React，Graphql模式类型声明

Question

我在React和阿波罗中使用graphql :generate来从graphql模式创建类型。

我还使用react阿波罗-钩子useQuery输出数据.

一切似乎都正常，我可以看到输出的数据

我的问题是，我不知道如何声明生成的类型，然后在vc代码中获取intelliSense。

查询

import { gql } from 'apollo-boost';

export const GET_ALL_RECIPES = gql`

  query RecipesData{
    recipe{
      _id
      name
      description
    }
  }
`

生成类型

/* tslint:disable */
/* eslint-disable */
// This file was automatically generated and should not be edited.

// ====================================================
// GraphQL query operation: RecipesData
// ====================================================

export interface RecipesData_recipe {
  __typename: "Recipe";
  _id: string | null;
  name: string | null;
  description: string | null;
}

export interface RecipesData {
  recipe: (RecipesData_recipe | null)[] | null;
}   

App.tsx

import React from 'react';
import './App.css';

import { useQuery } from 'react-apollo-hooks';
import { GET_ALL_RECIPES } from '../queries';

import { RecipesData_recipe } from '../generated/RecipesData';

const App: React.FC<RecipesData> = () => {

  const { data, loading } = useQuery(GET_ALL_RECIPES, {
    suspend: false
  })

  if (loading || !data) return <div>Loading</div>

  return (
    <ul>
      {data.recipe.map((recipe) =>
        <li key={recipe._id}>{recipe.name}</li>
      )}
    </ul>
  );
}

export default App; 

如何正确地将RecipesData或RecipesData_recipe在App.tsx中声明为使用intelliSense for recipe.name

Answer 1

文档总是有帮助的。只需导入生成的...Data接口和/或...Vars接口(如果您正在查询)，然后使用它来转换useQuery。

import { RecipesData } from '../generated/RecipesData';

const { data, loading } = useQuery<RecipesData>(GET_ALL_RECIPES, {
  suspend: false
});

然后简单地以更干净的方式返回组件：

return (
  <div>
    {loading ? (
      <p>Loading ...</p>
    ) : (
      <ul>
        {data && data.recipe.map(recipe => (
          <li key={recipe._id}>{recipe.name}</li>
        ))}
      </ul>
    )}
  </div>
);