如果 array2 没有相同的 object 键值，则从对象数组 (array1) 中删除 object

Question

I have two array of objects arr1 and arr2.

enter code here
const arr1 = [
   {name: 'Jake', age: 17},
   {name: 'Mike', age: 15},
   {name: 'Ryan', age: 18},
 ];

 const arr2 = [
   {name: 'Milo', age: 17},
   {name: 'Rem', age: 18},
   {name: 'Oliver', age: 19},
 ];

I want to remove objects in arr1 that does not have key value that arr2 has. I tried this, but no success:

const arr3 = this.arr1.filter(x => {
   let z = this.arr2.find(y => y.age== x.age)
   if (z) {
      return x;
   }
});

My expected array is this:

array3 = [
  {name: 'Jake', age: 17},
  {name: 'Ryan', age: 18},
];

Answer 1

You just have to remove this from the variables. Also, you can shorten the return statements like this:

 const arr1 = [ {name: 'Jake', age: 17}, {name: 'Mike', age: 15}, {name: 'Ryan', age: 18}, ]; const arr2 = [ {name: 'Milo', age: 17}, {name: 'Rem', age: 18}, {name: 'Oliver', age: 19}, ]; const arr3 = arr1.filter(x => z = arr2.find(y => y.age== x.age)); console.log(arr3)

However, if you want keep this you can declare the variables using var and then this.arr will be referring to window.arr and the code will still work as var creates a property on the global object while const or let do not:

 var arr1 = [ {name: 'Jake', age: 17}, {name: 'Mike', age: 15}, {name: 'Ryan', age: 18}, ]; var arr2 = [ {name: 'Milo', age: 17}, {name: 'Rem', age: 18}, {name: 'Oliver', age: 19}, ]; const arr3 = this.arr1.filter(x => z = this.arr2.find(y => y.age== x.age)); console.log(arr3)

Answer 2

The Array. some methods returns a boolean value which it slightly more appropriate. However find works too since non-null/undefined are treated as thruthy. If you arrays are truly in a class, then:

const arr3 = this.arr1.filter(x => this.arr2.some(y => y.age === x.age));

Otherwise need to remove this :

const arr3 =arr1.filter(x => arr2.some(y => y.age === x.age));