Required properties on array of objects

Question

I am trying to find out if two required properties are found in array of objects, I need a boolean that says true if at least those items are found and false if any of them is missing, any other properties are not an issue for the boolean as long as those two required properties are there.

So I got this, but I don't know how to produce a true boolean when both files are included

const data = [
  {
    name: 'required1'
  },
  {
    name: 'noRequired'
  },
  {
    name: 'required2'
  },
  {
    name: 'noRequired'
  },
]

const required = ['required1', 'required2']

const result = data.filter(({ name }) => !required.includes(name))


console.log(result)

.as-console-wrapper { max-height: 100% !important; top: 0; }

I have tried some and every but it does not work as it has to be the two included in the array as in 'every' but not the rest in the array as in 'some'

sid busa · Accepted Answer · 2021-07-16 11:35:46Z

3

function containsAll(needles, haystack){ 
  for(var i = 0; i < needles.length; i++){
     if($.inArray(needles[i], haystack) == -1) return false;
  }
  return true;
}

containsAll([34, 78, 89], [78, 67, 34, 99, 56, 89]); // true
containsAll([34, 78, 89], [78, 67, 99, 56, 89]); // false
containsAll([34, 78, 89], [78, 89]); // false

edited Jul 16, 2021 at 11:35

answered Jul 16, 2021 at 11:20

sid busa

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5 Comments

Álvaro Over a year ago

It does not work, if you remove required2 from data you still get true

sid busa Over a year ago

you need to check if required array data available or not in data array right ?? so if required2 remove then required1 is still available in data's array and you can use here forEach loop also instead of filter

Álvaro Over a year ago

['required1', 'required2' ] both need to be there, if so is true, regardless of the orher values. If any of those strings are missing, it should be false.

sid busa Over a year ago

okk got you wait i'll do some changes in code

sid busa Over a year ago

please use now if still not working then tell me

Andy · Accepted Answer · 2021-07-16 11:46:15Z

2

every does work, but you have to create a new array of names, and then check to see if those names are included in your required list. You can map over an array of objects to create an new array of names and use that for a comparison.

const data=[{name:"required1"},{name:"noRequired"},{name:"required2"},{name:"noRequired"}];

const required = ['required1', 'required2'];

const values = data.map(obj => obj.name);

const result = required.every(name => {
  return values.includes(name);
});

console.log(result);

edited Jul 16, 2021 at 11:46

answered Jul 16, 2021 at 11:36

Andy

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3 Comments

Álvaro Over a year ago

I like this approach but why not just do const values = data.map(({name}) => name)? and then just const result = required.every(name => values.includes(name))

marzelin Over a year ago

yeah, data is an array so Object.values(data) gives a shallow copy of the array which makes no use here.

Andy Over a year ago

Sorry! I've been up all night. You're both right. Sometimes I'm a moron.

Nikhil Patil · Accepted Answer · 2021-07-16 11:35:23Z

1

You can just check the difference between 2 arrays (required - data) and if the length is 0 then true else false

const data = [{
    name: 'required1'
  },
  {
    name: 'noRequired'
  },
  {
    name: 'required2'
  },
  {
    name: 'noRequired'
  },
]

const required = ['required1', 'required2']

const names = data.map(m => m.name);
const result = !required.some(f => !names.includes(f));

// Alternative -
//const diffCount = required.filter(f => !names.includes(f)).length;
//const result = diffCount == 0;


console.log(result)

answered Jul 16, 2021 at 11:35

Nikhil Patil

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2 Comments

marzelin Over a year ago

!some(!condition) is the same as every(condition)

Nikhil Patil Over a year ago

@marzelin, yes I realize that now.

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