Question

我有2个名为

的数组

a = [{"id":"1","mark":"96",...},{"id":"2","mark":"65",...}];

b = [{"student_id":"1","grade":"A"},{"student_id":"2","grade":"c"}];

我想比较2个数组，如果'id'和'student_id' 匹配，我想将等级值推送到' '阵列。然后我的结果将是：

预期结果：

c = [{"id":"1","mark":"96","grade":"A",...},
    {"id":"2","mark":"65","grade":"c".....}];

Answer 1

以下是n复杂性：

中如何做到这一点

＆＃13;

const a = [{
  "id": "1",
  "mark": "96"
}, {
  "id": "2",
  "mark": "65"
}];

const b = [{
  "student_id": "1",
  "grade": "A"
}, {
  "student_id": "2",
  "grade": "c"
}];

let map = a.reduce((acc, item) => {
  acc[item.id] = item;
  return acc;
}, {});

b.forEach(item => {
  if (map[item.student_id]) {
    map[item.student_id].grades = item.grade;
  }
});

console.log(Object.values(map));

＆＃13;

Answer 2

使用.find标识另一个数组中的匹配元素，并使用.reduce从中创建一个新对象：

const marks = [{"id":"1","mark":"96"},{"id":"2","mark":"65"}];
const students = [{"student_id":"1","grade":"A"},{"student_id":"2","grade":"c"}];

const result = marks.reduce((resultsSoFar, mark) => {
  const id = mark.id;
  const grade = students.find(({ student_id }) => student_id === id).grade;
  resultsSoFar.push({ ...mark, grade });
  return resultsSoFar;
}, []);
console.log(result);

Answer 3

您可以使用数组＆＃39; map()和forEach()来执行以下操作：

＆＃13;

var a = [{"id":"1","mark":"96"},{"id":"2","mark":"65"}];

var b = [{"student_id":"1","grade":"A"},{"student_id":"2","grade":"c"}];

var c = a.map(function(i){
  b.forEach(function(j){
    if(i.id===j.student_id){
      i.grade = j.grade;
    }
  });
  return i;
})

console.log(c);

＆＃13;

Answer 4

您可以使用for循环来遍历数组a和数组b，然后只要您在id和student_id值匹配时添加grade属性并将对象推送到新数组c中。你也可以break for循环，以避免在找到匹配后进一步不必要的循环。

＆＃13;

var a = [{"id":"1","mark":"96"},{"id":"2","mark":"65"}];
var b = [{"student_id":"1","grade":"A"},{"student_id":"2","grade":"c"}];
var c = [];

for(var i=0; i<a.length; i++){
  var objA = a[i];
  for(var j=0; j<b.length; j++){
    var objB = b[j];
    if(objA.id === objB.student_id){
      objA.grade = objB.grade;
      c.push(objA);
      break;
    }
  }
}

console.log(c);

＆＃13;

Answer 5

假设数组"grade"内没有a - 对象：

＆＃13;

let a = [{"id": "1", "mark": "96"}, {"id": "2", "mark": "65"}];
let b = [{"student_id": "1", "grade": "A"}, {"student_id": "2", "grade": "c"}];
let c =[];

for(var iterateA = 0, elA; elA = a[iterateA]; iterateA++) {
	for(var iterateB = 0, elB; elB = b[iterateB]; iterateB++) {
		if(elA.id === elB.student_id) {
			elA['grade'] = elB.grade;
			c.push(elA);
		}
	}
}
console.log(c);

＆＃13;

Answer 6

这应该有效

for(var i=0; i<a.length;i++) {
        for(var j = 0; j<b.length;j++) {
            if(a[i].id === b[j].student_id) {
                c.push({
                    'id': a[i].id,
                    'mark': a[i].mark,
                    'grade': b[j].grade
                });
            }
        }
    }

Answer 7

使用地图并找到

a = [{"id":"1","mark":"96"},{"id":"2","mark":"65"}];

b = [{"student_id":"1","grade":"A"},{"student_id":"2","grade":"c"}];

updatedA = a.map((val) => {
  let bVal = b.find((item) => item.student_id === val.id)
  if(bVal){
   val.grade = bVal.grade;
  }
  return val;
})

console.log(updatedA);

比较2个jsons数组并从中形成一个数组？

7 个答案: