了。我认为@ joesph-climber经过一些调整后的语法是正确的。

这对我来说很有用...

    class UserController {

        public async getUser(id: number): Promise<Array<{rows: IUser}>> {
            const [rows]: [Array<{rows: IUser}>] = await pool.query(
                "SELECT * FROM `users` LIMIT 3");        
            return rows;
        }
    }

这也可行，并且可能更容易理解。

    class UserController {

        public async getUser(id: number): Promise<IUser[]> {
            const [rows]: [IUser[]] = await pool.query(
                "SELECT * FROM `users` LIMIT 3");        
            return rows;
        }
    }

Answer 1

// the function should return a Promise that resolves to 
{ uid: number, uname: string }

// but typeof rows is 
{ rows: { uid: number, uname: string } }

执行该示例，我得到的结果是这样的:

[ TextRow { uid: 1, uname: 'foo', email: 'foo@mail.com' } ]

因此，pool.query返回一个以IUser数组作为第一个元素的数组。

Returning multiple users:

class UserController {
    // returns all users
    public async getUsers(): Promise<Array<IUser>> {
        const [rows]: [Array<IUser>] = await pool.query(
                "SELECT * FROM `user`", []);
        return rows; // rows is Array<IUser> so it matches the promise type
    }
}

Returning a specific user:

class UserController {
    public async getUser(id: number): Promise<IUser> { // Note the 
        const [rows]: [Array<IUser>] = 
            await pool.query("SELECT * FROM `user` WHERE `email` = ?",
                            ["foo@mail.com"]);
        // If email is unique as it would be expected it will 
        // return a single value inside an array
        // rows is still Array<IUser>
        return rows[0]; // rows[0] is an IUser 
    }
}