Graphql 中的动态函数名称

Question

我有以下查询

const COURSES = gql`
  query(
    $course: Course!
  ) {
    **english**(course: $course) {
      transfers(
        options: {
          desc: ["count_in", "count_out"]
        }
      ) {
        school {
          address
          symbol
          tokenType
        }
      }
    }
  }
`

注意第 5 行以 'english' 开头，假设我有 1000 门不同的课程，例如 ['english'、'politics'、'arts'、....第 1000 门课程).我怎样才能使这个课程名称动态化。我没有开发后端，所以改变模式对我来说不是一个选择。如何通过使课程名称动态化来避免重复。

为了清楚起见，我想将变量名放在第 5 行的 'english' 的位置，以便该行的行为如下所示

   english(course: $course) {
   politics(course: $course) {
   art(course: $course) {
   anyCourseName(course: $course) {

更新：我正在使用 useQuery 调用上面的查询并传递如下参数 $course

useQuery(
    COURSES,
    {
      variables: {
        course, //this serves as variable $course in above query
      },
    }
  )

Answer 1

const COURSES = (c) => gql`
  query(
    $course: Course!
  ) {
    ${c}(course: $course) {
      transfers(
        options: {
          desc: ["count_in", "count_out"]
        }
      ) {
        school {
          address
          symbol
          tokenType
        }
      }
    }
  }

const query = COURSES('art')

?

Answer 2

你可以简单地这样做：

const course = "english";

const COURSES = gql`
  query(
    $course: Course!
  ) {
    ${course}(course: $course) {
      transfers(
        options: {
          desc: ["count_in", "count_out"]
        }
      ) {
        school {
          address
          symbol
          tokenType
        }
      }
    }
  }
`

或者如果您有多个查询：

const courses = [
  "english",
  "politics",
  "art",
  "anyCourseName",
];

const queries = courses.reduce((acc, query) => (
  acc.concat(`${query}(course: $course) {
      transfers(
        options: {
          desc: ["count_in", "count_out"]
        }
      ) {
        school {
          address
          symbol
          tokenType
        }
      }
  }\n`)
), "");

console.log(queries);

然后在你的 gql 中使用它：

const COURSES = gql`
  query(
    $course: Course!
  )
  {
    ${queries}
  }
`;

Graphql 中的动态函数名称

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