-1

我正在尝试使用它们的 WordPress ID 从单个查询中的两个单独对象获取数据，但是我GraphQLError: The ID input is invalid. Make sure you set the proper idType for your input.在 WordPress 中使用 GraphQL IDE 它可以按预期获取所有数据，但是我在代码中遇到了该错误。例如，如果我将 idType 设置为字符串，我会得到Variable "$editorId" of type "String!" used in position expecting type "ID!".

gatsby-node.js > createPages 函数：

// Video Detail pages
  const {
    data: {
      cartel: { videoDetailPages },
    },
  } = await graphql(`
    query {
      cartel {
        videoDetailPages(first: 300) {
          nodes {
            id
            slug
            videoDetail {
              editor
              editorId
            }
          }
        }
      }
    }
  `);

  const videoDetailTemplate = path.resolve(`./src/templates/videoDetail.js`);

  videoDetailPages.nodes.forEach(page => {
    const editorSlug = page.videoDetail.editor.replace(' ', '-').toLowerCase();
    const { editorId } = page.videoDetail;

    createPage({
      // will be the url for the page
      path: `${editorSlug}/${page.slug}`,
      // specify the component template of your choice
      component: slash(videoDetailTemplate),
      // In the ^template's GraphQL query, 'id' will be available
      // as a GraphQL variable to query for this page's data.
      context: {
        id: page.id,
        editorId,
      },
    });
  });

页面模板查询：

export const query = graphql`
  query($id: ID!, $editorId: ID!) {
    cartel {
      videoDetailPage(id: $id) {
        videoDetail {
          client
          director
          duration
          editor
          productionCompany
          videoStill {
            altText
            sourceUrl
          }
          videoUrl
          title
        }
      }
    }
    cartel {
      editorDetailPage(id: $editorId) {
        editorDetail {
          editorVideos {
            pagePath
            image {
              altText
              sourceUrl
              title
            }
          }
        }
      }
    }
  }
`;

盖茨比信息：

  System:
    OS: macOS 10.15.7
    CPU: (12) x64 Intel(R) Core(TM) i9-8950HK CPU @ 2.90GHz
    Shell: 5.7.1 - /bin/zsh
  Binaries:
    Node: 10.23.0 - ~/.nvm/versions/node/v10.23.0/bin/node
    Yarn: 1.22.4 - /usr/local/bin/yarn
    npm: 6.14.8 - ~/.nvm/versions/node/v10.23.0/bin/npm
  Languages:
    Python: 2.7.16 - /usr/bin/python
  Browsers:
    Chrome: 91.0.4472.77
    Firefox: 87.0
    Safari: 14.1
  npmPackages:
    gatsby: ^2.24.36 => 2.32.13 
    gatsby-image: ^2.4.14 => 2.11.0 
    gatsby-plugin-accessibilityjs: ^1.0.3 => 1.0.3 
    gatsby-plugin-google-tagmanager: ^2.3.11 => 2.11.0 
    gatsby-plugin-manifest: ^2.4.22 => 2.12.1 
    gatsby-plugin-offline: ^2.2.7 => 2.2.10 
    gatsby-plugin-react-helmet: ^3.3.10 => 3.10.0 
    gatsby-plugin-remove-trailing-slashes: ^2.3.11 => 2.10.0 
    gatsby-plugin-sass: ^2.3.12 => 2.8.0 
    gatsby-plugin-sharp: ^2.6.25 => 2.14.4 
    gatsby-plugin-sitemap: ^2.4.11 => 2.12.0 
    gatsby-plugin-web-font-loader: ^1.0.4 => 1.0.4 
    gatsby-source-filesystem: ^2.3.24 => 2.11.1 
    gatsby-source-graphql: ^3.4.0 => 3.4.0 
    gatsby-transformer-sharp: ^2.5.12 => 2.12.1

我没有任何运气发现我做错了什么。

4

2 回答 2

0

你gatsby-node.js看起来很完美。您的问题是由您发送到模板 ( videoDetailTemplate) 的数据上下文的类型引起的。你告诉 GraphQL 两者id和editorId都是ID类型，而我猜它们应该是字符串。

我想改变这一行：

  query($id: ID!, $editorId: ID!) {

为此应该做的伎俩：

  query($id: String!, $editorId: String!) {

正如您从GraphQL 类型定义文档中看到的那样：

标ID量类型表示唯一标识符，通常用于重新获取对象或作为缓存的键。ID类型的序列化方式与String相同；但是，将其定义为 anID意味着它不适合人类阅读。

注意：找出 String 和 ID 类型之间的区别（“将其定义为 anID表示它不是人类可读的”）

您应该能够在插件设置中为每个字段配置类型：

module.exports = {
  plugins: [
    {
      resolve: "gatsby-source-graphql",
      options: {
        // Remote schema query type. This is an arbitrary name.
        typeName: "WPGraphQL",
        // Field name under which it will be available. Used in your Gatsby query. This is also an arbitrary name.
        fieldName: "wpcontent",
        // GraphQL endpoint, relative to your WordPress home URL.
        url: "https://example.com/blog/graphql",
      },
    },
  ],
}

此外，您应该能够在通过上下文发送时强制使用parseInt、Number、String等方法的类型。

然后where按照文档建议使用过滤器：

editorDetailPage(where: { id: $editorId})

videoDetailPage (where: { id: $id})

您能否提供有关您的实现的更多详细信息（插件、配置、版本等）？这似乎是一个错误：

资源：

于 2021-06-03T04:51:57.377 回答
-1

搞定了 需要对 editorDetailPages 列表使用 where 查询。非常感谢 Ferran Buireu 引导我朝着正确的方向前进。

gatsby-node.js > createPages 函数：

// Video Detail pages
  const {
    data: {
      cartel: { videoDetailPages },
    },
  } = await graphql(`
    query {
      cartel {
        videoDetailPages(first: 300) {
          nodes {
            id
            slug
            videoDetail {
              editor
              editorId
            }
          }
        }
      }
    }
  `);

  const videoDetailTemplate = path.resolve(`./src/templates/videoDetail.js`);

  videoDetailPages.nodes.forEach(page => {
    const editorSlug = page.videoDetail.editor.replace(' ', '-').toLowerCase();
    const { editorId } = page.videoDetail;

    createPage({
      // will be the url for the page
      path: `${editorSlug}/${page.slug}`,
      // specify the component template of your choice
      component: slash(videoDetailTemplate),
      // In the ^template's GraphQL query, 'id' will be available
      // as a GraphQL variable to query for this page's data.
      context: {
        id: page.id,
        editorId: parseInt(editorId, 10),
      },
    });
  });

页面模板查询：

export const query = graphql`
  query($id: ID!, $editorId: Int!) {
    cartel {
      videoDetailPage(id: $id) {
        videoDetail {
          client
          director
          duration
          editor
          productionCompany
          videoStill {
            altText
            sourceUrl
          }
          videoUrl
          title
        }
      }
    }
    cartel {
      editorDetailPages(where: { id: $editorId }) {
        nodes {
          editorDetail {
            editorVideos {
              pagePath
              image {
                altText
                sourceUrl
                title
              }
            }
          }
        }
      }
    }
  }
`;
于 2021-06-03T16:05:15.483 回答