如何在GrpahQL中仅在某些特定条件下使参数不为空

query GetUser($includeAddress: Boolean!) {
  user {
    name
    address @include(if: $includeAddress) {
      street
      city
      country
    }
  }
}

const { graphql } = require('graphql');

const schema = ... // 定义GraphQL模式

const query = `
  query GetUser($includeAddress: Boolean!) {
    user {
      name
      address @include(if: $includeAddress) {
        street
        city
        country
      }
    }
  }
`;

const variables = {
  includeAddress: true, // 设置includeAddress变量为真
};

graphql(schema, query, null, null, variables).then((result) => {
  console.log(result);
});