Check if object exists in array excluding specific property

Question

Is there a way to check if an object exists in an array while excluding one property? I don't know which properties will exist in the object so I can't compare properties manually.

Based on the code below, I would want to check if obj exists in arr excluding prop3 .

Example:

const obj = {
  "prop1": "value1",
  "prop2": "value2"
}

const arr = [{
    "prop1": "value1",
    "prop2": "value2",
    "prop3": "value3"
  },
  {
    "prop1": "value4",
    "prop2": "value5",
    "prop3": "value6"
  }
]

arr.indexOf(obj) > -1; // returns false, need something like this that returns true

Answer 1

User Array.findIndex() method to find index:

const keys = Object.keys(obj);
arr.findIndex((o) => {
  return keys.every(k => o[k] && o[k] === obj[k])
});

Answer 2

Use a combination of Array.findIndex , Object.keys & Array.every like so:

 const obj = { "prop1": "value1", "prop2": "value2" } const arr = [{ "prop1": "value1", "prop2": "value2", "prop3": "value3" }, { "prop1": "value4", "prop2": "value5", "prop3": "value6" } ] const index = arr.findIndex(item => { // For every item of array: // - Iterate over obj keys // - If every k/v pair of obj matches ak/v pair of item // then it's a match return Object.keys(obj).every(key => item[key] === obj[key]) }) console.log(index) 

Although if you just want to know if it exists or not, you should use Array.some instead of Array.findIndex .