Apollo GraphQL 客户端 + Next.JS 错误处理

Question

该组件很好地呈现错误状态，但异常在控制台中显示为未捕获，并且在浏览器的下一个开发中显示对话框。 有没有办法处理预期的错误来抑制这种行为？

import { useMutation, gql } from "@apollo/client";
import { useEffect } from "react";

const CONSUME_MAGIC_LINK = gql`
  mutation ConsumeMagicLink($token: String!) {
    consumeMagicLink(token: $token) {
      token
      member {
        id
      }
    }
  }
`;

export default function ConsumeMagicLink({ token }) {
  const [consumeMagicLink, { data, loading, error }] =
    useMutation(CONSUME_MAGIC_LINK);

  console.log("DATA", data, "loading:", loading, "error:", error);

  useEffect(() => {
    try {
      consumeMagicLink({ variables: { token } });
    } catch (e) {
      console.log(e);
    }
  }, []);

  var text = "Link has expired or has been used previously";

  if (data) text = "SUCCESS: REDIRECTING";
  if (loading) text = "Processing";
  if (error) text = "Link has expired or has been used previously";

  return (
    <div>
      <h2>{text}</h2>
    </div>
  );
}

控制台结果：

浏览器中显示的错误：

Answer 1

错误来自客户端而不是突变，因此您的 try-catch 无法捕获它。 要处理此问题，您可以将错误处理添加到客户端，例如：

const errorLink = onError(({ graphQLErrors, networkError }) => {
  if (graphQLErrors)
    graphQLErrors.forEach(({ message, locations, path }) =>
      console.log(
        `[GraphQL error]: Message: ${message}, Location: ${locations}, Path: ${path}`,
      ),
    );

  if (networkError) console.log(`[Network error]: ${networkError}`);
});

const httpLink = new HttpLink({
  uri: "some invalid link"
});

const client = new ApolloClient({
  link:from([httpLink,errorLink]),
  cache: new InMemoryCache()
})

当您遇到授权错误时，我建议您检查您的标题。

Answer 2

使用 nextjs，如何防止 modal 显示 Apollo Link 中出现的每一个错误？ 有人帮我吗？