Question

帮助我从以下JSON数据中获取所有学生的ID和大学ID的数组。我想拥有;

//to store id of all colleges.
var collegeid = [];

//store id of all students.
var studentsid = [];

//students id as per college:
var studentsIdInCollege1 = [];

var data = {  
  "college":[  
    {  
      "id":1,
      "school":"abc",
      "course":"cde",
      "students":[  
        {  
          "id":1,
          "name":"abc 123",
          "number":"156888"
        },
        {  
          "id":2,
          "name":"abc 123",
          "number":"156888"
        }
      ]
    },
    {  
      "id":2,
      "school":"xyz",
      "course":"lopl",
      "students":[  
        {  
          "id":3,
          "name":"abc 123",
          "number":"156888"
        },
        {  
          "id":4,
          "name":"abc 123",
          "number":"156888"
        }
      ]
    }
  ]
}

`

Answer 1

对于大学，只需map大学数组到每个对象的id。对于学生，使用[].concat(...分散学生ID的地图并获得一个扁平的对象：

const input = {  
  "college":[  
    {  
      "id":1,
      "school":"abc",
      "course":"cde",
      "students":[  
        {  
          "id":1,
          "name":"abc 123",
          "number":"156888"
        },
        {  
          "id":2,
          "name":"abc 123",
          "number":"156888"
        }
      ]
    },
    {  
      "id":2,
      "school":"xyz",
      "course":"lopl",
      "students":[  
        {  
          "id":3,
          "name":"abc 123",
          "number":"156888"
        },
        {  
          "id":4,
          "name":"abc 123",
          "number":"156888"
        }
      ]
    }
  ]
}

const collegeIds = input.college.map(({ id }) => id);
console.log('collegeIds: ' + collegeIds);
const studentIds = [].concat(...input.college.map(({ students }) => students.map(({ id }) => id)));
console.log('studentIds: ' + studentIds);

Answer 2

使用.map()获取大学ID：

let collegeIds = data.college.map(({ id }) => id);

对学生ID使用.reduce()：

let studentIds = data.college.reduce((a, c) => (
  a.concat(c.students.map(({ id }) => id))
), []);

使用.reduce()获取一个数组数组，其中每个数组都包含student id：

let studentInEachCollege = data.college.reduce((a, c) => (
  a.push(c.students.map(({ id }) => id)), a
), []);

这将允许您使用第一个大学的studentInEachCollege[0]索引，第二个大学studentInEachCollege[1]等索引访问每个大学生，依此类推。

<强>演示：

var data = {  
  "college":[  
    {  
      "id":1,
      "school":"abc",
      "course":"cde",
      "students":[  
        {  
          "id":1,
          "name":"abc 123",
          "number":"156888"
        },
        {  
          "id":2,
          "name":"abc 123",
          "number":"156888"
        }
      ]
    },
    {  
      "id":2,
      "school":"xyz",
      "course":"lopl",
      "students":[  
        {  
          "id":3,
          "name":"abc 123",
          "number":"156888"
        },
        {  
          "id":4,
          "name":"abc 123",
          "number":"156888"
        }
      ]
    }
  ]
};

let collegeIds = data.college.map(({ id }) => id);
let studentIds = data.college.reduce((a, c) => (
  a.concat(c.students.map(({ id }) => id))
), []);
let studentInEachCollege = data.college.reduce((a, c) => (
  a.push(c.students.map(({ id }) => id)), a
), []);

console.log(collegeIds);
console.log(studentIds);
console.log(studentInEachCollege);

<强>文档：

Answer 3

// College ids 
var cID = data.college.map((c) => { return c.id});

// Students
// It returns two arrays so you need to reduce it...
var students = data.college.map((c) => { return c.students});

var allStudents = students.reduce((a,b) => {return a.concat(b)}, []);

// IDs
var sID = allStudents.map((c) => { return c.id});

获取一系列学生的身份

3 个答案: