Question

鉴于用户的ID，我希望获得所有请求审阅者的拉请求。

以下操作不起作用，因为它只允许我获得该用户打开的拉取请求：

const state = {
  teachers: [
    { teacherId: 'ttt1', teacherName: 'Susan', isCool: true, teachesAt: 'sss1'},
    { teacherId: 'ttt2', teacherName: 'Karen', isCool: false, teachesAt: 'sss2'},
    { teacherId: 'ttt3', teacherName: 'Bob', isCool: true, teachesAt: 'sss3'},
    { teacherId: 'ttt4', teacherName: 'Mike', isCool: false, teachesAt: 'sss1'},
  ],

  schools: [
    { schoolId: 'sss1', schoolName: 'Washington'},
    { schoolId: 'sss2', schoolName: 'Lincoln'},
    { schoolId: 'sss3', schoolName: 'Jefferson'},
  ],

  students: [
    { schoolIdEnrolled: 'sss1', studentName: 'Billy'},
    { schoolIdEnrolled: 'sss1', studentName: 'Steven'},
    { schoolIdEnrolled: 'sss2', studentName: 'Bobby'},
    { schoolIdEnrolled: 'sss3', studentName: 'Mikey'},
    { schoolIdEnrolled: 'sss3', studentName: 'Sally'},
    { schoolIdEnrolled: 'sss3', studentName: 'Cindy'},
    { schoolIdEnrolled: 'sss3', studentName: 'Mandy'},
  ],
}


const teachersReducer = (teachers, id) => {
  const numOfteachers = teachers.reduce((result, current) => {
    if (current.teachesAt === id && current.isCool) {
      result++;
    }
    return result;
  }, 0);
  return numOfteachers;
}

const studentsReducer = (students, id) => {
  const numOfstudents = students.reduce((result, current) => {
    if (current.schoolIdEnrolled === id) {
      result++;
    }
    return result;
  }, 0);
  return numOfstudents;
}


const resultData = state.schools.reduce((result, currentSchool) => {
  const currentId = currentSchool.schoolId;
  const numOfTeachers = teachersReducer(state.teachers, currentId);
  const numOfStudents = studentsReducer(state.students, currentId);
  return {
    ...result, 
    [currentSchool.schoolName]: {
        ...result[currentSchool],
        numOfTeachers,
        numOfStudents
    }
  }
}, {});

console.log(resultData);

有办法做到这一点吗？

谢谢！

Answer 1

您可以从用户ID获取用户名，并使用review-requested执行搜索，以匹配要求审核的用户：

获取用户名：

{
  node(id: "MDQ6VXNlcjk2OTQ3") {
    ... on User {
      login
    }
  }
}

获取要求审核用户的公开PR：

{
  search(query: "type:pr state:open review-requested:refack", type: ISSUE, first: 100) {
    issueCount
    pageInfo {
      endCursor
      startCursor
    }
    edges {
      node {
        ... on PullRequest {
          repository {
            nameWithOwner
          }
          number
          url
        }
      }
    }
  }
}

您可以将查询字符串定义为变量try it in the explorer

如何使用Github的v4 GraphQL API查询用户的审核请求？

1 个答案: