Find object by property in an array of JavaScript objects inside another array

Question

I have an array called mainarray which has three objects inside it. Inside each object there's another array called innerarray which has its own three objects.

How do I get the first property of every second object inside each of the innerarray? Is this possible?

mainarray: [{
    innerarray: [{
        property1: value1,
        property2: value2,
        property3: value3,
    }, {
        propertyiwant: value1,
        property2: value2,
        property3: value3,
    }, {
        property1: value1,
        property2: value2,
        property3: value3,
    }]
}, {
    innerarray: [{
        property1: value1,
        property2: value2,
        property3: value3,
    }, {
        propertyiwant: value1,
        property2: value2,
        property3: value3,
    }, {
        property1: value1,
        property2: value2,
        property3: value3,
    }]
}, { 
    innerarray: [{
        property1: value1,
        property2: value2,
        property3: value3,
    }, {
        propertyiwant: value1,
        property2: value2,
        property3: value3,
    }, {
        property1: value1,
        property2: value2,
        property3: value3,
    }]
}]

This question is totally different than From an array of objects, extract value of a property as array. As here i am trying to get property of an object inside an array which is inside another array.

1983 · Accepted Answer · 2016-06-10 15:09:24Z

2

mainarray.map(item => item.innerarray[1].propertyiwant);

or, if you use underscore and es5 only

_.map(mainarray, function(item){
    return item.innerarray[1].propertyiwant
});

edited Jun 10, 2016 at 15:09

1983

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answered Jun 10, 2016 at 14:08

Foker

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2 Comments

1983 Over a year ago

Well, Array.prototype.map is in ES5 :-)

Foker Over a year ago

yep:) i just presented approach with arrow functions for example:)

Redu · Accepted Answer · 2016-06-10 14:41:46Z

2

You can try the following code. It will bring you every second (in the sense of 2nd, 4th, 6th... etc) object in each subarray if they exist. If you are interested in only the second object then it's a matter of replacing the whole e.innerarray.reduce(... part with e.innerarray[2]. This will get you the whole desired 2nd object and then you can access any property of it that you like.

var mainarray = [{
    innerarray: [{
        property1: "value1",
        property2: "value2",
        property3: "value3",
    }, {
        propertyiwant: "value1",
        property2: "value2",
        property3: "value3",
    }, {
        property1: "value1",
        property2: "value2",
        property3: "value3",
    }]
}, {
    innerarray: [{
        property1: "value1",
        property2: "value2",
        property3: "value3",
    }, {
        propertyiwant: "value1",
        property2: "value2",
        property3: "value3",
    }, {
        property1: "value1",
        property2: "value2",
        property3: "value3",
    }]
}, { 
    innerarray: [{
        property1: "value1",
        property2: "value2",
        property3: "value3",
    }, {
        propertyiwant: "value1",
        property2: "value2",
        property3: "value3",
    }, {
        property1: "value1",
        property2: "value2",
        property3: "value3",
    }]
}],
mapped = mainarray.map(e => e.innerarray.reduce((p,c,i) => (i%2 && p.push(c),p),[]));
console.log(mapped);

edited Jun 10, 2016 at 14:41

answered Jun 10, 2016 at 14:36

Redu

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5 Comments

krimz Over a year ago

Hey thanks for your time, i tried what you said: mapped = mainarray.map(e => e.innerarray[2].propertyiwant) but it throws me an error of "cant read property of undefined" . i also tried mainarray.map(e => e.innerarray[2]['propertyiwant']) but still doesnt work

1983 Over a year ago

Arrays start at index 0, so you want innerarray[1].

Redu Over a year ago

@krimz repl.it/C0aO basically this is exactly what Foker had suggested.

1983 Over a year ago

krimz, both this and @Foker's answer worked fine when I tested them. You should give your expected output, and show what you've tried so far.

krimz Over a year ago

@FizzyTea hey! you're right, they work just fine with these examples. I was looking for something else, i was just not clear enough

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