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我有一个基本的 pubsub 使用样板和 graphql-yoga 在这里工作： https://github.com/ryanking1809/prisma2_subscriptions https://codesandbox.io/s/github/ryanking1809/prisma2_subscriptions/tree/sql-lite

使用发布突变：

const Mutation = objectType({
  name: 'Mutation',
  definition(t) {
   //...
    t.field('publish', {
      type: 'Post',
      nullable: true,
      args: {
        id: idArg(),
      },
      resolve: async (parent, { id }, ctx) => {
        const post = await ctx.photon.posts.update({
          where: { id },
          data: { published: true },
          include: { author: true }
        });
        ctx.pubsub.publish("PUBLISHED_POST", {
          publishedPost: post
        });
        return post
      },
    })
  },
})

和订阅 - 我只是回来true确保withFilter（来自graphql-yoga）工作。

const Subscription = objectType({
    name: "Subscription",
    definition(t) {
        t.field("publishedPostWithEmail", {
            type: "Post",
            args: {
                authorEmail: stringArg({ required: false })
            },
            subscribe: withFilter(
                (parent, { authorEmail }, ctx) => ctx.pubsub.asyncIterator("PUBLISHED_POST"),
                (payload, { authorEmail }) => true
            )
        });
    }
});

返回以下内容publish（您可以将这些复制并粘贴到代码和框中 - 这很整洁！）

mutation {
  publish(
    id: "cjzwz39og0000nss9b3gbzb7v"
  ) {
    id,
    title,
    author {
      email
    }
  }
}

subscription {
  publishedPostWithEmail(authorEmail:"prisma@subscriptions.com") {
    title,
    content,
    published
  }
}

{
  "errors": [
    {
      "message": "Cannot return null for non-nullable field Subscription.publishedPostWithEmail.",
      "locations": [
        {
          "line": 2,
          "column": 3
        }
      ],
      "path": [
        "publishedPostWithEmail"
      ]
    }
  ],
  "data": null
}

由于某种原因，它正在返回data: null。当我登录payload.publishedPosts过滤器功能时，似乎一切都在那里。

{ id: 'cjzwqcf2x0001q6s97m4yzqpi',
  createdAt: '2019-08-29T13:34:26.648Z',
  updatedAt: '2019-08-29T13:54:19.479Z',
  published: true,
  title: 'Check Author',
  content: 'Do you save the author?',
  author:
   { id: 'sdfsdfsdfsdf',
     email: 'prisma@subscriptions.com',
     name: 'Prisma Sub' } }

有什么我想念的吗？

4

1 回答 1

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终于想通了！

订阅函数需要以 pubsub 中的 key 命名。因此，如果您有如下发布功能：

ctx.pubsub.publish("PUBLISHED_POST", {
          publishedPost: post
        });

那么你必须命名你的订阅publishedPost

        t.field("publishedPost", {
            type: "Post",
            args: {
                authorEmail: stringArg({ required: false })
            },
            subscribe: withFilter(
                (parent, { authorEmail }, ctx) =>
                    ctx.pubsub.asyncIterator("PUBLISHED_POST"),
                (payload, { authorEmail }) => payload.publishedPost.author.email === authorEmail
            )
        });

如果您为订阅命名，publishedPostWithEmail则不会返回任何数据

        t.field("publishedPostWithEmail", {
                //...
        });

有趣的是，如果你有 2 把钥匙

ctx.pubsub.publish("PUBLISHED_POST", {
          publishedPost2: post,
          publishedPost3: post
        });

然后，如果您命名您的订阅，publishedPost2则publishedPost3结果中会省略。

奇怪的是，如果您订阅 2 条消息，您将获得所有数据

ctx.pubsub.publish("PUBLISHED_POST", {
          publishedPost: post,
          publishedPost2: post
        });
        ctx.pubsub.publish("PUBLISHED_POST_X", {
          publishedPostX: post,
          publishedPostY: post
        });

ctx.pubsub.asyncIterator([
                        "PUBLISHED_POST",
                        "PUBLISHED_POST_X"
                    ]),

返回publishedPost, publishedPost2, publishedPostX,publishedPostY

因此，您可以通过使用单个项目订阅数组来解决上述问题，并且订阅的名称变得无关紧要。

        t.field("publishedPostXYZ", {
            type: "Post",
            args: {
                authorEmail: stringArg({ required: false })
            },
            subscribe: withFilter(
                (parent, { authorEmail }, ctx) =>
                    ctx.pubsub.asyncIterator([
                        "PUBLISHED_POST"
                    ]),
                (payload, { authorEmail }) => {
                    return payload.publishedPost.author.email === authorEmail;
                }
            )
        });

所以看起来这可能是一个错误

于 2019-08-30T23:19:38.857 回答