小编Pet*_*lik的帖子

relayStylePagination()我已经根据apollo文档（https://www.apollographql.com/docs/react/pagination/cursor-based/#relay-style-cursor-pagination）通过以下方式实现：

索引.js：

const httpLink=new HttpLink({
  uri:'https://api.github.com/graphql',
  headers:{
    authorization: 'Bearer -'
  }
})

const cache = new InMemoryCache({
  typePolicies: {
    Query: {
      fields: {
        repositories:relayStylePagination()
      },
    },
  },
});

const client=new ApolloClient({
  link:httpLink,
  cache
})

ReactDOM.render(
  <ApolloProvider client={client}>
  <React.StrictMode>
    <App />
  </React.StrictMode>
  </ApolloProvider>,
  document.getElementById('root')
);

应用程序.js：

const  App=()=> {

  const {loading,error,data,fetchMore}=useQuery(GET_REPOSITORIES_OF_CURRENT_USER,{
    variables:{login:"rwieruch"}
  })

  if (loading) return <h1>Loading...</h1>
  if (error) return <p>Error...</p>
  console.log(data.user.repositories.edges)
  console.log(data)

  const pageInfo=data.user.repositories.pageInfo
  console.log(pageInfo)
  return(
    <div>
      <RepositoryList repositories={data.user.repositories} onLoadMore={()=>
           {return fetchMore({
              variables:{
                after: data.user.repositories.pageInfo.endCursor,
              }
            })} …

我正在尝试实现 signInWithRedirect() 函数，但是 - 奇怪的是，它不起作用。在下面的代码中，Check 2永远Check 3不会被记录 - 并且不会引发错误：

const signInWithGoogle = async () => {
        try {
            console.log("Check 1")
            await signInWithRedirect(auth, googleAuthprovider);

            console.log("Check 2")
            const result = await getRedirectResult(auth);

            console.log("Check 3");
            if (result) {
                console.log(result.user);
            }

        } catch (e) {
            console.log(e.message.slice(10));
        }
    };

另外，如果我使用完全相同的 Google 帐户通过 signInWithPopup() 方法登录，一切都会按预期进行：

const signInWithGoogle = async () => {
        const result = await signInWithPopup(auth, googleAuthprovider)

            const user = result.user;
            console.log(user)
    };

我真的很感激一些帮助。谢谢！