无法解决 Next.js Typescript 代码中的 prisma 类型错误

Question

我在修复 Typescript 错误时遇到了麻烦。这是代码：

import { PrismaClient, dog, cat, bird } from '@prisma/client'
export type AnyPrismaClientType = PrismaClient["dog"] | PrismaClient["cat"]  | PrismaClient["bird"]

export function makePrismaFromName(name: string): AnyPrismaClientType {
    const prismaClient = new PrismaClient()
    var client;

    switch (name) {
        case "dog": {
            client = prismaClient.dog;
            break;
        }
        case "cat": {
            client = prismaClient.cat;
            break;
        }
        case "bird": {
            client = prismaClient.bird;
            break;
        }
        default: {
            const msg = `unrecognized animal type ${String(name)}`;
            throw (msg);
        }
    }

    return client;
}

const myClient = makePrismaFromName('dog');

const inactiveAccountCount = await prismaClient.count( // --> This is the line giving me the type error.
    {
        where: {
            stray: false,
            clean: true,
        }
    }
);

这是错误：

This expression is not callable.
  Each member of the union type '(<T extends dogCountArgs>(args?: Subset<T, DogCountArgs<DefaultArgs>> | undefined) =>
  PrismaPromise<T extends Record_2<"select", any> ? T["select"] extends true ? number : { [P in keyof T["select"]]: P extends
  keyof DogCountAggregateOutputType ? DogCountAggregateOutputType[P] : never; } : number>) | (<T ...' has signatures, but none
  of those signatures are compatible with each other.

Answer 1

您在这里面临的问题是，所有这些返回类型 (

dog, cat, or bird

) 都会给 TS 带来混乱，它无法确定使用哪种类型来运行 count 方法。

您正在寻找的解决方案是使用泛型，您的

makePrismaFromName

函数会将这个泛型定义为某种类型的扩展，为了简洁起见，这里称为 Animal ，并且返回类型与提供的泛型匹配。这可以这样实现：

import { PrismaClient } from '@prisma/client'

type Animal = 'dog' | 'cat' | 'bird';

export function makePrismaFromName<T extends Animal>(name: T): PrismaClient[T] {
    const prismaClient = new PrismaClient();

    switch (name) {
        case "dog":
            return prismaClient.dog as PrismaClient[T];
        case "cat":
            return prismaClient.cat as PrismaClient[T];
        case "bird":
            return prismaClient.bird as PrismaClient[T];
        default:
            throw new Error(`Unrecognized animal type: ${name}`);
    }
}

请注意，此处的类型断言是为了帮助 TS 根据给定的泛型找出类型。

您的调用函数和其他部分将保持不变。