Filter for array of objects that contain all key value pairs of one object in another array of objec-CodePudding

Given I have this array of objects:

    let array1 = [
        {id: 1, name: "Test Item 1", price: 100, type: 'item'},
        {id: 1, name: "Test Fee 1", price: 200, type: 'fee'},
        {id: 3, name: "Test 3", price: 300, type: 'item'},
    ]

Now i want to filter so I only have the objects where every key value of the following array match:

    let array2 = [
        {id: 1, type: "item"},
        {id: 3, type: "item"},
    ]

This should return array1[0] and array1[2]. My current solution has been this, but it wont work if array2 has multiple object keys to compare to:

array1.filter(row => {
    return array2.find(r => {
        return Object.entries(row).some(([key, value]) => r[key] === value);
    })
});

CodePudding user response：

array1.filter((row) =>
  array2.some((template) =>
    Object.keys(template).every((key) => template[key] === row[key])
  )
);

I think the answer in comment above is cleaner than mine.

CodePudding user response：

In array1.filter() you can iterate all objects in array2 and compare keys using Object.keys()

let array1 = [
  {id: 1, name: "Test Item 1", price: 100, type: 'item'},
  {id: 1, name: "Test Fee 1", price: 200, type: 'fee'},
  {id: 3, name: "Test 3", price: 300, type: 'item'},
]

 let array2 = [
   {id: 1, type: "item"},
   {id: 3, type: "item"},
 ]

const res = array1.filter(row => {
  for (const obj of array2) {
    if (Object.keys(obj).every(k => obj[k] === row[k])) {
      return row
    }
  }
})

console.log(res)

CodePudding user response：

Lodash if you don't mind:

const array1 = [{id: 1, name: "Test Item 1", price: 100, type: 'item'},{id: 1, name: "Test Fee 1", price: 200, type: 'fee'},{id: 3, name: "Test 3", price: 300, type: 'item'}];
const array2 = [{id: 1, type: "item"},{id: 3, type: "item"}];

const fn = _.overSome(array2.map(_.matches));
const result = _.filter(array1, fn);

console.log(result);

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