'React, Graphql schema type declaration
I'm using graphql in React and Apollo:generate to create types from the graphql schema.
I'm also using react-apollo-hooks useQuery to output the data.
Everything seems to be working and I can see the data outputted
My problem is I can't see how to declare the types generated to then get intelliSense in vc code.
Query
import { gql } from 'apollo-boost';
export const GET_ALL_RECIPES = gql`
query RecipesData{
recipe{
_id
name
description
}
}
`
Generated Types
/* 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;
How do I correctly declare the RecipesData
or RecipesData_recipe
in App.tsx to use the intelliSense for recipe.name
Solution 1:[1]
Docs are always helpful. Simply import the generated ...Data interface and/or ...Vars interface (if you're querying) and then use it to cast the useQuery
.
import { RecipesData } from '../generated/RecipesData';
const { data, loading } = useQuery<RecipesData>(GET_ALL_RECIPES, {
suspend: false
});
Then simply return the component in a more cleaner way:
return (
<div>
{loading ? (
<p>Loading ...</p>
) : (
<ul>
{data && data.recipe.map(recipe => (
<li key={recipe._id}>{recipe.name}</li>
))}
</ul>
)}
</div>
);
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
Solution | Source |
---|---|
Solution 1 | Mohamed Mohamed |