Apollo GraphQL 连接失败

Question

我的根组件已经用 ApolloProvider 标签包装，但错误消息告诉我它不是。

错误信息

Invariant Violation: Could not find "client" in the context or passed in as an option. Wrap the root component in an <ApolloProvider>, or pass an ApolloClient instance in via options.

This error is located at:
    in App (created by ExpoRoot)

问题是我的根组件已经被 ApolloProvider 标签包裹

React 本机代码

IMPORT 语句

import {
  ApolloClient,
  InMemoryCache,
  useQuery,
  ApolloProvider,
  gql,
} from "@apollo/client";

与 GraphQL 的连接

const client = new ApolloClient({
  uri: "https://www.outvite.me/gql/gql",
  cache: new InMemoryCache(),
  defaultOptions: { watchQuery: { fetchPolicy: 'cache-and-network' } },
})

测试查询

const USER_QUERY = gql`
  query USER {
    users {
      nodes {
        edge {
          username
        }
      }
    }
  }
`

默认应用程序

这是抛出错误的地方

const { data, loading } = useQuery(USER_QUERY) 是回溯显示的行

export default function App() {
    const { data, loading } = useQuery(USER_QUERY)
    return (
        <ApolloProvider client={client}>
           <View>
             <Text style={styles.text}>Open</Text>
             <Text style={styles.text}>Another text</Text>
           </View>
           <Button title="Toggle Sidebar" onPress={() => toggleSidebarView()} />
           <Button title="Change theme" onPress={() => toggleColorTheme()} />
        </ApolloProvider>
    );
}

Answer 1

如果我没记错的话，useQuery 挂钩仅在您位于已包装在 ApolloProvider 中的组件中时才有效，因此您可能想做这样的事情

export default function MainApp() {
    const { data, loading } = useQuery(USER_QUERY)
    return (
      <View>
        ... use 'data' in here somewhere...
      </View>
    );
}

然后顶级 App 组件看起来像

export default function App() {
    return (
      <ApolloProvider client={client}>
        <MainApp />
      </ApolloProvider>
    );
}

Apollo GraphQL 连接失败

Apollo GraphQL failing connection

graphql

apollo

apollo-client

react-apollo

react-apollo-hooks

React 本机代码