可以计算 type-graphql 中的结果列表元素

Posted 2023-03-08

【中文标题】可以计算 type-graphql 中的结果列表元素

【英文标题】：possible to count result list elements in type-graphql

【发布时间】：2019-09-20 05:57:15

【问题描述】：

我是 typeGraphQL 的新手，我的问题是可以计算结果列表元素

我有模型类型书籍

import  Field, ID, ObjectType, Root  from "type-graphql";
import  Model, Column  from "sequelize-typescript";


@ObjectType()
export default class Books extends Model<Books> 


    @Field(() => ID)
    @Column
    id: number;

    @Field(() => String)
    @Column
    bookName: string;

解析器中的查询

@Query(() => [Books])
async getBooks()

    return Books.findAll()

在 graphiQL 中运行查询时

getTest

    id
    bookName

得到回应

"getBooks" : 
[
    
        "id": "1",
        "bookName": "bookOne"
    ,
    
        "id": "2",
        "bookName": "bookTwo"
    
]

但我需要添加附加字段，例如获取所有收到项目的总数。 如何正确地做到这一点？ 现在我们必须创建一个单独的查询，但是对于大样本来说这很不方便，并且会导致复杂和大查询的重复

@Query(() => [Books])
async countBooks()

    return Books.findAll().length

我尝试为元素数量和模型本身创建一个单独的联合类型

@ObjectType()
export default class UnionType extends Model<UnionType> 


    @Field(() => [Books])
    books: Books[];

    @Field(() => Int)
    totalCount(@Root() parent : UnionType) : number 
    
        return parent.books.length;
    

并在解析器中运行下一个查询

@Query(() => [UnionType])
async getBooksAndCountAll()

    let union : any =         

    union.books = Books.findAll();
    union.totalCount = union.books.length;

    return union;

但在运行查询时出现 graphiQL 错误 error "message": "Expected Iterable, but did not find one for field Query.getBooks.", 据我了解，它没有传输模型期望的数据

我尝试使用 createUnionType

import  createUnionType  from "type-graphql";
const SearchResultUnion = createUnionType(
    name: "Books", // the name of the GraphQL union
    types: [Books, CountType], // array of object types classes
  );

UnionType 在哪里

@ObjectType()
export default class CountType extends Model<CountType> 

    @Field(() => Int,  nullable : true )
    totalCount: number;

解析器中的查询

    @Query(returns => [SearchResultUnion])
    async getBooks(
    ): Promise<Array<typeof SearchResultUnion>>

        return new Promise((resolve, reject) => 
            Books.findAll()
            .then(books => 

                let totalCount = books.length;

                return [...books, ...totalCount];
            );
        );
    

但字符串中出现错误return [...books, ...totalCount]; на ...totalCount Type 'number' must have a '[Symbol.iterator]()' method that returns an iterator.

如果你没有通过...totalCount请求有效，但是没有分别已经没有totalCount

getTest
__typename
  ... on Books
  
    id
    bookName
  

请求

 "getBooks": [
    
        "id": "1",
        "bookName": "bookOne"
    ,
    
        "id": "2",
        "bookName": "bookTwo"
    
  ]

因此，我需要一个请求

getTest

    totalCount
    __typename
      ... on Books
        id
        bookName
      

有可能吗？

【参考方案1】：

https://github.com/19majkel94/的答案 谢谢，Michał Lytek

@ObjectType()
export default class GetBooksResponse 
    @Field(() => [Books])
    books: Books[];

    @Field(() => Int)
    totalCount : number;


@Query(() => GetBooksResponse)
async getBooks() 
  const books = await Books.findAll();
  return  books, totalCount: books.length ;