Question

我需要通过匹配ID从两个数组中收集数据，并将该数据推入新数组中。

我试图通过循环来做这些事情，但是它不能准确地推送。在第一个循环中，我迭代了学生的数组，在第二个循环中，我针对了课程数组的内部循环。

var students = [
    {
        _id: 'std1',
        name: 'Bill Gates',
        age: 30
    },
    {
        _id: 'std2',
        name: 'Steve Jobs',
        age: 30
    },
    {
        _id: 'std3',
        name: 'Suzanne Collins',
        age: 30
    }];

var courses = [
    {
        name: 'Maths',
        std_id: ['std3', 'std1']
    },
    {
        name: 'Computer Science',
        std_id: ['std3', 'std2']
    },
    {
        name: 'Commerce',
        std_id: ['std1']
    },
]

var sorted = [];
for(var i=0; i<students.length; i++){
    for(var j=0; j<courses.length; j++){
        if(students[i]._id == courses[j].std_id){
            sorted.push({'name' : students[i].name, 'studen ID' :students[i]._id, 'courses' :[courses[j].name]})

        }
    }

}

console.log(sorted);

{
    course: ['Maths','Commerce']   
    name: 'Bill Gates',
    studentId: 'std1'
}

Answer 1

尝试一下：

var students = [
    {
        _id: 'std1',
        name: 'Bill Gates',
        age: 30
    },
    {
        _id: 'std2',
        name: 'Steve Jobs',
        age: 30
    },
    {
        _id: 'std3',
        name: 'Suzanne Collins',
        age: 30
    }];

var courses = [
    {
        name: 'Maths',
        std_id: ['std3', 'std1']
    },
    {
        name: 'Computer Science',
        std_id: ['std3', 'std2']
    },
    {
        name: 'Commerce',
        std_id: ['std1']
    },
]

let sortedArr = [];

for(let i = 0; i < students.length; i++) {
    let stCourse = [];
    for(let j = 0; j < courses.length; j++) {
        if(courses[j].std_id.includes(students[i]._id)) {
            stCourse.push(courses[j].name)
        }
    }
    sortedArr.push({name: students[i].name, 'studen ID': students[i]._id, courses: stCourse})
}

console.log(sortedArr)

Answer 2

var students = [
  {
    _id: 'std1',
    name: 'Bill Gates',
    age: 30
  },
  {
    _id: 'std2',
    name: 'Steve Jobs',
    age: 30
  },
  {
    _id: 'std3',
    name: 'Suzanne Collins',
    age: 30
  }];

var courses = [
  {
    name: 'Maths',
    std_id: ['std3', 'std1']
  },
  {
    name: 'Computer Science',
    std_id: ['std3', 'std2']
  },
  {
    name: 'Commerce',
    std_id: ['std1']
  },
]

const studentWithCourses = students.map(student => {
  let stdCourses = courses.filter(course => course.std_id.includes(student._id))
  return { ...student, courses: stdCourses.map(c => c.name) }
})

console.log(studentWithCourses)

尝试使用此代码段。我使用了地图和过滤器方法。我为每个学生根据学生证过滤课程。

Answer 3

您可以这样做

const result = students.map(student => {
    const {_id, name} = student;
    return {
        name: name,
        id: _id,
        courses: courses.reduce((a, course) => {
            course.std_id.includes(_id) && a.push(course.name);
            return a;
        }, [])
    }
});

Answer 4

您可以使用减速器（Array.reduce，see MDN）。

该代码段还包含一个map / filter替代项。

const arrays2Combine = getArrays();

// reduce
const combined = arrays2Combine.courses.reduce ( (students, val) => {
  students.forEach(student => {
    if (val.std_id.includes(student._id)) {
      student.courses = (student.courses || []).concat(val.name);
    }
  })
  return students;
}, arrays2Combine.students);

console.log(`**Reduced:\n`, combined);

// Alternative using map/filter
const combined2 = arrays2Combine.students.map( student => {
  student.courses = arrays2Combine.courses
    .filter(c => c.std_id.includes(student._id))
    .map(course => course.name);
  return student; 
});


setTimeout(() => {
    console.clear(); 
    console.log(`**Map/filter alternative\n`, combined2);
  },
  5000
);


function getArrays() {
return { students: 
    [
      {
          _id: 'std1',
          name: 'Bill Gates',
          age: 30
      },
      {
          _id: 'std2',
          name: 'Steve Jobs',
          age: 30
      },
      {
          _id: 'std3',
          name: 'Suzanne Collins',
          age: 30
      }
    ],
      courses: [
      {
          name: 'Maths',
          std_id: ['std3', 'std1']
      },
      {
          name: 'Computer Science',
          std_id: ['std3', 'std2']
      },
      {
          name: 'Commerce',
          std_id: ['std1']
      }
    ]
  };
}

Answer 5

希望对您有帮助

  var students = [
        {
            _id: 'std1',
            name: 'Bill Gates',
            age: 30
        },
        {
            _id: 'std2',
            name: 'Steve Jobs',
            age: 30
        },
        {
            _id: 'std3',
            name: 'Suzanne Collins',
            age: 30
        }];

    var courses = [
        {
            name: 'Maths',
            std_id: ['std3', 'std1']
        },
        {
            name: 'Computer Science',
            std_id: ['std3', 'std2']
        },
        {
            name: 'Commerce',
            std_id: ['std1']
        },
    ]

    var sorted = [];
    let array=[];
    for(var i=0; i<students.length; i++){

        for(var j=0; j<courses.length; j++){

                if(courses[j].std_id.indexOf(students[i]._id)>-1){
                        array.push(courses[j].name);
                 }

         }

       sorted.push({'name' : students[i].name, 'studen ID' :students[i]._id, 
       'courses' :array})
        array=[];


    }

有两个对象的数组，在这些数组中，如果ID相同，我想匹配ID。我需要具有相同ID的合并信息的新数组

5 个答案: