prisma2订阅返回数据: null

Question

我有一个基本的酒吧在这里工作使用样板和图形-瑜伽：subscriptions subscriptions/tree/sql-lite

有一个公开的突变：

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
      },
    })
  },
})

还有一个订阅--我只是返回true以确保withFilter (来自graphql-yoga)正常工作。

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
            )
        });
    }
});

在publish上返回以下内容(您可以将它们复制并粘贴到代码框中--这很好！)

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。当我在filter函数中记录payload.publishedPosts时，似乎所有东西都在那里。

{ 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' } }

我遗漏了什么吗？

Answer 1

终于弄明白了！

订阅函数需要以pubsub中的键命名。因此，如果您有如下所示的发布函数：

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

然后，您必须将订阅publishedPost命名为

        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
            )
        });

如果将订阅命名为publishedPostWithEmail，则不返回任何数据。

        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，如果您将订阅命名为publishedPost2，则将从结果中删除publishedPost3。

奇怪的是，如果你订阅了2条消息，你就会得到所有的数据。

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，您可以通过订阅带有单个项的数组来解决上述问题，订阅的名称就变得无关紧要了。

        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个错误