Compare all values of an Array of Objects with an Object

Question

I have the following data:

 const myArr = [{ id: 0, company: "microsoft", location: "berlin" }, { id: 1, company: "google", location: "london" }, { id: 2, company: "twitter", location: "berlin" }]; let myObj = { company: ["google", "twitter"], location: ["london"] } 

and given that the myObj.company entries are changing (irrelevant how) I am trying to create a function that filters results and only returns Objects that satisfy the location and company criteria.

In the example above, what we need returned is :

{ 
  id: 1,
  company: "google",
  location: "london"
}

If myObj was

let myObj = {
  company: ["google", "twitter"],
  location: []
}

then the returned result should be

{ 
  id: 1,
  company: "google",
  location: "london"
},
{ 
  id: 2,
  company: "twitter",
  location: "berlin"
}

Answer 1

Use Array#filter method with Array#includes method(for old browser support use Array#indexOf method)

myArr.filter(o => (myObj.company.length == 0 || myObj.company.includes(o.company)) && (myObj.location.length == 0 || myObj.location.includes(o.location)))

 const myArr = [{ id: 0, company: "microsoft", location: "berlin" }, { id: 1, company: "google", location: "london" }, { id: 2, company: "twitter", location: "berlin" }]; let myObj = { company: ["google", "twitter"], location: ["london"] } console.log( myArr.filter(o => (myObj.company.length == 0 || myObj.company.includes(o.company)) && (myObj.location.length == 0 || myObj.location.includes(o.location))) ) myObj = { company: ["google", "twitter"], location: [] } console.log( myArr.filter(o => (myObj.company.length == 0 || myObj.company.includes(o.company)) && (myObj.location.length == 0 || myObj.location.includes(o.location))) ) 

---

UPDATE : In case there an n number of properties collection then you need to make some variation, where you can use Object.keys and Array#every methods.

var keys = Object.keys(myObj);

myArr.filter(o => keys.every(k => myObj[k].length == 0 || myObj[k].includes(o[k])))

 const myArr = [{ id: 0, company: "microsoft", location: "berlin" }, { id: 1, company: "google", location: "london" }, { id: 2, company: "twitter", location: "berlin" }]; let myObj = { company: ["google", "twitter"], location: ["london"] } var keys = Object.keys(myObj); console.log( myArr.filter(o => keys.every(k => myObj[k].length == 0 || myObj[k].includes(o[k]))) ) myObj = { company: ["google", "twitter"], location: [] } keys = Object.keys(myObj); console.log( myArr.filter(o => keys.every(k => myObj[k].length == 0 || myObj[k].includes(o[k]))) ) 

Answer 2

Use Array.prototype.filter , Array.prototype.every and Object.keys to get the desired result like this: (note that obj could have any number of keys, it flexible that way)

 const myArr = [{ id: 0, company: "microsoft", location: "berlin" }, { id: 1, company: "google", location: "london" }, { id: 2, company: "twitter", location: "berlin" }]; let myObj = { company: ["google", "twitter"], location: ["london"] } function find(arr, obj) { // get only the keys from obj that their corresponding array is not empty var keys = Object.keys(obj).filter(k => obj[k].length !== 0); // return a filtered array of objects that ... return arr.filter(o => { // ... met the creteria (for every key k in obj, the current object o must have its value of the key k includded in the array obj[k]) return keys.every(k => { return obj[k].indexOf(o[k]) != -1; }); }); } console.log(find(myArr, myObj)); 

Answer 3

You could filter with iterating the keys of criteria and check the content for equality of if the length of the array is not zero.

 const filterBy = (array, criteria) => array.filter(o => Object.keys(criteria).every(k => criteria[k].some(c => c === o[k]) || !criteria[k].length) ); const myArr = [{ id: 0, company: "microsoft", location: "berlin" }, { id: 1, company: "google", location: "london" }, { id: 2, company: "twitter", location: "berlin" }]; console.log(filterBy(myArr, { company: ["google", "twitter"], location: ["london"] })); console.log(filterBy(myArr, { company: ["google", "twitter"], location: [] })); 

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With Array#includes instead of Array#some

 const filterBy = (array, criteria) => array.filter(o => Object.keys(criteria).every(k => criteria[k].includes(o[k]) || !criteria[k].length) ); const myArr = [{ id: 0, company: "microsoft", location: "berlin" }, { id: 1, company: "google", location: "london" }, { id: 2, company: "twitter", location: "berlin" }]; console.log(filterBy(myArr, { company: ["google", "twitter"], location: ["london"] })); console.log(filterBy(myArr, { company: ["google", "twitter"], location: [] })); 

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