Filter array of objects with another array of object without knowing which properties are in the objects

Question

I have an array of objects that I would like to filter but I do not know the properties that I have in the objects (neither the data nor the filter array):

myArray = [
   { id: "100", area: "01", country: "AT"},
   { id: "101", area: "02", country: "DE"}, 
   { id: "102", area: "01", country: "DE"},
   { id: "103", area: "03", country: "CH"},
   { id: "104", area: "01", country: "AT"}
]

and my filter is:

myFilter = { area: "01", country: "AT" }

I would like to have this returned:

myArrayFiltered = [
   { id: "100", area: "01", country: "AT"},
   { id: "104", area: "01", country: "AT"}
]

How would I go about doing this? Just to be perfectly clear, I do not know the properties that I will be searching for i.e. I do not know which properties myArray or myFilter will have. If the filter contains a key that is not in myArray I would expect an empty array!

PS: I do also not know how many entries are going to be filters by i.e. depending what my users enter I could just get:

myFilter = [ {area: "01"} ]

Gearoid · Accepted Answer · 2022-08-25 13:50:43Z

3

myArray.filter((item) => 
 Object.entries(myFilter).every(([key, val]) => item[key] === val))

answered Aug 25, 2022 at 13:50

Gearoid

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Comments

Manjeet Singh · Accepted Answer · 2022-08-25 13:46:05Z

2

var myArray  = 
[
    { id: "100", area: "01", country: "AT"},
    { id: "101", area: "02", country: "DE"}, 
    { id: "102", area: "01", country: "DE"},
    { id: "103", area: "03", country: "CH"},
    { id: "104", area: "01", country: "AT"}
 ]

 var  myFilter = { area: "01", country: "AT" }

    var filteredArray = myArray.filter(function(item) {
        return Object.keys(myFilter).every(function(c) {
            return item[c] === myFilter[c];
        });
    } );
    
    console.log(filteredArray)

answered Aug 25, 2022 at 13:46

Manjeet Singh

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Comments

Thomas Hunt · Accepted Answer · 2022-08-25 13:57:18Z

0

myArray.filter((item) => Object.keys(item).every((key) => !myFilter[key]|| myFilter[key] === item[key]))

edited Aug 25, 2022 at 13:57

answered Aug 25, 2022 at 13:56

Thomas Hunt

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1 Comment

Mark Rotteveel Over a year ago

Please don't post only code as answer, but also provide an explanation what your code does and how it solves the problem of the question. Answers with an explanation are usually more helpful and of better quality, and are more likely to attract upvotes.

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